RD Sharma XII Vol 1 2020 2021 Solutions for Class 12 Commerce Maths Chapter 15 - Tangents And Normals

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Page No 15.10:

Question 1:

Find the slopes of the tangent and the normal to the following curves at the indicted points:

(i) $y = \sqrt{x^{3}} at x = 4$
(ii) $y = \sqrt{x} at x = 9$
(iii) y = x³ − x at x = 2
(iv) y = 2x² + 3 sin x at x = 0
(v) x = a (θ − sin θ), y = a(1 − cos θ) at θ = −π/2
(vi) x = a cos³ θ, y = a sin³ θ at θ = π/4
(vii) x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2
(viii) y = (sin 2x + cot x + 2)² at x = π/2
(ix) x² + 3y + y² = 5 at (1, 1)
(x) xy = 6 at (1, 6)

Answer:

$(i) y = \sqrt{x^{3}} = x^{\frac{3}{2}} \Rightarrow \frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x} When x =4, y = \sqrt{x^{3}} = \sqrt{64} = 8 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(4, 8)} = \frac{3}{2} \sqrt{4} = 3 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(4, 8)}} = \frac{- 1}{3}$

$(i i) y = \sqrt{x} = x^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2} x^{\frac{- 1}{2}} = \frac{1}{2 \sqrt{x}} When x =9, y = \sqrt{x} = \sqrt{9} = 3 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(9, 3)} = \frac{1}{2 \sqrt{9}} = \frac{1}{6} Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(9, 3)}} = \frac{- 1}{(\frac{1}{6})} =-6$

$(i i i) y = x^{3} - x \Rightarrow \frac{d y}{d x} = 3 x^{2} - 1 When x =2, y = x^{3} - x = 2^{3} - 2 = 6 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(2, 6)} =3 {(2)}^{2} -1=11 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(2, 6)}} = \frac{- 1}{11}$

$(i v) y = 2 x^{2} + 3 \sin x \Rightarrow \frac{d y}{d x} = 4 x + 3 \cos x When x =0, y = 2 x^{2} + 3 \sin x = 2 {(0)}^{2} + 3 \sin 0 = 0 Now, Slope of the tangent= {(\frac{d y}{d x})}_{(0, 0)} =4 (0) + 3 cos 0=3 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(0, 0)}} = \frac{- 1}{3}$

$(v) x = a (θ - \sin θ) \Rightarrow \frac{d x}{d θ} = a (1 - \cos θ) y = a (1 + \cos θ) \Rightarrow \frac{d y}{d θ} = a (- \sin θ) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (- \sin θ)}{a (1 - \cos θ)} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = - c o t \frac{θ}{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{- π}{2}} =-cot (\frac{\frac{- π}{2}}{2}) =-cot (\frac{- π}{4}) =1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{- π}{2}}} = \frac{- 1}{1} =-1$

$(v i) x = a \cos^{3} θ \Rightarrow \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ y = a \sin^{3} θ \Rightarrow \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} = - \tan θ Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{4}} =-tan \frac{π}{4} =-1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{π}{4}}} = \frac{- 1}{- 1} =1$

$(v i i) x = a (θ - \sin θ) \Rightarrow \frac{d x}{d θ} = a (1 - \cos θ) y = a (1 - \cos θ) \Rightarrow \frac{d y}{d θ} = a (\sin θ) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (\sin θ)}{a (1 - \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = \cot \frac{θ}{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{2}} =cot (\frac{\frac{π}{2}}{2}) =cot (\frac{π}{4}) =1 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{π}{2}}} = \frac{- 1}{1} =-1$

$(v i i i) y = {(\sin 2 x + \cot x + 2)}^{2} \Rightarrow \frac{d y}{d x} = 2 (\sin 2 x + \cot x + 2) (2 \cos 2 x - \cos e c^{2} x) Now, Slope of the tangent= {(\frac{d y}{d x})}_{x = \frac{π}{2}} = 2 [\sin 2 (\frac{π}{2}) + \cot (\frac{π}{2}) + 2] [2 \cos 2 (\frac{π}{2}) - {cosec}^{2} (\frac{π}{2})] = 2 (0 + 0 + 2) (- 2 - 1) = - 12 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{x = \frac{π}{2}}} = \frac{- 1}{- 12} = \frac{1}{12}$

$(i x) x^{2} + 3 y + y^{2} = 5 On differentiating both sides w.r.t. x, we get 2 x + 3 \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (3 + 2 y) = - 2 x \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{3 + 2 y} Now, Slope of the tangent= {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 2 x}{3 + 2 y} = \frac{- 2}{3 + 2} = \frac{- 2}{5} Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(1, 1)}} = \frac{- 1}{(\frac{- 2}{5})} = \frac{5}{2}$

$(x) x y = 6 On differentiating both sides w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow x \frac{d y}{d x} = - y \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Now, Slope of the tangent= {(\frac{d y}{d x})}_{(1, 6)} = \frac{- y}{x} = \frac{- 6}{1} =-6 Slope of the normal= \frac{- 1}{{(\frac{d y}{d x})}_{(1, 6)}} = \frac{- 1}{- 6} = \frac{1}{6}$

Page No 15.10:

Question 2:

Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.

Answer:

$Given : x y + a x + b y = 2 . . . (1) On differentiating both sides w.r.t. x, we get x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (x + b) = - a - y \Rightarrow \frac{d y}{d x} = \frac{- a - y}{x + b} Now, {(\frac{d y}{d x})}_{(1, 1)} = 2 \Rightarrow \frac{- a - 1}{1 + b} = 2 \Rightarrow - a - 1 = 2 + 2 b \Rightarrow - a = 3 + 2 b \Rightarrow a = - (3 + 2 b) On substituting a = - (3 + 2 b), x =1 and y =1 in eq. (1), we get 1 - (3 + 2 b) + b = 2 \Rightarrow 1 - 3 - 2 b + b = 2 \Rightarrow b = - 4 and a = - (3 + 2 b) = - (3 - 8) = 5 ∴ a = 5 and b = - 4$

Page No 15.10:

Question 3:

If the tangent to the curve y = x³ + ax + b at (1, − 6) is parallel to the line x − y + 5 = 0, find a and b.

Answer:

$Given: x - y + 5 = 0 \Rightarrow y = x + 5 \Rightarrow \frac{d y}{d x} = 1 Now, y = x^{3} + a x + b . . . (1) \Rightarrow \frac{d y}{d x} = 3 x^{2} + a Slope of the tangent at (1, - 6) = Slope of the given line \Rightarrow {(\frac{d y}{d x})}_{(1, - 6)} = 1 \Rightarrow 3 + a = 1 \Rightarrow a = - 2 On substituting a= - 2, x =1 and y =-6 in eq. (1), we get - 6 = 1 - 2 + b \Rightarrow b = - 5 ∴ a = - 2 and b = - 5$

Page No 15.10:

Question 4:

Find a point on the curve y = x³ − 3x where the tangent is parallel to the chord joining (1, −2) and (2, 2).

Answer:

Let (x₁, y₁) be the required point.

$Slope of the chord = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 + 2}{2 - 1} = 4 y = x^{3} - 3 x \Rightarrow \frac{d y}{d x} = 3 x^{2} - 3 . . . (1) Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} {=3x}_{1}^{2} -3 It is given that the tangent and the chord are parallel. ∴ Slope of the tangent = Slope of the chord \Rightarrow 3 {x_{1}}^{2} - 3 = 4 \Rightarrow 3 {x_{1}}^{2} = 7 \Rightarrow {x_{1}}^{2} = \frac{7}{3} \Rightarrow x_{1} = \pm \sqrt{\frac{7}{3}} = \sqrt{\frac{7}{3}} or - \sqrt{\frac{7}{3}} Case 1 When x_{1} = \sqrt{\frac{7}{3}} On substituting this in eq. (1), we get y_{1} = {(\sqrt{\frac{7}{3}})}^{3} - 3 (\sqrt{\frac{7}{3}}) = \frac{7}{3} \sqrt{\frac{7}{3}} - 3 \sqrt{\frac{7}{3}} = \frac{- 2}{3} \sqrt{\frac{7}{3}} ∴ (x_{1}, y_{1}) = (\sqrt{\frac{7}{3}}, \frac{- 2}{3} \sqrt{\frac{7}{3}}) Case 2 When x_{1} = - \sqrt{\frac{7}{3}} On substituting this in eq. (1), we get y_{1} = {(- \sqrt{\frac{7}{3}})}^{3} - 3 (- \sqrt{\frac{7}{3}}) = \frac{- 7}{3} \sqrt{\frac{7}{3}} + 3 \sqrt{\frac{7}{3}} = \frac{2}{3} \sqrt{\frac{7}{3}} ∴ (x_{1}, y_{1}) = (- \sqrt{\frac{7}{3}}, \frac{2}{3} \sqrt{\frac{7}{3}})$

Page No 15.10:

Question 5:

Find the points on the curve y = x³ − 2x² − 2x at which the tangent lines are parallel to the line y = 2x − 3.

Answer:

Let (x₁, y₁) be the required point.
Given:

$y = 2 x - 3 ∴ Slope of the line = \frac{d y}{d x} = 2 y = x^{3} - 2 x^{2} - 2 x Since (x_{1} y_{1}) lies on curve, y_{1} = {x_{1}}^{3} - 2 {x_{1}}^{2} - 2 x_{1} . . . (1) \Rightarrow {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} - 4 x_{1} - 2 It is given that the tangent and the given line are parallel. ∴ Slope of the tangent = Slope of the given line 3 {x_{1}}^{2} - 4 x_{1} - 2 = 2 \Rightarrow 3 {x_{1}}^{2} - 4 x_{1} - 4 = 0 \Rightarrow 3 {x_{1}}^{2} - 6 x_{1} + 2 x_{1} - 4 = 0 \Rightarrow 3 x_{1} (x_{1} - 2) + 2 (x_{1} - 2) = 0 \Rightarrow (x_{1} - 2) (3 x_{1} + 2) = 0 \Rightarrow x_{1} = 2 or x_{1} = \frac{- 2}{3} Case 1 When x_{1} = 2 On s ubstituting the value of x_{1} in eq. (1), we get y_{1} = 8 - 8 - 4 = - 4 ∴ (x_{1}, y_{1}) = (2, - 4) Case 2 When x_{1} = \frac{- 2}{3} On s ubstituting the value of x_{1} in eq. (1), we get y_{1} = \frac{- 8}{27} - \frac{8}{9} + \frac{4}{3} = \frac{- 8 - 24 + 36}{27} = \frac{4}{27} ∴ (x_{1}, y_{1}) = (\frac{- 2}{3}, \frac{4}{27})$

Page No 15.10:

Question 6:

Find the points on the curve y² = 2x³ at which the slope of the tangent is 3.

Answer:

Let (x₁, y₁) be the required point.
Given:
$y^{2} = 2 x^{3} Since (x_{1} y_{1}) lies on a curve, {y_{1}}^{2} = 2 {x_{1}}^{3} . . . . (1) \Rightarrow 2 y \frac{d y}{d x} = 6 x^{2} \Rightarrow \frac{d y}{d x} = \frac{6 x^{2}}{2 y} = \frac{3 x^{2}}{y} Slope of the tangent at (x, y) = \frac{3 {x_{1}}^{2}}{y_{1}} S lope of the tangent=3 [Given] ∴ \frac{3 {x_{1}}^{2}}{y_{1}} = 3 . . . . (2) \Rightarrow y_{1} = {x_{1}}^{2} On substituting the value of y_{1} in eq. (1), we get {x_{1}}^{4} = 2 {x_{1}}^{3} \Rightarrow {x_{1}}^{3} (x_{1} - 2) = 0 \Rightarrow x_{1} = 0, 2 Case 1 When x_{1} = 0, y_{1} = x^{2} = 0 . Thus, we get the point (0, 0) . But, it does not satisfy eq. (2). So, we can ignore (0, 0). Case 2 When x_{1} = 2, y_{1} = {x_{1}}^{2} = 4 . Thus, we get the point (2, 4) .$

Page No 15.10:

Question 7:

Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis.

Answer:

Let the required point be (x₁, y₁).
Slope of the tangent at this point = tan 45 $°$ = 1
Given:
$x y + 4 = 0 . . . (1) Since the point satisfies the above equation, x_{1} y_{1} + 4 = 0 . . . (2) On differentiating equation (2) both sides with respect to x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x, y)} = \frac{- y_{1}}{x_{1}} Slope of the tangent =1 [Given] ∴ \frac{- y_{1}}{x_{1}} = 1 \Rightarrow x_{1} = - y_{1} On substituting the value of x_{1} in eq. (2), we get - {y_{1}}^{2} + 4 = 0 \Rightarrow {y_{1}}^{2} = 4 \Rightarrow y_{1} = \pm 2 Case 1 When y_{1} = 2, x_{1} = - y_{1} = - 2 ∴ (x_{1}, y_{1}) = (-2, 2) Case 2 When y_{1} = - 2, x_{1} = - y_{1} = 2 ∴ (x_{1}, y_{1}) = (2, -2)$

Page No 15.10:

Question 8:

Find the point on the curve y = x² where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let the required point be (x₁, y₁).
Given:
$y = x^{2} Point (x_{1}, y_{1}) lies on a curve . ∴ y_{1} = {x_{1}}^{2} . . . (1) Now, y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} S lope of the tangent = x coordinate of the point [Given] ∴ 2 x_{1} = x_{1} This happens only when x_{1} = 0. On putting x_{1} = 0 in eq . (1), we get y_{1} = {x_{1}}^{2} = 0^{2} = 0 Thus, the required point is (0, 0) .$

Page No 15.10:

Question 9:

At what points on the circle x² + y² − 2x − 4y + 1 = 0, the tangent is parallel to x-axis?

Answer:

Let the required point be (x₁, y₁).
We know that the slope of the x-axis is 0.
Given:

$x^{2} + y^{2} - 2 x - 4 y + 1 = 0 (x_{1}, y_{1}) lies on a curve . ∴ {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 4 y_{1} + 1 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 4) = 2 - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y - 4} = \frac{1 - x}{y - 2} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1} - 2} ...(2) Slope of the tangent = 0 [Given] ∴ \frac{1 - x_{1}}{y_{1} - 2} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 On substituting the value of x_{1} in eq. (1), we get 1 + {y_{1}}^{2} - 2 - 4 y_{1} + 1 = 0 \Rightarrow {y_{1}}^{2} - 4 y_{1} = 0 \Rightarrow y_{1} (y_{1} - 4) = 0 \Rightarrow y_{1} = 0, 4 Thus, the required points are (1, 0) and (1, 4).$

Page No 15.10:

Question 10:

At what point of the curve y = x² does the tangent make an angle of 45° with the x-axis?

Answer:

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.10:

Question 11:

Find the points on the curve y = 3x² − 9x + 8 at which the tangents are equally inclined with the axes.

Answer:

Let (x₁, y₁) be the required point.
It is given that the tangent at this point is equally inclined to the axes. It means that the angle made by the tangent with the x-axis is $\pm$ 45 $°$ .
∴ Slope of the tangent = tan ( $\pm$ 45) = $\pm$ 1 ...(1)

$Since, the point lies on the curve . Hence, y_{1} = 3 {x_{1}}^{2} - 9 x_{1} + 8 Now, y = 3 x^{2} - 9 x + 8 \Rightarrow \frac{d y}{d x} = 6 x - 9 Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =6 x_{1} -9 ...(2) From eq. (1) and eq. (2), we get 6 x_{1} - 9 = \pm 1 \Rightarrow 6 x_{1} - 9 = 1 or 6 x_{1} - 9 = - 1 \Rightarrow 6 x_{1} = 10 or 6 x_{1} = 8 \Rightarrow x_{1} = \frac{10}{6} = \frac{5}{3} or x_{1} = \frac{8}{6} = \frac{4}{3} Also, y_{1} = 3 {(\frac{5}{3})}^{2} - 9 (\frac{5}{3}) + 8 or y_{1} = 3 {(\frac{4}{3})}^{2} - 9 (\frac{4}{3}) + 8 \Rightarrow y_{1} = \frac{25}{3} - \frac{45}{3} + 8 or y_{1} = \frac{16}{3} - \frac{36}{3} + 8 \Rightarrow y_{1} = \frac{4}{3} or y_{1} = \frac{4}{3} Thus, the required points are (\frac{5}{3}, \frac{4}{3}) and (\frac{4}{3}, \frac{4}{3}) .$

Page No 15.10:

Question 12:

At what points on the curve y = 2x² − x + 1 is the tangent parallel to the line y = 3x + 4?

Answer:

Let (x₁, y₁) be the required point.
The slope of line y = 3x + 4 is 3.

$Since, the point lies on the curve . Hence, y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 Now, y = 2 x^{2} - x + 1 \frac{d y}{d x} = 4 x - 1 Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =4 x_{1} -1 Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] ∴ 4 x_{1} - 1 = 3 \Rightarrow 4 x_{1} = 4 \Rightarrow x_{1} = 1 and y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 = 2 - 1 + 1 = 2 Thus, the required point is (1, 2) .$

Page No 15.10:

Question 13:

Find the point on the curve y = 3x² + 4 at which the tangent is perpendicular to the line whose slop is $- \frac{1}{6}$ .

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $\frac{- 1}{6}$
∴ Slope of the line perpendicular to it = 6

$Since, the point lies on the curve . Hence, y_{1} = 3 {x_{1}}^{2} + 4 Now, y = 3 x^{2} + 4 ∴ \frac{d y}{d x} = 6 x Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =6 x_{1} Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] ∴ 6 x_{1} = 6 \Rightarrow x_{1} = 1 and y_{1} = 3 {x_{1}}^{2} + 4 = 3 + 4 = 7 Thus, the required point is (1, 7) .$

Page No 15.11:

Question 14:

Find the points on the curve x² + y² = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

Answer:

Let (x₁, y₁) represent the required point.
The slope of line 2x + 3y = 7 is $\frac{- 2}{3}$ .

$Since, the point lies on the curve . Hence, {x_{1}}^{2} + {y_{1}}^{2} = 13 . . . (1) Now, x^{2} + y^{2} = 13 On differentiating both sides w.r.t. x, we get 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- x_{1}}{y_{1}} Slope of the tangent at (x_{1}, y_{1}) = Slope of the given line [Given] \Rightarrow \frac{- x_{1}}{y_{1}} = \frac{- 2}{3} \Rightarrow x_{1} = \frac{2 y_{1}}{3} . . . (2) From eq. (1), we get {(\frac{2 y_{1}}{3})}^{2} + {y_{1}}^{2} = 13 \Rightarrow \frac{13 {y_{1}}^{2}}{9} = 13 \Rightarrow {y_{1}}^{2} = 9 \Rightarrow y_{1} = \pm 3 \Rightarrow y_{1} = 3 or y_{1} = - 3 and x_{1} = 2 or x_{1} = - 2 [From eq. (2)] Thus, the required points are (2, 3) and (- 2, - 3) .$

Page No 15.11:

Question 15:

Find the points on the curve 2a²y = x³ − 3ax² where the tangent is parallel to x-axis.

Answer:

Let (x₁, y₁) represent the required points.
The slope of the x-axis is 0.
Here,
$2 a^{2} y = x^{3} - 3 a x^{2} Since, the point lies on the curve . Hence, 2 a^{2} y_{1} = {x_{1}}^{3} - 3 a {x_{1}}^{2} . . . (1) Now, 2 a^{2} y = x^{3} - 3 a x^{2} On differentiating both sides w.r.t. x, we get 2 a^{2} \frac{d y}{d x} = 3 x^{2} - 6 a x \Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 6 a x}{2 a^{2}} Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{3 {x_{1}}^{2} - 6 a x_{1}}{2 a^{2}} Given: Slope of the tangent at (x_{1}, y_{1}) = Slope of the x -axis \Rightarrow \frac{3 {x_{1}}^{2} - 6 a x_{1}}{2 a^{2}} = 0 \Rightarrow 3 {x_{1}}^{2} - 6 a x_{1} = 0 \Rightarrow x_{1} (3 x_{1} - 6 a) = 0 \Rightarrow x_{1} = 0 or x_{1} = 2 a Also, 2 a^{2} y_{1} = 0 or 2 a^{2} y_{1} = 8 a^{3} - 12 a^{3} [From eq. (1)] \Rightarrow y_{1} = 0 or y_{1} = - 2 a Thus, the required points are (0, 0) and (2 a, - 2 a) .$

Page No 15.11:

Question 16:

At what points on the curve y = x² − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $\frac{- 1}{2}$
Slope of the line perpendicular to this line = 2

$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 4 x_{1} + 5 . . . (1) Now, y = x^{2} - 4 x + 5 ∴ \frac{d y}{d x} = 2 x - 4 Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =2 x_{1} -4 Slope of the tangent at (x_{1}, y_{1}) =Slope of the given line [Given] ∴ 2 x_{1} - 4 = 2 \Rightarrow 2 x_{1} = 6 \Rightarrow x_{1} = 3 Also, y_{1} = 9 - 12 + 5 = 2 [From eq. (1)] Thus, the required point is (3, 2) .$

Page No 15.11:

Question 17:

Find the points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ at which the tangents are parallel to the (i) x-axis (ii) y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{4} + \frac{{y_{1}}^{2}}{25} = 1 . . . (1) Now, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 ∴ \frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2} \Rightarrow \frac{d y}{d x} = \frac{- 25 x}{4 y} Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 25 x_{1}}{4 y_{1}} Slope of the tangent at (x_{1}, y_{1}) =Slope of the x -axis [Given] ∴ \frac{- 25 x_{1}}{4 y_{1}} = 0 \Rightarrow x_{1} = 0 Also, 0 + \frac{{y_{1}}^{2}}{25} = 1 [From eq. (1)] \Rightarrow {y_{1}}^{2} = 25 \Rightarrow y_{1} = \pm 5 Thus, the required points are (0, 5) and (0, - 5) .$

(ii) The slope of the y-axis is $\infty$ .
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{4} + \frac{{y_{1}}^{2}}{25} = 1 . . . (1) Now, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 ∴ \frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2} \Rightarrow \frac{d y}{d x} = \frac{- 25 x}{4 y} Now, Slope of the tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 25 x_{1}}{4 y_{1}} Slope of the tangent at (x_{1}, y_{1}) =Slope of the y -axis [Given] ∴ \frac{- 25 x_{1}}{4 y_{1}} = \infty \Rightarrow \frac{4 y_{1}}{- 25 x_{1}} = 0 \Rightarrow y_{1} = 0 Also, \frac{{x_{1}}^{2}}{4} = 1 [From eq. (1)] \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Thus, the required points are (2, 0) and (- 2, 0) .$

Page No 15.11:

Question 18:

Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Answer:

Let (x₁, y₁) be the required point.

$Since the point lie on the curve . Hence {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 3 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 = 0 ∴ \frac{d y}{d x} = \frac{2 - 2 x}{2 y} = \frac{1 - x}{y} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} S lope of the tangent = 0 (Given) ∴ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 From (1), we get {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 \Rightarrow 1 + {y_{1}}^{2} - 2 - 3 = 0 \Rightarrow {y_{1}}^{2} - 4 = 0 \Rightarrow y_{1} = \pm 2 Hence, the points are (1, 2) and (1, - 2) .$

Page No 15.11:

Question 19:

Find the points on the curve $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$ at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Answer:

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{9} + \frac{{y_{1}}^{2}}{16} = 1 . . . (1) Now, \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \Rightarrow \frac{y}{16} \frac{d y}{d x} = \frac{- x}{9} \Rightarrow \frac{d y}{d x} = \frac{- 16 x}{9 y} Now, Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 16 x_{1}}{9 y_{1}} Slope of the tangent at (x, y) = Slope of the x -axis [Given] ∴ \frac{- 16 x_{1}}{9 y_{1}} = 0 \Rightarrow x_{1} = 0 0 + \frac{{y_{1}}^{2}}{16} = 1 [From eq. (1)] Also, {y_{1}}^{2} = 16 y_{1} = \pm 4 Thus, the required points are (0, 4) and (0, - 4) .$

(ii) The slope of the y-axis is $\infty$ .
Let (x₁, y₁) be the required point.
Given:
$Since, the point lies on the curve . Hence, \frac{{x_{1}}^{2}}{9} + \frac{{y_{1}}^{2}}{16} = 1 . . . (1) \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \Rightarrow \frac{y}{16} \frac{d y}{d x} = \frac{- x}{9} \Rightarrow \frac{d y}{d x} = \frac{- 16 x}{9 y} Now, Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 16 x_{1}}{9 y_{1}} Slope of the tangent at (x_{1}, y_{1}) = Slope of the y -axis [Given] ∴ \frac{- 16 x_{1}}{9 y_{1}} = \infty \Rightarrow \frac{9 y_{1}}{- 16 x_{1}} = 0 \Rightarrow y_{1} = 0 \Rightarrow \frac{{x_{1}}^{2}}{9} + 0 = 1 [From eq. (1)] \Rightarrow {x_{1}}^{2} = 9 \Rightarrow x_{1} = \pm 3 Thus, the required points are (3, 0) and (- 3, 0) .$

Page No 15.11:

Question 20:

Who that the tangents to the curve y = 7x³ + 11 at the points x = 2 and x = −2 are parallel.

Answer:

$Given : y = 7 x^{3} + 11 ∴ \frac{d y}{d x} = 21 x^{2} Now, Slope of the tangent at (x = 2) = {(\frac{d y}{d x})}_{x = 2} = 21 {(2)}^{2} = 84 Slope of the tangent at (x = - 2) = {(\frac{d y}{d x})}_{x = - 2} = 21 {(- 2)}^{2} = 84$

Both slopes are the same. Hence, the tangents at points x = 2 and x = −2 are parallel.

Page No 15.11:

Question 21:

Find the points on the curve y = x³ where the slope of the tangent is equal to the x-coordinate of the point.

Answer:

Let (x₁, y₁) be the required point.
x coordinate of the point is x₁.

$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{3} . . . (1) Now, y = x^{3} \Rightarrow \frac{d y}{d x} = 3 x^{2} Slope of tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} Given that Slope of tangent at (x_{1}, y_{1}) = x co-ordinate of the point \Rightarrow 3 {x_{1}}^{2} = x_{1} \Rightarrow x_{1} (3 x_{1} - 1) = 0 \Rightarrow x_{1} = 0 or x_{1} = \frac{1}{3} \Rightarrow y_{1} = 0^{3} or y_{1} = {(\frac{1}{3})}^{3} (From (1)) \Rightarrow y_{1} = 0 or y_{1} = \frac{1}{27} So, the points are (x_{1}, y_{1}) = (0, 0), (\frac{1}{3}, \frac{1}{27})$

Page No 15.27:

Question 1:

Find the equation of the tangent to the curve $\sqrt{x} + \sqrt{y} = a$ , at the point $(\frac{a^{2}}{4}, \frac{a^{2}}{4})$ .

Answer:

$\sqrt{x} + \sqrt{y} = a Differentiating both sides w.r.t. x, \Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- \sqrt{y}}{\sqrt{x}} Given (x_{1}, y_{1}) = (\frac{a^{2}}{4}, \frac{a^{2}}{4}) Slope of tangent, m = {(\frac{d y}{d x})}_{(\frac{a^{2}}{4}, \frac{a^{2}}{4})} = \frac{- \sqrt{\frac{a^{2}}{4}}}{\sqrt{\frac{a^{2}}{4}}} =-1 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{a^{2}}{4} = - 1 (x - \frac{a^{2}}{4}) \Rightarrow y - \frac{a^{2}}{4} = - x + \frac{a^{2}}{4} \Rightarrow x + y = \frac{a^{2}}{2}$

Page No 15.27:

Question 2:

Find the equation of the normal to y = 2x³ − x² + 3 at (1, 4).

Answer:

$y = 2 x^{3} - x^{2} + 3 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 6 x^{2} - 2 x Slope of tangent = {(\frac{d y}{d x})}_{(1, 4)} =6 {(1)}^{2} -2 (1) =4 Slope of normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{4} Given (x_{1}, y_{1}) = (1, 4) Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 4 = \frac{- 1}{4} (x - 1) \Rightarrow 4 y - 16 = - x + 1 \Rightarrow x + 4 y = 17$

Page No 15.27:

Question 3:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) y = x⁴ − bx³ + 13x² − 10x + 5 at (0, 5) [NCERT]
(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at x = 1 [NCERT, CBSE 2011]
(iii) y = x² at (0, 0) [NCERT]
(iv) y = 2x² − 3x − 1 at (1, −2)
(v) $y^{2} = \frac{x^{3}}{4 - x} at (2, - 2)$
(vi) y = x² + 4x + 1 at x = 3 [CBSE 2004]
(vii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (a \cos θ, b \sin θ)$
(viii) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (a \sec θ, b \tan θ)$
(ix) y² = 4ax at $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(x) $c^{2} (x^{2} + y^{2}) = x^{2} y^{2} at (\frac{c}{\cos θ}, \frac{c}{\sin θ})$
(ix) xy = c² at $(c t, \frac{c}{t})$
(xii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (x_{1}, y_{1})$
(xiii) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (x_{0}, y_{0})$ [NCERT]
(xiv) $x^{\frac{2}{3}} + y^{\frac{2}{3}}$ = 2 at (1, 1) [NCERT]
(xv) x² = 4y at (2, 1)
(xvi) y² = 4x at (1, 2) [NCERT]
(xvii) 4x² + 9y² = 36 at (3cosθ, 2sinθ) [CBSE 2011]
(xviii) y² = 4ax at (x₁, y₁) [CBSE 2012]
(xix) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (\sqrt{2} a, b)$ [CBSE 2014]

Answer:

(i)
$y = x^{4} - b x^{3} + 13 x^{2} - 10 x + 5 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x^{3} - 3 b x^{2} + 26 x - 10 Slope of tangent, m = {(\frac{d y}{d x})}_{(0, 5)} =-10 Given (x_{1}, y_{1}) = (0, 5) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 5 = - 10 (x - 0) \Rightarrow y - 5 = - 10 x \Rightarrow y + 10 x - 5 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 5 = \frac{1}{10} (x - 0) \Rightarrow 10 y - 50 = x \Rightarrow x - 10 y + 50 = 0$

(ii)
$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 When x = 1, y = 1 - 6 + 13 - 10 + 5 = 3 So, (x_{1}, y_{1}) = (1, 3) Now, y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 3)} =4-18+ 26 - 10 = 2 Equation of tangent is, y - y_{1} = 2 (x - x_{1}) \Rightarrow y - 3 = 2 (x - 1) \Rightarrow y - 3 = 2 x - 2 \Rightarrow 2 x - y + 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 3 = \frac{- 1}{2} (x - 1) \Rightarrow 3 y - 6 = - x + 1 \Rightarrow x + 3 y - 7 = 0$

(iii)
$y = x^{2} Differentiating both sides w.r.t. x, \frac{d y}{d x} = 2 x Given (x_{1}, y_{1}) = (0, 0) Slope of tangent, m = {(\frac{d y}{d x})}_{(0, 0)} =2 (0) =0 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = 0 (x - 0) \Rightarrow y = 0$

$Equation of normal is, \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 0 = \frac{- 1}{0} (x - 0) \Rightarrow x = 0$

(iv)
$y =2 x^{2} -3 x -1 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 4 x - 3 Given (x_{1}, y_{1}) = (1, - 2) Slope of tangent, m = {(\frac{d y}{d x})}_{(1, - 2)} =4-3=1 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 2 = 1 (x - 1) \Rightarrow y + 2 = x - 1 \Rightarrow x - y - 3 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y + 2 = - 1 (x - 1) \Rightarrow y + 2 = - x + 1 \Rightarrow x + y + 1 = 0$

(v)
$y^{2} = \frac{x^{3}}{4 - x} Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = \frac{(4 - x) (3 x^{2}) - x^{3} (- 1)}{{(4 - x)}^{2}} = \frac{12 x^{2} - 3 x^{3} + x^{3}}{{(4 - x)}^{2}} = \frac{12 x^{2} - 2 x^{3}}{{(4 - x)}^{2}} \frac{d y}{d x} = \frac{12 x^{2} - 2 x^{3}}{2 y {(4 - x)}^{2}} Given (x_{1}, y_{1}) = (2, - 2) Slope of tangent, m = {(\frac{d y}{d x})}_{(2, - 2)} = \frac{48 - 16}{- 16} =-2 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 2 = - 2 (x - 2) \Rightarrow y + 2 = - 2 x + 4 \Rightarrow 2 x + y - 2 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y + 2 = \frac{1}{2} (x - 2) \Rightarrow 2 y + 4 = x - 2 \Rightarrow x - 2 y - 6 = 0$

(vi)
$y = x^{2} + 4 x + 1 Differentiating both sides w.r.t. x, \frac{d y}{d x} = 2 x + 4 When x =3, y = 9 + 12 + 1 = 22 So, (x_{1}, y_{1}) = (3, 22) Slope of tangent, m = {(\frac{d y}{d x})}_{x = 3} =10 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 22 = 10 (x - 3) \Rightarrow y - 22 = 10 x - 30 \Rightarrow 10 x - y - 8 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 22 = \frac{- 1}{10} (x - 3) \Rightarrow 10 y - 220 = - x + 3 \Rightarrow x + 10 y - 223 = 0$

(vii)
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \Rightarrow \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{- 2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(a \cos θ, b \sin θ)} = \frac{- a \cos θ (b^{2})}{b \sin θ (a^{2})} = \frac{- b \cos θ}{a \sin θ} Given (x_{1}, y_{1}) = (a \cos θ, b \sin θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \sin θ = \frac{- b \cos θ}{a \sin θ} (x - a \cos θ) \Rightarrow a y \sin θ - a b \sin^{2} θ = - b x \cos θ + a b \cos^{2} θ \Rightarrow b x \cos θ + a y \sin θ = a b Dividing by ab, \Rightarrow \frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b \sin θ = \frac{a \sin θ}{b \cos θ} (x - a \cos θ) \Rightarrow b y \cos θ - b^{2} \sin θ \cos θ = a x \sin θ - a^{2} \sin θ \cos θ \Rightarrow a x \sin θ - b y \cos θ = (a^{2} - b^{2}) \sin θ \cos θ Dividing by \sin θ \cos θ, a x \sec θ - b y cosec θ = (a^{2} - b^{2})$

(viii)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \Rightarrow \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(a \sec θ, b \tan θ)} = \frac{a \sec θ (b^{2})}{b \tan θ (a^{2})} = \frac{b}{a \sin θ} Given (x_{1}, y_{1}) = (a \sec θ, b \tan θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan θ = \frac{b}{a \sin θ} (x - a \sec θ) \Rightarrow a y \sin θ - a b \frac{\sin^{2} θ}{\cos θ} = b x - \frac{a b}{\cos θ} \Rightarrow \frac{a y \sin θ \cos θ - a b \sin^{2} θ}{\cos θ} = \frac{b x \cos θ - a b}{\cos θ} \Rightarrow a y \sin θ \cos θ - a b \sin^{2} θ = b x \cos θ - a b \Rightarrow b x \cos θ - a y \sin θ \cos θ = a b (1 - \sin^{2} θ) \Rightarrow b x \cos θ - a y \sin θ \cos θ = a b \cos^{2} θ Dividing by ab \cos^{2} θ, \Rightarrow \frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b \tan θ = \frac{- a \sin θ}{b} (x - a \sec θ) \Rightarrow y b - b^{2} \tan θ = - a x \sin θ + a^{2} \tan θ \Rightarrow a x \sin θ + b y = (a^{2} + b^{2}) \tan θ Dividing by tan θ, a x \cos θ + b y \cot θ = (a^{2} + b^{2})$

(ix)
$y^{2} =4 ax Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} Given (x_{1}, y_{1}) = (\frac{a}{m^{2}}, \frac{2 a}{m}) Slope of tangent = {(\frac{d y}{d x})}_{(\frac{a}{m^{2}}, \frac{2 a}{m})} = \frac{2 a}{(\frac{2 a}{m})} = m Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{2 a}{m} = m (x - \frac{a}{m^{2}}) \Rightarrow \frac{m y - 2 a}{m} = m (\frac{m^{2} x - a}{m^{2}}) \Rightarrow m y - 2 a = m^{2} x - a \Rightarrow m^{2} x - m y + a = 0$

$Equation of normal is, y - y_{1} = \frac{1}{Slope of tangent} (x - x_{1}) \Rightarrow y - \frac{2 a}{m} = \frac{- 1}{m} (x - \frac{a}{m^{2}}) \Rightarrow \frac{m y - 2 a}{m} = \frac{- 1}{m} (\frac{m^{2} x - a}{m^{2}}) \Rightarrow m^{3} y - 2 a m^{2} = - m^{2} x + a \Rightarrow m^{2} x + m^{3} y - 2 a m^{2} - a = 0$

(x)
$c^{2} (x^{2} + y^{2}) = x^{2} y^{2} Differentiating both sides w.r.t. x, \Rightarrow 2 x c^{2} + 2 y c^{2} \frac{d y}{d x} = x^{2} 2 y \frac{d y}{d x} + 2 x y^{2} \Rightarrow \frac{d y}{d x} (2 y c^{2} - 2 x^{2} y) = 2 x y^{2} - 2 x c^{2} \Rightarrow \frac{d y}{d x} = \frac{x y^{2} - x c^{2}}{y c^{2} - x^{2} y} Slope of tangent, m = {(\frac{d y}{d x})}_{(\frac{c}{\cos θ}, \frac{c}{\sin θ})} = \frac{\frac{c^{3}}{\cos θ \sin^{2} θ} - \frac{c^{3}}{\cos θ}}{\frac{c^{3}}{\sin θ} - \frac{c^{3}}{\cos^{2} θ \sin θ}} = \frac{\frac{1 - \sin^{2} θ}{\cos θ \sin^{2} θ}}{\frac{\cos^{2} θ - 1}{\cos^{2} θ \sin θ}} = \frac{c o s^{2} θ}{\cos θ \sin^{2} θ} \times \frac{\cos^{2} θ \sin θ}{- \sin^{2} θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} Given (x_{1}, y_{1}) = (\frac{c}{\cos θ}, \frac{c}{\sin θ}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (x - \frac{c}{\cos θ}) \Rightarrow \frac{y \sin θ - c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (\frac{x \cos θ - c}{\cos θ}) \Rightarrow \sin^{2} θ (y \sin θ - c) = - \cos^{2} θ (x \cos θ - c) \Rightarrow y \sin^{3} θ - c \sin^{2} θ = - x \cos^{3} θ + c \cos^{2} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c (s i n^{2} θ + \cos^{2} θ) \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{c}{\sin θ} = \frac{\sin^{3} θ}{\cos^{3} θ} (x - \frac{c}{\cos θ}) \Rightarrow \cos^{3} θ (y - \frac{c}{\sin θ}) = \sin^{3} θ (x - \frac{c}{\cos θ}) \Rightarrow y \cos^{3} θ - \frac{c \cos^{3} θ}{\sin θ} = x \sin^{3} θ - \frac{c \sin^{3} θ}{\cos θ} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = \frac{c \sin^{3} θ}{\cos θ} - \frac{c \cos^{3} θ}{\sin θ} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c (\frac{\sin^{4} θ - \cos^{4} θ}{\cos θ \sin θ}) \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c [\frac{(\sin^{2} θ + \cos^{2} θ) (\sin^{2} θ - \cos^{2} θ)}{\cos θ \sin θ}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- (\cos^{2} θ - \sin^{2} θ)}{2 \cos θ \sin θ}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- \cos (2 θ)}{\sin (2 θ)}] \Rightarrow \sin^{3} θ - y \cos^{3} θ = - 2 c c o t (2 θ) \Rightarrow \sin^{3} θ - y \cos^{3} θ + 2 c c o t (2 θ) = 0$

(xi)
${xy=c}^{2} Differentiating both sides w.r.t. x, x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} Given (x_{1}, y_{1}) = (c t, \frac{c}{t}) Slope of tangent, m = {(\frac{d y}{d x})}_{(c t, \frac{c}{t})} = \frac{- \frac{c}{t}}{c t} = \frac{- 1}{t^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{c}{t} = \frac{- 1}{t^{2}} (x - c t) \Rightarrow \frac{y t - c}{t} = \frac{- 1}{t^{2}} (x - c t) \Rightarrow y t^{2} - c t = - x + c t \Rightarrow x + y t^{2} = 2 c t$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{c}{t} = t^{2} (x - c t) \Rightarrow y t - c = t^{3} x - c t^{4} \Rightarrow x t^{3} - y t = c t^{4} - c$

(xii)
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{- 2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- x_{1} b^{2}}{y_{1} a^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{- x_{1} b^{2}}{y_{1} a^{2}} (x - x_{1}) \Rightarrow y y_{1} a^{2} - {y_{1}}^{2} a^{2} = - x x_{1} b^{2} + {x_{1}}^{2} b^{2} \Rightarrow x x_{1} b^{2} + y y_{1} a^{2} = {x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2} . . . (1) Since (x_{1}, y_{1}) lies on the given curve.Therefore, \frac{{x_{1}}^{2}}{a^{2}} + \frac{{y_{1}}^{2}}{b^{2}} = 1 \Rightarrow \frac{{x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2}}{a^{2} b^{2}} = 1 \Rightarrow {x_{1}}^{2} b^{2} + {y_{1}}^{2} a^{2} = a^{2} b^{2} Substituting this in (1), we get x x_{1} b^{2} + y y_{1} a^{2} = a^{2} b^{2} {Dividing this by a}^{2} b^{2}, \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{y_{1} a^{2}}{x_{1} b^{2}} (x - x_{1}) \Rightarrow y x_{1} b^{2} - x_{1} y_{1} b^{2} = x y_{1} a^{2} - x_{1} y_{1} a^{2} \Rightarrow x y_{1} a^{2} - y x_{1} b^{2} = x_{1} y_{1} a^{2} - x_{1} y_{1} b^{2} \Rightarrow x y_{1} a^{2} - y x_{1} b^{2} = x_{1} y_{1} (a^{2} - b^{2}) Dividing by x_{1} y_{1} \frac{a^{2} x}{x_{1}} - \frac{b^{2} y}{y_{1}} = a^{2} - b^{2}$

(xiii)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{0}, y_{0})} = \frac{x_{0} b^{2}}{y_{0} a^{2}} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{0} = \frac{x_{0} b^{2}}{y_{0} a^{2}} (x - x_{0}) \Rightarrow y y_{0} a^{2} - {y_{0}}^{2} a^{2} = x x_{0} b^{2} - {x_{0}}^{2} b^{2} x x_{0} b^{2} - y y_{0} a^{2} = {x_{0}}^{2} b^{2} - {y_{0}}^{2} a^{2} . . . (1) Since (x_{0}, y_{0}) lies on the given curve, \Rightarrow \frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = 1 \Rightarrow {x_{0}}^{2} b^{2} - {y_{0}}^{2} a^{2} = a^{2} b^{2} Substituting this in (1), we get \Rightarrow x x_{0} b^{2} - y y_{0} a^{2} = a^{2} b^{2} Dividing this by a^{2} b^{2} \frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = 1$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{0} = \frac{- y_{0} a^{2}}{x_{0} b^{2}} (x - x_{0}) \Rightarrow y x_{0} b^{2} - x_{0} y_{0} b^{2} = - x y_{0} a^{2} + x_{0} y_{0} a^{2} \Rightarrow x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} a^{2} + x_{0} y_{0} b^{2} \Rightarrow x y_{0} a^{2} + y x_{0} b^{2} = x_{0} y_{0} (a^{2} + b^{2}) Dividing by x_{0} y_{0} \frac{a^{2} x}{x_{0}} + \frac{b^{2} y}{y_{0}} = a^{2} + b^{2}$

(xiv)
$x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2 Differentiating both sides w.r.t. x, \frac{2}{3} x^{\frac{- 1}{3}} + \frac{2}{3} y^{\frac{- 1}{3}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x^{\frac{- 1}{3}}}{y^{\frac{- 1}{3}}} = \frac{- y^{\frac{1}{3}}}{x^{\frac{1}{3}}} Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{1} =-1 Given (x_{1}, y_{1}) = (1, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = - 1 (x - 1) \Rightarrow y - 1 = - x + 1 \Rightarrow x + y - 2 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = 1 (x - 1) \Rightarrow y - 1 = x - 1 \Rightarrow y - x = 0$

(xv)
$x^{2} = 4 y Differentiating both sides w.r.t. x, 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2} Slope of tangent, m = {(\frac{d y}{d x})}_{(2, 1)} = \frac{2}{2} =1 Given (x_{1}, y_{1}) = (2, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = 1 (x - 2) \Rightarrow y - 1 = x - 2 \Rightarrow x - y - 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = - 1 (x - 2) \Rightarrow y - 1 = - x + 2 \Rightarrow x + y - 3 = 0$

(xvi)
$y^{2} = 4 x Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} Slope of tangent, m = {(\frac{d y}{d x})}_{(1, 2)} = \frac{2}{2} =1 Given (x_{1}, y_{1}) = (1, 2) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = 1 (x - 1) \Rightarrow y - 2 = x - 1 \Rightarrow x - y + 1 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 = - 1 (x - 1) \Rightarrow y - 2 = - x + 1 \Rightarrow x + y - 3 = 0$

(xvii) Equation of tangent:
$4 x^{2} + 9 y^{2} = 36 Differentiating both sides w.r.t. x, 8 x + 18 y \frac{d y}{d x} = 0 \Rightarrow 18 y \frac{d y}{d x} = - 8 x \Rightarrow \frac{d y}{d x} = \frac{- 8 x}{18 y} = \frac{- 4 x}{9 y} Slope of tangent, m = {(\frac{d y}{d x})}_{(3 \cos θ, 2 \sin θ)} = \frac{- 12 \cos θ}{18 \sin θ} = \frac{- 2 \cos θ}{3 \sin θ} Given (x_{1}, y_{1}) = (3 \cos θ, 2 \sin θ) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 \sin θ = \frac{- 2 \cos θ}{3 \sin θ} (x - 3 \cos θ) \Rightarrow 3 y \sin θ - 6 \sin^{2} θ = - 2 x \cos θ + 6 \cos^{2} θ \Rightarrow 2 x \cos θ + 3 y \sin θ = 6 (\cos^{2} θ + \sin^{2} θ) \Rightarrow 2 x \cos θ + 3 y \sin θ = 6$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 \sin θ = \frac{3 \sin θ}{2 \cos θ} (x - 3 \cos θ) \Rightarrow 2 y \cos θ - 4 \sin θ \cos θ = 3 x \sin θ - 9 \sin θ \cos θ \Rightarrow 3 x \sin θ - 2 y \cos θ - 5 \sin θ \cos θ = 0$

(xviii)
$y^{2} =4 ax Differentiating both sides w.r.t. x, 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} At (x_{1}, y_{1}) Slope of tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{2 a}{y_{1}} = m Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - y_{1} = \frac{2 a (x - x_{1})}{y_{1}} \Rightarrow y y_{1} - {y_{1}}^{2} = 2 a x - 2 a x_{1} \Rightarrow y y_{1} - 4 a x_{1} = 2 a x - 2 a x_{1} \Rightarrow y y_{1} = 2 a x + 2 a x_{1} \Rightarrow y y_{1} = 2 a (x + x_{1})$

$Equation of normal is, y - y_{1} = \frac{1}{Slope of tangent} (x - x_{1}) \Rightarrow y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - y_{1} = \frac{- y_{1}}{2 a} (x - x_{1})$

(xix)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Differentiating both sides w.r.t. x, \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} Slope of tangent, m = {(\frac{d y}{d x})}_{(\sqrt{2} a,b)} = \frac{\sqrt{2} a b^{2}}{b a^{2}} = \frac{\sqrt{2} b}{a} Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b = \frac{\sqrt{2} b}{a} (x - \sqrt{2} a) \Rightarrow a y - a b = \sqrt{2} b x - 2 a b \Rightarrow \sqrt{2} b x - a y = a b \Rightarrow \frac{\sqrt{2} x}{a} - \frac{y}{b} = 1$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - b = \frac{- a}{\sqrt{2} b} (x - \sqrt{2} a) \Rightarrow \sqrt{2} b y - \sqrt{2} b^{2} = - a x + \sqrt{2} a^{2} \Rightarrow a x + \sqrt{2} b y = \sqrt{2} b^{2} + \sqrt{2} a^{2} \Rightarrow \frac{a x}{\sqrt{2}} + b y = a^{2} + b^{2}$

Page No 15.28:

Question 4:

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Answer:

$x = θ + \sin θ and y = 1 + \cos θ \frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{1 + \cos θ} Slope of tangent= {(\frac{d y}{d x})}_{θ = \frac{π}{4}} = \frac{- \sin \frac{π}{4}}{1 + \cos \frac{π}{4}} = \frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{-1}{\sqrt{2} + 1} = \frac{-1}{\sqrt{2} + 1} × \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = 1 - \sqrt{2} (x_{1}, y_{1}) = (\frac{π}{4} + \sin \frac{π}{4}, 1 + \cos \frac{π}{4}) = (\frac{π}{4} + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - (1 + \frac{1}{\sqrt{2}}) = (1 - \sqrt{2}) [x - (\frac{π}{4} + \frac{1}{\sqrt{2}})] \Rightarrow y - 1 - \frac{1}{\sqrt{2}} = (1 - \sqrt{2}) [x - \frac{π}{4} - \frac{1}{\sqrt{2}}]$

Page No 15.28:

Question 5:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθ, y = 1 + cosθ at θ = $\frac{π}{2}$
(ii) $x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = \frac{1}{2}$
(iii) x = at², y = 2at at t = 1
(iv) x = asect, y = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi) x = 3cosθ − cos³θ, y = 3sinθ − sin³θ [NCERT EXEMPLAR]

Answer:

(i)
$x = θ + \sin θ and y = 1 + \cos θ \frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{1 + \cos θ} Slope of tangent, m = {(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{- \sin \frac{π}{2}}{1 + \cos \frac{π}{2}} = \frac{- 1}{1 + 0} =-1 Now, (x_{1}, y_{1}) = (\frac{π}{2} + \sin \frac{π}{2}, 1 + \cos \frac{π}{2}) = (\frac{π}{2} + 1, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = - 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = - 2 x + π + 2 \Rightarrow 2 x + 2 y - π - 4 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = 2 x - π - 2 \Rightarrow 2 x - 2 y = π$

(ii) Equation of tangent:
$x = \frac{2 a t^{2}}{1 + t^{2}} and y = \frac{2 a t^{3}}{1 + t^{2}} \frac{d x}{d t} = \frac{(1 + t^{2}) (4 a t) - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} = \frac{4 a t}{{(1 + t^{2})}^{2}} and \frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t^{2} - 2 a t^{3} (2 t)}{{(1 + t^{2})}^{2}} = \frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}}}{\frac{4 a t}{{(1 + t^{2})}^{2}}} = \frac{6 a t^{2} + 2 a t^{4}}{4 a t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{1}{2}} = \frac{\frac{3 a}{2} + \frac{a}{8}}{2 a} = \frac{\frac{12 a + a}{8}}{2 a} = \frac{13 a}{8} × \frac{1}{2 a} = \frac{13}{16} Now, (x_{1}, y_{1}) = (\frac{2 a t^{2}}{1 + t^{2}}, \frac{2 a t^{3}}{1 + t^{2}}) = (\frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}}) = (\frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}}) = (\frac{2 a}{5}, \frac{a}{5}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{13}{16} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{13}{16} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{13}{16} (5 x - 2 a) \Rightarrow 80 y - 16 a = 65 x - 26 a \Rightarrow 65 x - 80 y - 10 a = 0 \Rightarrow 13 x - 16 y - 2 a = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{- 16}{13} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{- 16}{13} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{- 16}{13} (5 x - 2 a) \Rightarrow 65 y - 13 a = - 80 x + 32 a \Rightarrow 80 x + 65 y - 45 a = 0 \Rightarrow 16 x + 13 y - 9 a = 0$

(iii)
$x = a t^{2} and y = 2 a t \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = 1} = \frac{1}{1} =1 Now, (x_{1}, y_{1}) = (a, 2 a) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = 1 (x - a) \Rightarrow y - 2 a = x - a \Rightarrow x - y + a = 0$

Equation of normal:

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = - 1 (x - a) \Rightarrow y - 2 a = - x + a \Rightarrow x + y = 3 a$

(iv)
$x = a \sec t and y = b \tan t \frac{d x}{d t} = a \sec t \tan t and \frac{d y}{d t} = b \sec^{2} t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b \sec^{2} t}{a \sec t \tan t} = \frac{b}{a} \cos e c t Slope of tangent, m = {(\frac{d y}{d x})}_{t = t} = \frac{b}{a} \cos e c t Now, (x_{1}, y_{1}) = (a \sec t, b \tan t) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{b}{a} \cos e c t (x - a s e c t) \Rightarrow y - \frac{b \sin t}{\cos t} = \frac{b}{a \sin t} (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{b}{a \sin t} (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{b}{a \sin t} (x \cos t - a) \Rightarrow a y \sin t \cos t - a b \sin^{2} t = b x \cos t - a b \Rightarrow b x \cos t - a y \sin t \cos t - a b (1 - \sin^{2} t) = 0 \Rightarrow b x \cos t - a y \sin t \cos t = a b \cos^{2} t Dividing by \cos^{2} t, b x \sec t - a y \tan t = a b$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{- a}{b} sin t (x - a \sec t) \Rightarrow y - b \frac{\sin t}{\cos t} = \frac{- a}{b} sin t (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{- a}{b} sin t (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{- a}{b} s in t (x \cos t - a) \Rightarrow b y \cos t - b^{2} \sin t = - a x \sin t \cos t + a^{2} \sin t \Rightarrow a x \sin t \cos t + b y \cos t = (a^{2} + b^{2}) \sin t Dividing both sides by sin t, a x \cos t + b y \cot t = a^{2} + b^{2}$

(v)
$x = a (θ + \sin θ) and y = a (1 - \cos θ) \frac{d x}{d θ} = a (1 + \cos θ) and \frac{d y}{d θ} = a \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a \sin θ}{a (1 + \cos θ)} = \frac{\sin θ}{(1 + \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} = \tan \frac{θ}{2} . . . (1) Slope of tangent, m = {(\frac{d y}{d x})}_{θ} = \tan \frac{θ}{2} Now, (x_{1}, y_{1}) = [a (θ + \sin θ), a (1 - \cos θ)] Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - a (1 - \cos θ) = \tan \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \tan \frac{θ}{2} \sin θ \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} 2 \sin \frac{θ}{2} \cos \frac{θ}{2} (From (1)) \Rightarrow y - 2 a \sin^{2} \frac{θ}{2} = (x - a θ) \tan \frac{θ}{2} - 2 a \sin^{2} \frac{θ}{2} \Rightarrow y = (x - a θ) \tan \frac{θ}{2}$

$Equation of normal is, y - a (1 - \cos θ) = - \cot \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow \tan \frac{θ}{2} [y - a (2 \sin^{2} \frac{θ}{2})] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} [y - a \{2 (1 - \cos^{2} \frac{θ}{2})\}] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a (2 \sin \frac{θ}{2} \cos \frac{θ}{2}) = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a \sin θ = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) = - x + a θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + x - a θ = 0$

(vi)
$x = 3 \cos θ - \cos^{3} θ and y = 3 \sin θ - \sin^{3} θ \frac{d x}{d θ} = - 3 \sin θ + 3 \cos^{2} θ \sin θ = 3 \sin θ (\cos^{2} θ - 1) = 3 \sin θ (- \sin^{2} θ) = - 3 \sin^{3} θ and \frac{d y}{d θ} = 3 \cos θ - 3 \sin^{2} θ \cos θ = 3 \cos θ (1 - \sin^{2} θ) = 3 \cos θ (\cos^{2} θ) = 3 \cos^{3} θ ∴ \frac{d y}{d x} = \frac{3 \cos^{3} θ}{- 3 \sin^{3} θ} = - \tan^{3} θ . . . (1) Slope of tangent, m = {(\frac{d y}{d x})}_{θ} = - \tan^{3} θ Now, (x_{1}, y_{1}) = [3 \cos θ - \cos^{3} θ, 3 \sin θ - \sin^{3} θ] Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 3 \sin θ - \sin^{3} θ = - \tan^{3} θ [x - 3 \cos θ - \cos^{3} θ] \Rightarrow y - 3 \sin θ - \sin^{3} θ = \frac{- \sin^{3} θ}{\cos^{3} θ} [x - 3 \cos θ - \cos^{3} θ] \Rightarrow y \cos^{3} θ - 3 \sin θ \cos^{3} θ - \sin^{3} θ \cos^{3} θ = - x \sin^{3} θ + 3 \sin^{3} θ \cos θ + \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ + 3 \sin θ \cos^{3} θ + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ (\sin^{2} θ + \cos^{2} θ) + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x \sin^{3} θ + y \cos^{3} θ = 3 \sin^{3} θ \cos θ + 2 \sin^{3} θ \cos^{3} θ \Rightarrow x + y \cot^{3} θ = 3 \cos θ + 2 \cos^{3} θ$

$Equation of normal is, y - (3 \cos θ - \cos^{3} θ) = \frac{- 1}{m} [x - (3 \sin θ - \sin^{3} θ)] \Rightarrow y - 3 \cos θ + \cos^{3} θ = \frac{- 1}{\tan^{3} θ} [x - 3 \sin θ + \sin^{3} θ] \Rightarrow y - 3 \cos θ + \cos^{3} θ = \frac{- \cos^{3} θ}{\sin^{3} θ} [x - 3 \sin θ + \sin^{3} θ] \Rightarrow y \sin^{3} θ - 3 \sin^{3} θ \cos θ + \sin^{3} θ \cos^{3} θ = - x \cos^{3} θ + 3 \cos^{3} θ \sin θ - \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ + 3 \cos^{3} θ \sin θ - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ (\sin^{2} θ + \cos^{2} θ) - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = 3 \sin^{3} θ \cos θ - 2 \cos^{3} θ \sin^{3} θ \Rightarrow x co t^{3} θ + y = 3 \cos θ - 2 \cos^{3} θ$

Page No 15.28:

Question 6:

Find the equation of the normal to the curve x² + 2y² − 4x − 6y + 8 = 0 at the point whose abscissa is 2.

Answer:

Abscissa means the horizontal co-ordiante of a point.
Given that abscissa = 2.
i.e., x = 2

$x^{2} + 2 y^{2} - 4 x - 6 y + 8 = 0 . . . (1) Differentiating both sides w.r.t. x, 2 x + 4 y \frac{d y}{d x} - 4 - 6 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (4 y - 6) = 4 - 2 x \Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{4 y - 6} = \frac{2 - x}{2 y - 3} When x =2, from (1), we get 4 + 2 y^{2} - 8 - 6 y + 8 = 0 \Rightarrow 2 y^{2} - 6 y + 4 = 0 \Rightarrow y^{2} - 3 y + 2 = 0 \Rightarrow (y - 1) (y - 2) = 0 \Rightarrow y = 1 or y = 2 Case-1: y = 1 Slope of tangent = {(\frac{d y}{d x})}_{(2, 1)} = \frac{0}{- 1} =0 (x_{1}, y_{1}) = (2, 1) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = \frac{- 1}{0} (x - 2) \Rightarrow x - 2 = 0 \Rightarrow x = 2 Case-2: y = 2 Slope of tangent = {(\frac{d y}{d x})}_{(2, 2)} = \frac{0}{1} =0 (x_{1}, y_{1}) = (2, 2) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 2 = \frac{- 1}{0} (x - 2) \Rightarrow x - 2 = 0 \Rightarrow x = 2$

In both cases, the equation of normal is x = 2

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Question 7:

Find the equation of the normal to the curve ay² = x³ at the point (am², am³).

Answer:

$a y^{2} = x^{3} Differentiating both sides w.r.t. x, 2 a y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{2 a y} Slope of tangent = {(\frac{d y}{d x})}_{(a m^{2}, a m^{3})} = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} = \frac{3 m}{2} Given (x_{1}, y_{1}) = (a m^{2}, a m^{3}) Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - a m^{3} = \frac{- 2}{3 m} (x - a m^{2}) \Rightarrow 3 m y - 3 a m^{4} = - 2 x + 2 a m^{2} \Rightarrow 2 x + 3 m y - a m^{2} (2 + 3 m^{2}) = 0$

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Question 8:

The equation of the tangent at (2, 3) on the curve y² = ax³ + b is y = 4x − 5. Find the values of a and b.

Answer:

The slope of the given line y = 4x − 5 is 4
$y^{2} = a x^{3} + b . . . (1) 2 y \frac{d y}{d x} = 3 a x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 a x^{2}}{2 y} Slope of tangent= {(\frac{d y}{d x})}_{(2, 3)} = \frac{12 a}{6} =2a Given that Slope of tangent= slope of given line 2 a = 4 \Rightarrow a = 2 Substituting this and x = 2, y = 3 in (1), we get 9 = 16 + b \Rightarrow b = - 7 Hence, a = 2 and b = - 7$

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Question 9:

Find the equation of the tangent line to the curve y = x² + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

Answer:

Let (x₀, y₀) be the point of intersection of both the curve and the tangent.
$y = x^{2} + 4 x - 16 Since, (x_{0}, y_{0}) lies on curve . Therefore y_{0} = {x_{0}}^{2} + 4 x_{0} - 16 . . . (1) Now, y = x^{2} + 4 x - 16 \Rightarrow \frac{d y}{d x} = 2 x + 4 Slope of tangent = {(\frac{d y}{d x})}_{(x_{0}, y_{0})} =2 x_{0} +4 Given that The tangent is parallel to the line So, Slope of tangent=slope of the given line 2 x_{0} + 4 = 3 \Rightarrow 2 x_{0} = - 1 \Rightarrow x_{0} = \frac{- 1}{2} From (1), y_{0} = \frac{1}{4} - 2 - 16 = \frac{- 71}{4} Now, slope of tangent, m =3 (x_{0}, y_{0}) = (\frac{- 1}{2}, \frac{- 71}{4}) Equation of tangent is y - y_{0} = m (x - x_{0}) \Rightarrow y + \frac{71}{4} = 3 (x + \frac{1}{2}) \Rightarrow \frac{4 y + 71}{4} = 3 (\frac{2 x + 1}{2}) \Rightarrow 4 y + 71 = 12 x + 6 \Rightarrow 12 x - 4 y - 65 = 0$

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Question 10:

Find an equation of normal line to the curve y = x³ + 2x + 6 which is parallel to the line x + 14y + 4 = 0.

Answer:

Let (x₁, y₁) be a point on the curve where we need to find the normal.
Slope of the given line = $\frac{- 1}{14}$
$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 Now, y = x^{3} + 2 x + 6 \Rightarrow \frac{d y}{d x} = 3 x^{2} + 2 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} =3 {x_{1}}^{2} +2 Slope of the normal= \frac{- 1}{slope of the tangent} = = \frac{- 1}{3 {x_{1}}^{2} + 2} Given that, slope of the normal=slope of the given line \Rightarrow \frac{- 1}{3 {x_{1}}^{2} + 2} = \frac{- 1}{14} \Rightarrow 3 {x_{1}}^{2} + 2 = 14 \Rightarrow 3 {x_{1}}^{2} = 12 \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Case-1: x_{1} = 2 y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 = 8 + 4 + 6 = 18 ∴ (x_{1}, y_{1}) = (2, 18) Slope of the normal, m = \frac{- 1}{14} Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 18 = \frac{- 1}{14} (x - 2) \Rightarrow 14 y - 252 = - x + 2 \Rightarrow x + 14 y - 254 = 0 Case-2: x_{1} = - 2 y_{1} = {x_{1}}^{3} + 2 x_{1} + 6 = - 8 - 4 + 6 = - 6 ∴ (x_{1}, y_{1}) = (- 2, - 6) Slope of the normal, m = \frac{- 1}{14} Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 6 = \frac{- 1}{14} (x + 2) \Rightarrow 14 y + 84 = - x - 2 \Rightarrow x + 14 y + 86 = 0$

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Question 11:

Determine the equation(s) of tangent (s) line to the curve y = 4x³ − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Answer:

Let (x₁, y₁) be a point on the curve where we need to find the tangent(s).
Slope of the given line = $\frac{- 1}{9}$

Since, tangent is perpendicular to the given line,
Slope of the tangent = $\frac{- 1}{(\frac{- 1}{9})} = 9$
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to this curve. Since, the point lies on the curve . Hence, y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 Now, y = 4 x^{3} - 3 x + 5 \Rightarrow \frac{d y}{d x} = 12 x^{2} - 3 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 12 {x_{1}}^{2} - 3 Given that, slope of the tangent=slope of the perpendicular line \Rightarrow 12 {x_{1}}^{2} - 3 = 9 \Rightarrow 12 {x_{1}}^{2} = 12 \Rightarrow {x_{1}}^{2} = 1 \Rightarrow x_{1} = \pm 1 Case-1: x_{1} = 1 y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 = 4 - 3 + 5 = 6 ∴ (x_{1}, y_{1}) = (1, 6) Slope of the tangent =9 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 6 = 9 (x - 1) \Rightarrow y - 6 = 9 x - 9 \Rightarrow 9 x - y - 3 = 0 Case-2: x_{1} = - 1 y_{1} = 4 {x_{1}}^{3} - 3 x_{1} + 5 = - 4 + 3 + 5 = 4 ∴ (x_{1}, y_{1}) = (- 1, 4) Slope of the tangent =9 Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 4 = 9 (x + 1) \Rightarrow y - 4 = 9 x + 9 \Rightarrow 9 x - y + 13 = 0$

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Question 12:

Find the equation of a normal to the curve y = x log_ex which is parallel to the line 2x − 2y + 3 = 0.

Answer:

Slope of the given line is 1
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve. Since, the point lies on the curve . Hence, y_{1} = x_{1} \log_{e} x_{1} . . . (1) Now, y = x \log_{e} x \Rightarrow \frac{d y}{d x} = x \times \frac{1}{x} + \log_{e} x (1) = 1 + \log_{e} x Slope of tangent= 1 + \log_{e} x_{1} Slope of normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{1 + \log_{e} x_{1}} Given that Slope of normal = slope of the given line \frac{- 1}{1 + \log_{e} x_{1}} = 1 \Rightarrow - 1 = 1 + \log_{e} x_{1} \Rightarrow - 2 = \log_{e} x_{1} \Rightarrow x_{1} = e^{- 2} = \frac{1}{e^{2}} Now, y_{1} = e^{- 2} (- 2) = \frac{- 2}{e^{2}} [From (1)] ∴ (x_{1}, y_{1}) = (\frac{1}{e^{2}}, \frac{- 2}{e^{2}}) Equation of normal is, y + \frac{2}{e^{2}} = 1 (x - \frac{1}{e^{2}}) \Rightarrow y + \frac{2}{e^{2}} = x - \frac{1}{e^{2}} \Rightarrow x - y = \frac{3}{e^{2}} \Rightarrow x - y = 3 e^{- 2}$

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Question 13:

Find the equation of the tangent line to the curve y = x² − 2x + 7 which is (i) parallel to the line 2x − y + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.

Answer:

(i) Slope of the given line is 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve y = x^{2} - 2 x + 7 Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 7 . . . (1) Now, y = x^{2} - 2 x + 7 \Rightarrow \frac{d y}{d x} = 2 x - 2 Slope of tangent at point (x_{1}, y_{1}) = 2 x_{1} - 2 Given that Slope of tangent= Slope of the given line \Rightarrow 2 x_{1} - 2 = 2 \Rightarrow 2 x_{1} = 4 \Rightarrow x_{1} = 2 Now, y_{1} = 4 - 4 + 7 = 7 ∴ (x_{1}, y_{1}) = (2, 7) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 7 = 2 (x - 2) \Rightarrow y - 7 = 2 x - 4 \Rightarrow 2 x - y + 3 = 0$

(ii) Slope of the given line is 3
Slope of the line perpendicular to this line = $\frac{- 1}{3}$

$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve. Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 7 . . . (1) Now, y = x^{2} - 2 x + 7 \Rightarrow \frac{d y}{d x} = 2 x - 2 Slope of tangent at (x_{1}, y_{1}) = 2 x_{1} - 2 Given that Slope of tangent at (x_{1}, y_{1}) = Slope of the perpendicular line \Rightarrow 2 x_{1} - 2 = \frac{- 1}{3} \Rightarrow 6 x_{1} - 6 = - 1 \Rightarrow 6 x_{1} = 5 \Rightarrow x_{1} = \frac{5}{6} Now, y_{1} = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36} ∴ (x_{1}, y_{1}) = (\frac{5}{6}, \frac{217}{36}) Equation of tangent is, y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6}) \Rightarrow \frac{36 y - 217}{36} = \frac{- 6 x + 5}{18} \Rightarrow 36 y - 217 = - 12 x + 10 \Rightarrow 12 x + 36 y - 227 = 0$

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Question 14:

Find the equations of all lines having slope 2 and that are tangent to the curve $y = \frac{1}{x - 3}, x \neq 3$ .

Answer:

Slope of given tangent = 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to this curve. Since, the point lies on the curve . Hence, y_{1} = \frac{1}{x_{1} - 3} Now, y = \frac{1}{x - 3} ⇒ \frac{d y}{d x} = \frac{- 1}{{(x - 3)}^{2}} Slope of tangent = (\frac{d y}{d x}) = \frac{- 1}{{(x_{1} - 3)}^{2}} Given that Slope of the tangent = 2 \Rightarrow \frac{- 1}{{(x_{1} - 3)}^{2}} = 2 \Rightarrow {(x_{1} - 3)}^{2} = - 2 \Rightarrow x_{1} - 3 = \sqrt{- 2}, which does not exist because 2 is negative.$
So, there does not exist any such tangent.

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Question 15:

Find the equations of all lines of slope zero and that are tangent to the curve $y = \frac{1}{x^{2} - 2 x + 3}$ .

Answer:

Slope of the given tangent is 0.

$Let (x_{1}, y_{1}) be a point where the tangent is drawn to the curve (1). Since, the point lies on the curve . Hence, y_{1} = \frac{1}{{x_{1}}^{2} - 2 x_{1} + 3} . . . (1) Now, y = \frac{1}{x^{2} - 2 x + 3} \Rightarrow \frac{d y}{d x} = \frac{(x^{2} - 2 x + 3) (0) - (2 x - 2) 1}{{(x^{2} - 2 x + 3)}^{2}} = \frac{- 2 x + 2}{{(x^{2} - 2 x + 3)}^{2}} Slope of tangent= \frac{- 2 x_{1} + 2}{{({x_{1}}^{2} - 2 x_{1} + 3)}^{2}} Given that Slope of tangent = slope of the given line \Rightarrow \frac{- 2 x_{1} + 2}{{({x_{1}}^{2} - 2 x_{1} + 3)}^{2}} = 0 \Rightarrow - 2 x_{1} + 2 = 0 \Rightarrow 2 x_{1} = 2 \Rightarrow x_{1} = 1 Now, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} [From (1)] ∴ (x_{1}, y_{1}) = (1, \frac{1}{2}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{1}{2} = 0 (x - 1) \Rightarrow y = \frac{1}{2}$

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Question 16:

Find the equation of the tangent to the curve $y = \sqrt{3 x - 2}$ which is parallel to the 4x − 2y + 5 = 0.

Answer:

Slope of the given line is 2
$Let (x_{1}, y_{1}) be the point where the tangent is drawn to the curve y = \sqrt{3 x - 2} Since, the point lies on the curve . Hence, y_{1} = \sqrt{3 x_{1} - 2} . . . (1) Now, y = \sqrt{3 x - 2} \Rightarrow \frac{d y}{d x} = \frac{3}{2 \sqrt{3 x - 2}} Slope of tangent at (x_{1}, y_{1}) = \frac{3}{2 \sqrt{3 x_{1} - 2}} Given that Slope of tangent = slope of the given line \Rightarrow \frac{3}{2 \sqrt{3 x_{1} - 2}} = 2 \Rightarrow 3 = 4 \sqrt{3 x_{1} - 2} \Rightarrow 9 = 16 (3 x_{1} - 2) \Rightarrow \frac{9}{16} = 3 x_{1} - 2 \Rightarrow 3 x_{1} = \frac{9}{16} + 2 = \frac{9 + 32}{16} = \frac{41}{16} \Rightarrow x_{1} = \frac{41}{48} Now, y_{1} = \sqrt{\frac{123}{48} - 2} = \sqrt{\frac{27}{48}} = \sqrt{\frac{9}{16}} = \frac{3}{4} [From (1)] ∴ (x_{1}, y_{1}) = (\frac{41}{48}, \frac{3}{4}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{3}{4} = 2 (x - \frac{41}{48}) \Rightarrow \frac{4 y - 3}{4} = 2 (\frac{48 x - 41}{48}) \Rightarrow 24 y - 18 = 48 x - 41 \Rightarrow 48 x - 24 y - 23 = 0$

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Question 17:

Find the equation of the tangent to the curve x² + 3y − 3 = 0, which is parallel to the line y = 4x − 5.

Answer:

Suppose (x₁, y₁) be the point of contact of tangent.
We can find the slope of the given line by differentiating the equation w.r.t x
So, Slope of the line = 4

$Since, (x_{1}, y_{1}) lies on the curve . Therefore, {x_{1}}^{2} + 3 y_{1} - 3 = 0 . . . (1) Now, x^{2} + 3 y - 3 = 0 \Rightarrow 2 x + 3 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{3} Slope of tangent, m = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2 x_{1}}{3} Given that tangent is parallel to the line, So Slope of tangent, m = slope of the given line \frac{- 2 x_{1}}{3} = 4 \Rightarrow x_{1} = - 6 36 + 3 y_{1} - 3 = 0 (From (1)) \Rightarrow 3 y_{1} = - 33 \Rightarrow y_{1} = - 11 (x_{1}, y_{1}) = (- 6, - 11) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y + 11 = 4 (x + 6) \Rightarrow y + 11 = 4 x + 24 \Rightarrow 4 x - y + 13 = 0$

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Question 18:

Prove that ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$ touches the straight line $\frac{x}{a} + \frac{y}{b} = 2$ for all n ∈ N, at the point (a, b).

Answer:

$Now, {(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2 \frac{n}{a} {(\frac{x}{a})}^{n - 1} + \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = 0 \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = \frac{- n}{a} {(\frac{x}{a})}^{n - 1} \frac{d y}{d x} = \frac{- n}{a} {(\frac{x}{a})}^{n - 1} \times \frac{b}{n} {(\frac{b}{y})}^{n - 1} = \frac{- b}{a} {(\frac{b x}{a y})}^{n - 1} Slope of tangent= {(\frac{d y}{d x})}_{(a, b)} = \frac{- b}{a} {(\frac{b * a}{a * b})}^{n - 1} = \frac{- b}{a} ... (2) The equation of tangent is y - b = \frac{- b}{a} (x - a) \Rightarrow y a - a b = - x b + a b \Rightarrow x b + y a = 2 a b \Rightarrow \frac{x}{a} + \frac{y}{b} = 2$

So, the given line touches the given curve at the given point.

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Question 19:

Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at $t = \frac{π}{4}$ .

Answer:

$x = \sin 3 t and y = \cos 2 t \frac{d x}{d t} = 3 \cos 3 t and \frac{d y}{d t} = - 2 \sin 2 t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 \sin 2 t}{3 \cos 3 t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{π}{4}} =- \frac{- 2 \sin (\frac{π}{2})}{3 \cos (\frac{3 π}{4})} = \frac{- 2}{\frac{- 3}{\sqrt{2}}} = \frac{2 \sqrt{2}}{3} x_{1} = \sin (3 \times \frac{π}{4}) = \frac{1}{\sqrt{2}} and y_{1} = \cos (2 \times \frac{π}{4}) = 0 So, (x_{1}, y_{1}) = (\frac{1}{\sqrt{2}}, 0) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = \frac{2 \sqrt{2}}{3} (x - \frac{1}{\sqrt{2}}) \Rightarrow 3 y = 2 \sqrt{2} x - 2 \Rightarrow 2 \sqrt{2} x - 3 y - 2 = 0$

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Question 20:

At what points will be tangents to the curve y = 2x³ − 15x² + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.

Answer:

Slope of x - axis is 0
Let (x₁, y₁) be the required point.
$y = 2 x^{3} - 15 x^{2} + 36 x - 21 Since (x_{1}, y_{1}) lies on the curve . Therefore y_{1} = 2 {x_{1}}^{3} - 15 {x_{1}}^{2} + 36 x_{1} - 21 . . . (1) Now, y = 2 x^{3} - 15 x^{2} + 36 x - 21 \Rightarrow \frac{d y}{d x} = 6 x^{2} - 30 x + 36 Slope of tangent at (x_{1}, y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 6 {x_{1}}^{2} - 30 x_{1} + 36 Given that Slope of tangent at (x, y) = slope of the x -axis 6 {x_{1}}^{2} - 30 x_{1} + 36 = 0 \Rightarrow {x_{1}}^{2} - 5 x_{1} + 6 = 0 \Rightarrow (x_{1} - 2) (x_{1} - 3) = 0 \Rightarrow x_{1} = 2 or x_{1} = 3 Case-1: x_{1} = 2 y_{1} = 16 - 60 + 72 - 21 = 7 (From (1)) (x_{1}, y_{1}) = (2, 7) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 7 = 0 (x - 2) \Rightarrow y = 7 Case-2: x_{1} = 3 y_{1} = 54 - 135 + 108 - 21 = 6 (From (1)) (x_{1}, y_{1}) = (3, 6) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 6 = 0 (x - 3) \Rightarrow y = 6$

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Question 21:

Find the equation of the tangents to the curve 3x² – y² = 8, which passes through the point (4/3, 0).

Answer:

We have,

3x² – y² = 8 ...(i)

Differentiating both sides w.r.t x, we get

$6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow 2 y \frac{d y}{d x} = 6 x \Rightarrow \frac{d y}{d x} = \frac{6 x}{2 y} \Rightarrow \frac{d y}{d x} = \frac{3 x}{y}$

Let tangent at (h, k) pass through $(\frac{4}{3}, 0)$ .

Since, (h, k) lies on (i), we get

$3 h^{2} - k^{2} = 8 . . . (ii)$

Slope of tangent at (h, k) = $\frac{3 h}{k}$

The equation of tangent at (h, k) is given by,

$(y - k) = \frac{3 h}{k} (x - h) . . . (iii)$

Since, the tangent passess through $(\frac{4}{3}, 0)$ .

$∴ (0 - k) = \frac{3 h}{k} (\frac{4}{3} - h) \Rightarrow - k = \frac{4 h}{k} - \frac{3 h^{2}}{k} \Rightarrow - k^{2} = 4 h - 3 h^{2}$

$\Rightarrow 8 - 3 h^{2} = 4 h - 3 h^{2} [From (ii)] \Rightarrow 8 = 4 h \Rightarrow h = 2$

Using (ii), we get

$12 - k^{2} = 8 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$

So, the points on curve (i) at which tangents pass through $(\frac{4}{3}, 0)$ are $(2, \pm 2)$ .

Now, from (iii), the equation of tangents are
$(y - 2) = \frac{6}{2} (x - 2), or, 3 x - y - 4 = 0, and (y + 2) = \frac{6}{- 2} (x - 2), or, 3 x + y - 4 = 0$

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Question 1:

Find the angle of intersection of the following curves:

(i) y² = x and x² = y [NCERT EXEMPLAR]
(ii) y = x² and x² + y² = 20
(iii) 2y²= x³ and y² = 32x
(iv) x²+ y² − 4x − 1 = 0 and x² + y² − 2y − 9 = 0
(v) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and x² + y² = ab
(vi) x² + 4y² = 8 and x² − 2y² = 2
(vii) x² = 27y and y² = 8x
(viii) x² + y² = 2x and y² = x
(ix) y = 4 − x² and y = x² [NCERT EXEMPLAR]

Answer:

â€‹ $(i) Given curves are, y^{2} = x . . . (1) x^{2} = y . . . (2) From these two equations, we get {(x^{2})}^{2} = x \Rightarrow x^{4} - x = 0 \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y = 1 ∴ (x, y) = (0, 0) or (1, 1) Differenntiating (1) w.r.t. x, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (3) Differenntiating (2) w.r.t . x, 2 x = \frac{d y}{d x} . . . (4) Case -1: (x, y) = (0, 0) The tangent to curve is parallel to x - axis . Hence, the angle between the tangents to two curve at (0, 0) is a right angle . ∴ θ = \frac{π}{2} Case -2 : (x, y) = (1, 1) From (3) we have, m_{1} = \frac{1}{2} From (4) we have, m_{2} = 2 (1) = 2 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2}| = \frac{3}{4} \Rightarrow θ = \tan^{- 1} (\frac{3}{4})$

$(i i) Given curves are, y = x^{2} . . . (1) x^{2} + y^{2} = 20 . . . (2) From these two equations we get y + y^{2} = 20 \Rightarrow y^{2} + y - 20 = 0 \Rightarrow (y + 5) (y - 4) = 0 \Rightarrow y = - 5 or y = 4 Substituting the values of y in (1) we get, x^{2} = - 5 or x^{2} = 4 \Rightarrow x = \pm 2 and x^{2} = -5  has no real solution So, (x, y) = (2, 4) or (- 2, 4) Differenntiating (1) w.r.t. x, \frac{d y}{d x} = 2 x . . . (3) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} . . . (4) Case -1: (x, y) = (2, 4) From (3) we have, m_{1} = 2 (2) = 4 From (4) we have, m_{2} = \frac{- 2}{4} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{4 + \frac{1}{2}}{1 + 4 (\frac{- 1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2}) Case -2: (x, y) = (- 2, 4) From (3) we have, m_{1} = 2 (- 2) = - 4 From (4) we have, m_{2} = \frac{2}{4} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 4 - \frac{1}{2}}{1 - 4 (\frac{1}{2})}| = \frac{9}{2} \Rightarrow θ = \tan^{- 1} (\frac{9}{2})$

$(iii) Given curves are, 2 y^{2} = x^{3} . . . (1) y^{2} = 32 x . . . (2) From these two equations we get 2 (32 x) = x^{3} \Rightarrow 64 x = x^{3} \Rightarrow x (x^{2} - 64) = 0 \Rightarrow x = 0, 8, - 8 Substituting the value of x in (2) we get, y_{1} = 0, 16, - 16 ∴ (x_{1}, y_{1}) = (0, 0), (8, 16) or (8, - 16) Differentiating (1) w.r.t. x, 4 y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 y \frac{d y}{d x} = 32 \Rightarrow \frac{d y}{d x} = \frac{16}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (3) we have, m_{1} = \frac{0}{0} ∴ We cannot determine θ in this case. Case - 2 : (x, y) = (8, 16) From (3) we have, m_{1} = \frac{192}{64} = 3 From (4) we have, m_{2} = \frac{16}{16} = 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{3 - 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case- 3 : (x_{1}, y_{1}) = (8, - 16) From (3) we have, m_{1} = \frac{192}{- 64} = - 3 From (4) we have, m_{2} = \frac{16}{- 16} = - 1 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 3 + 1}{1 + 3}| = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(i v) Given curves are, x^{2} + y^{2} - 4 x - 1 = 0 . . . (1) x^{2} + y^{2} - 2 y - 9 = 0 . . . (2) From (3)  we get x^{2} + y^{2} = 4 x + 1 Substituting this in (2), 4 x + 1 - 2 y - 9 = 0 \Rightarrow 4 x - 2 y = 8 \Rightarrow 2 x - y = 4 \Rightarrow y = 2 x - 4 . . . (3) Substituting this in (1), x^{2} + {(2 x - 4)}^{2} - 4 x - 1 = 0 \Rightarrow x^{2} + 4 x^{2} + 16 - 16 x - 4 x - 1 = 0 \Rightarrow 5 x^{2} - 20 x + 15 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x = 3 or x = 1 Substituting the values of x in (3), we get, y = 2 or y = - 2 ∴ (x, y) = (3, 2), (1, - 2) Differentiating (1) w.r.t. x, 2 x + 2 y \frac{d y}{d x} - 4 = 0 \Rightarrow \frac{d y}{d x} = \frac{4 - 2 x}{2 y} = \frac{2 - x}{y} . . . (4) Differenntiating (2) w.r.t . x, 2 x + 2 y \frac{d y}{d x} - 2 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 2) = - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 x}{2 - 2 y} = \frac{x}{1 - y} . . . (5) Case - 1: (x, y) = (3, 2) From (4), we get, m_{1} = \frac{2 - 3}{2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{3}{1 - 2} = - 3 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} + 3}{1 + \frac{3}{2}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4} Case - 2: (x, y) = (1, - 2) From (4), we get, m_{1} = \frac{2 - 1}{- 2} = \frac{- 1}{2} From (5), we get, m_{2} = \frac{1}{1 + 2} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - \frac{1}{3}}{1 - \frac{1}{6}}| = 1 \Rightarrow θ = \tan^{- 1} (1) = \frac{π}{4}$

$(v) Given curves are, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . (1) x^{2} + y^{2} = a b . . . (2) Multiplying (2) by \frac{1}{a^{2}}, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = \frac{b}{a} . . . (3) Subtracting (1) from (3), we get \frac{y^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{b}{a} - 1 \Rightarrow y^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b - a}{a} \Rightarrow y^{2} = \frac{b - a}{a} \times \frac{a^{2} b^{2}}{(b + a) (b - a)} = \frac{a b^{2}}{b + a} \Rightarrow y = \pm b \sqrt{\frac{a}{b + a}} Substituting this in (3), \frac{x^{2}}{a^{2}} + \frac{a b^{2}}{(b + a) (a^{2})} = \frac{b}{a} \Rightarrow (a + b) x^{2} + a b^{2} = a b^{2} + a^{2} b \Rightarrow x^{2} = \frac{a^{2} b}{a + b} \Rightarrow x = \pm a \sqrt{\frac{b}{a + b}} ∴ (x, y) = (\pm a \sqrt{\frac{b}{a + b}}, \pm b \sqrt{\frac{a}{b + a}}) Now, (x, y) = (a \sqrt{\frac{b}{a + b}}, b \sqrt{\frac{a}{b + a}}) Differentiating (1) w.r.t. x, we get, \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{a^{2} y} \Rightarrow m_{1} = \frac{- a b^{2} \sqrt{\frac{b}{a + b}}}{a^{2} b \sqrt{\frac{a}{b + a}}} = \frac{- b \sqrt{b}}{a \sqrt{a}} Differenntiating (2) w.r.t . x, we get, 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} \Rightarrow m_{2} = \frac{- a \sqrt{\frac{b}{a + b}}}{b \sqrt{\frac{a}{b + a}}} = \frac{- a \sqrt{b}}{b \sqrt{a}} We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- b \sqrt{b}}{a \sqrt{a}} + \frac{a \sqrt{b}}{b \sqrt{a}}}{1 + (\frac{b \sqrt{b}}{a \sqrt{a}}) (\frac{a \sqrt{b}}{b \sqrt{a}})}| = \frac{\frac{- b^{2} \sqrt{a b} + a^{2} \sqrt{a b}}{a^{2} b}}{\frac{a^{2} b + a b^{2}}{a^{2} b}} = \frac{\sqrt{a b} (a + b) (a - b)}{a^{2} b} \times \frac{a^{2} b}{a b (a + b)} = \frac{a - b}{\sqrt{a b}} \Rightarrow θ = \tan^{- 1} (\frac{a - b}{\sqrt{a b}}) {Similarly,  we can prove that θ=tan}^{- 1} (\frac{a - b}{\sqrt{a b}}) for all possibilities of (x, y)$

$(v i) Given curves are, x^{2} + 4 y^{2} = 8 . . . (1) x^{2} - 2 y^{2} = 2 . . . (2) From (1) and (2) we get 6 y^{2} = 6 \Rightarrow y = 1 or y_{1} = - 1 Substituting the values of y in (1) x = 2, - 2 or x = 2, - 2 So, (x, y) = (2, 1), (2, - 1), (- 2, 1), (- 2, - 1) Differenntiating (1) w.r.t. x, 2 x + 8 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{4 y} . . . (3) Differenntiating (2) w.r.t. x, 2 x - 4 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{2 y} . . . (4) Case -1: (x, y) = (2, 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -2: (x, y) = (2, - 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -3 : (x, y) = (- 2, 1) From (3), we get, m_{1} = \frac{1}{2} From (4), we get, m_{2} = - 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} + 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3) Case -4 : (x, y) = (- 2, - 1) From (3), we get, m_{1} = \frac{- 1}{2} From (4), we get, m_{2} = 1 We have, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}}| = 3 \Rightarrow θ = \tan^{- 1} (3)$

$(vii) Given curves are, x^{2} = 27 y . . . (1) y^{2} = 8 x . . . (2) From (2)  we get x = \frac{y^{2}}{8} Substituting this in (1), {(\frac{y^{2}}{8})}^{2} = 27 y \Rightarrow y^{4} = 1728 y \Rightarrow y (y^{3} - 12^{3}) = 0 \Rightarrow y = 0 or y = 12 Substituting the v a l u e s o f y in (2), w e g e t, \Rightarrow x = 0 or x = 18 \Rightarrow (x, y) = (0, 0), (18, 12) Differentiating (1) w.r.t. x, 2 x = 27 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2 x}{27} . . . (3) Differenntiating (2) w.r.t . x, 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y} . . . (4) Case - 1: (x, y) = (0, 0) From (4) we have, m_{2} is undefined ∴ We cannot find θ Case - 2: (x, y) = (18, 12) From (3) we have, m_{1} = \frac{36}{27} = \frac{4}{3} From (4) we have, m_{2} = \frac{4}{12} = \frac{1}{3} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}}| = \frac{9}{13} \Rightarrow θ = \tan^{- 1} (\frac{9}{13})$

$(v i i i) Given curves are, x^{2} + y^{2} = 2 x . . . (1) y^{2} = x . . . (2) F rom these two equations we get x^{2} + x = 2 x \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x = 1 Substituting the values of x in (2) we get, y = 0 or y =±1 ∴ (x, y) = (0, 0), (1, 1), (1, - 1) Differenntiating (1) w.r.t. x, we get, 2 x + 2 y \frac{d y}{d x} = 2 \Rightarrow \frac{d y}{d x} = \frac{1 - x}{y} . . . (3) Differenntiating (2) w.r.t. x,we get, 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} . . . (4) Case -1: (x, y) = (0, 0) From (3) we get, m_{1} is undefined . ∴ We can not find θ Case -2: Let (x, y) = (1, 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 - \frac{1}{2}}{1 + 0}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2}) Case -3: Let (x, y) = (1, - 1) From (3) we get, m_{1} = 0 From (4) we get, m_{2} = \frac{- 1}{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{0 + \frac{1}{2}}{1}| = \frac{1}{2} \Rightarrow θ = \tan^{- 1} (\frac{1}{2})$

$(ix) Given curves are, y = 4 - x^{2} . . . . . (1) y = x^{2} . . . . . (2) From (1) and (2), we get 4 - x^{2} = x^{2} \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2} Substituting the v a l u e s o f x in (2), w e g e t, \Rightarrow y = 2 \Rightarrow (x, y) = (\sqrt{2},2), (- \sqrt{2}, 2) Differentiating (1) w.r.t. x, \frac{d y}{d x} = - 2 x . . . . . (3) Differentiating (2) w.r.t . x, \frac{d y}{d x} = 2 x . . . . . (4) Case 1: (x, y) = (\sqrt{2}, 2) From (3), we have, m_{1} = - 2 \sqrt{2} From (4) we have, m_{2} = 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- 2 \sqrt{2} - 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7}) Case 1: (x, y) = (- \sqrt{2}, 2) From (3), we have, m_{1} = 2 \sqrt{2} From (4) we have, m_{2} = - 2 \sqrt{2} Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{2 \sqrt{2} + 2 \sqrt{2}}{1 - 8}| = \frac{4 \sqrt{2}}{7} \Rightarrow θ = \tan^{- 1} (\frac{4 \sqrt{2}}{7})$

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Question 2:

Show that the following set of curves intersect orthogonally.

(i) y = x³ and 6y = 7 − x²
(ii) x³ − 3xy² = −2 and 3x²y − y³ = 2
(iii) x² + 4y² = 8 and x² − 2y² = 4

Answer:

$(i) y = x^{3} . . . (1) 6 y = 7 - x^{2} . . . (2) From (1) and (2) we get 6 x^{3} = 7 - x^{2} \Rightarrow 6 x^{3} + x^{2} - 7 = 0 x =1 satisfies this. Dividing this by x -1 ,we get 6 x^{2} + 7 x + 7 = 0, {Discriminant = 7}^{2} -4 (6) (7) =-119<0 So, this has no real roots. When x =1, y = x^{3} =1 (From (1)) So, (x, y) = (1, 1) Differentiating (1) w.r.t. x, \frac{d y}{d x} = 3 x^{2} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = 3 Differenntiating (2) w.r.t . x, 6 \frac{d x}{d x} = - 2 x \Rightarrow \frac{d x}{d x} = \frac{- 2 x}{6} = \frac{- x}{3} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{3} Now, m_{1} \times m_{2} = 3 \times \frac{- 1}{3} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1$
So, the given curves intersect orthogonally.

$(i i) Let the given curves intersect at (x_{1}, y_{1}) x^{3} - 3 x y^{2} = - 2 . . . (1) 3 x^{2} y - y^{3} = 2 . . . (2) Differentiating (1) w.r.t. x, 3 x^{2} - 3 y^{2} - 6 x y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 3 y^{2}}{6 x y} = \frac{x^{2} - y^{2}}{2 x y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{{x_{1}}^{2} - {y_{1}}^{2}}{2 x_{1} y_{1}} Differenntiating (2) w.r.t . x, 3 x^{2} \frac{d y}{d x} + 6 x y - 3 y^{2} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (3 x^{2} - 3 y^{2}) = - 6 x y \Rightarrow \frac{d y}{d x} = \frac{- 6 x y}{3 x^{2} - 3 y^{2}} = \frac{- 2 x y}{x^{2} - y^{2}} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2 x_{1} y_{1}}{{x_{1}}^{2} - {y_{1}}^{2}} Now, m_{1} \times m_{2} = \frac{{x_{1}}^{2} - {y_{1}}^{2}}{2 x_{1} y_{1}} \times \frac{- 2 x_{1} y_{1}}{{x_{1}}^{2} - {y_{1}}^{2}} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1$
So, the given curves intersect orthogonally.

$(iii) x^{2} + 4 y^{2} = 8 . . . (1) x^{2} - 2 y^{2} = 4 . . . (2) From (1) and (2) we get 6 y^{2} = 4 \Rightarrow y^{2} = \frac{2}{3} \Rightarrow y = \frac{\sqrt{2}}{\sqrt{3}} or y = \frac{- \sqrt{2}}{\sqrt{3}} From (1), x^{2} + \frac{8}{3} = 8 \Rightarrow x^{2} = \frac{16}{3} \Rightarrow x = \pm \frac{4}{\sqrt{3}} So, (x, y) = (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}), (\frac{4}{\sqrt{3}}, \frac{- \sqrt{2}}{\sqrt{3}}), (\frac{- 4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}), (\frac{- 4}{\sqrt{3}}, - \frac{\sqrt{2}}{\sqrt{3}}) Consider point (x_{1}, y_{1}) = (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}) Differentiating (1) w.r.t. x, 2 x + 8 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{4 y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}})} = \frac{- \frac{4}{\sqrt{3}}}{4 \frac{\sqrt{2}}{\sqrt{3}}} = \frac{- 1}{\sqrt{2}} Differenntiating (2) w.r.t . x, 2 x - 4 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{2 y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}})} = \frac{\frac{4}{\sqrt{3}}}{2 \frac{\sqrt{2}}{\sqrt{3}}} = \sqrt{2} Now, m_{1} \times m_{2} = \frac{- 1}{\sqrt{2}} \times \sqrt{2} \Rightarrow m_{1} \times m_{2} = - 1 Since, m_{1} \times m_{2} = - 1 Hence,, the curves are orthogonal at (\frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}) . Similarly, we can see that the curves are orthogonal in each possibility of (x_{1}, y_{1}) .$

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Question 3:

Show that the following curves intersect orthogonally at the indicated points:

(i) x² = 4y and 4y + x² = 8 at (2, 1)
(ii) x² = y and x³ + 6y = 7 at (1, 1)
(iii) y² = 8x and 2x² + y² = 10 at $(1, 2 \sqrt{2})$

Answer:

$(i) x^{2} = 4 y . . . (1) 4 y + x^{2} = 8 . . . (2) Given point is (2, 1) Differentiating (1) w.r.t. x, 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(2, 1)} = \frac{2}{2} = 1 Differenntiating (2) w.r.t . x, 4 \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{2} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(2, 1)} = \frac{- 2}{2} = - 1 Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

$(i i) x^{2} = y . . . (1) x^{3} + 6 y = 7 . . . (2) Given point is (1, 1) Differentiating (1) w.r.t. x, 2 x = \frac{d y}{d x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = 2 (1) = 2 Differenntiating (2) w.r.t . x, 3 x^{2} + 6 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x^{2}}{2} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{2} Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

$(i i i) y^{2} = 8 x . . . (1) 2 x^{2} + y^{2} = 10 . . . (2) Given point is (1, 2 \sqrt{2}) Differentiating (1) w.r.t. x, 2 y \frac{d y}{d x} = 8 \Rightarrow \frac{d y}{d x} = \frac{4}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 2 \sqrt{2})} = \frac{4}{2 \sqrt{2}} = \sqrt{2} Differenntiating (2) w.r.t . x, 4 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 2 \sqrt{2})} = \frac{- 2}{2 \sqrt{2}} = \frac{- 1}{\sqrt{2}} Since, m_{1} \times m_{2} = - 1$
Hence, the given curves intersect orthogonally at the given point.

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Question 4:

Show that the curves 4x = y² and 4xy = k cut at right angles, if k² = 512.

Answer:

$Given : 4 x = y^{2} . . . (1) 4 x y = k . . . (2) From (1) and (2), we get y^{3} = k \Rightarrow y = k^{\frac{1}{3}} From (1), we get 4 x = k^{\frac{2}{3}} \Rightarrow x = \frac{k^{\frac{2}{3}}}{4} On differentiating (1) w.r.t. x, we get 4 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{4}, k^{\frac{1}{3}})} = \frac{2}{k^{\frac{1}{3}}} = 2 k^{\frac{- 1}{3}} On differentiating (2) w.r.t . x, we get 4 x \frac{d y}{d x} + 4 y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{4}, k^{\frac{1}{3}})} = \frac{- k^{\frac{1}{3}}}{(\frac{k^{\frac{2}{3}}}{4})} = - 4 k^{\frac{- 1}{3}} It is given that the curves intersect at right angles. ∴ m_{1} \times m_{2} = - 1 \Rightarrow 2 k^{\frac{- 1}{3}} \times - 4 k^{\frac{- 1}{3}} = - 1 \Rightarrow 8 k^{\frac{- 2}{3}} = 1 \Rightarrow k^{\frac{- 2}{3}} = \frac{1}{8} \Rightarrow k^{\frac{2}{3}} = 8 Cubing on both sides, we get k^{2} = 512$

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Question 5:

Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8.

Answer:

$Given : 2 x = y^{2} . . . (1) 2 x y = k . . . (2) From (1) and (2), we get y^{3} = k \Rightarrow y = k^{\frac{1}{3}} From (1), we get 2 x = k^{\frac{2}{3}} \Rightarrow x = \frac{k^{\frac{2}{3}}}{2} On differentiating (1) w.r.t. x, we get 2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{2}, k^{\frac{1}{3}})} = \frac{1}{k^{\frac{1}{3}}} = k^{\frac{- 1}{3}} On differentiating (2) w.r.t . x, we get 2 x \frac{d y}{d x} + 2 y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(\frac{k^{\frac{2}{3}}}{2}, k^{\frac{1}{3}})} = \frac{- k^{\frac{1}{3}}}{(\frac{k^{\frac{2}{3}}}{2})} = - 2 k^{\frac{- 1}{3}} It is given that the curves intersect orthogonally . ∴ m_{1} \times m_{2} = - 1 \Rightarrow k^{\frac{- 1}{3}} \times - 2 k^{\frac{- 1}{3}} = - 1 \Rightarrow 2 k^{\frac{- 2}{3}} = 1 \Rightarrow k^{\frac{- 2}{3}} = \frac{1}{2} \Rightarrow k^{\frac{2}{3}} = 2 Cubing on both sides, we get k^{2} = 8$

Page No 15.40:

Question 6:

Prove that the curves xy = 4 and x² + y² = 8 touch each other. [NCERT EXEMPLAR]

Answer:

$Given : x y = 4 . . . . . (1) x^{2} + y^{2} = 8 . . . . . (2) From (1), we get x = \frac{4}{y} Substituting x = \frac{4}{y} in (2), we get {(\frac{4}{y})}^{2} + y^{2} = 8 \Rightarrow \frac{16}{y^{2}} + y^{2} = 8 \Rightarrow 16 + y^{4} = 8 y^{2} \Rightarrow y^{4} - 8 y^{2} + 16 = 0 \Rightarrow {(y^{2} - 4)}^{2} = 0 \Rightarrow y^{2} - 4 = 0 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2 Substituting y = \pm 2, we get x = \pm 2 So, the given curves touch each other at two points (2, 2) and (- 2, - 2) .$

Page No 15.40:

Question 7:

Prove that the curves y² = 4x and x² + y² $-$ 6x + 1 = 0 touch each other at the point (1, 2). [NCERT EXEMPLAR]

Answer:

$Given : y^{2} = 4 x . . . . . (1) and x^{2} + y^{2} - 6 x + 1 = 0 . . . . . (2) From (1) and (2), we get x^{2} + 4 x - 6 x + 1 = 0 \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 Substititing x = 1 in (1), we get y^{2} = 4 \Rightarrow y = \pm 2 So, the two given curves touch each other at two points (1, 2) and (1, - 2) .$

Page No 15.41:

Question 8:

Find the condition for the following set of curves to intersect orthogonally:

(i) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 and x y = c^{2}$
(ii) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$

Answer:

$(i) \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . (1) x y = c^{2} . . . (2) Let the curves intersect orthogonally at (x_{1}, y_{1}) . On d ifferentiating (1) on both sides w.r.t. x, we get \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{a^{2} y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{x_{1} b^{2}}{a^{2} y_{1}} On d ifferentiating (2) on both sides w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- y_{1}}{x_{1}} It is given that the curves intersect orhtogonally at (x_{1}, y_{1}) . ∴ m_{1} \times m_{2} = - 1 \Rightarrow \frac{x_{1} b^{2}}{a^{2} y_{1}} \times \frac{- y_{1}}{x_{1}} = - 1 \Rightarrow a^{2} = b^{2}$

(ii) The condition for the curves $a x^{2} + b y^{2} = 1 and a' x^{2} + b' y^{2} = 1$ to intersect orthogonally is given below:
$\frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'} So, the condition for the curves \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 to intersect orthogonally is \frac{1}{\frac{1}{a^{2}}} - \frac{1}{\frac{1}{b^{2}}} = \frac{1}{\frac{1}{A^{2}}} - \frac{1}{\frac{- 1}{B^{2}}} \Rightarrow a^{2} - b^{2} = A^{2} + B^{2}$

Page No 15.41:

Question 9:

Show that the curves $\frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1$ intersect at right angles.

Answer:

$We have, \frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 . . . (1) and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1 . . . (2) Now we can find the slope of both the curve by differentiating w . r . t x \Rightarrow \frac{2 x}{a^{2} + λ_{1}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{1}} = 0 and \frac{2 x}{a^{2} + λ_{2}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{2}} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} \Rightarrow m_{1} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and m_{2} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} Subtracting (2) from (1), we get, x^{2} (\frac{1}{a^{2} + λ_{1}} - \frac{1}{a^{2} + λ_{2}}) + y^{2} (\frac{1}{b^{2} + λ_{1}} - \frac{1}{b^{2} + λ_{2}}) = 0 \Rightarrow \frac{x^{2}}{y^{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{1}{\frac{λ_{1} - λ_{2}}{(a^{2} + λ_{1}) (a^{2} + λ_{2})}} N o w, m_{1} \times \times m_{2} = \frac{x^{2}}{y^{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{(a^{2} + λ_{1}) (a^{2} + λ_{2})}{λ_{1} - λ_{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = - 1 hence, (1) and (2) cuts orthogonally .$

Page No 15.41:

Question 10:

If the straight line xcos $α$ + ysin $α$ = p touches the curve $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , then prove that a²cos² $α$ $-$ b²sin² $α$ = p².

Answer:

$We have, x \cos α + y \sin α = p . . . . . (i) \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . . . (ii) As, the straight line (i) touches the curve (ii) . So, the straight line (i) is tangent to the curve (ii) . Also, the slope of the straight line, m = \frac{- \cos α}{\sin α} And, the slope of the tangent to the curve = \frac{d y}{d x} = \frac{b^{2}}{a^{2}} \times \frac{x}{y} So, \frac{b^{2}}{a^{2}} \times \frac{x}{y} = \frac{- \cos α}{\sin α} \Rightarrow x b^{2} \sin α = - y a^{2} \cos α \Rightarrow x = \frac{- y a^{2} \cos α}{b^{2} \sin α} . . . . . (iii) Substituting the value of x in (i), we get x \cos α + y \sin α = p \Rightarrow \frac{- y a^{2} \cos^{2} α}{b^{2} \sin α} + y \sin α = p \Rightarrow \frac{- y a^{2} \cos^{2} α + y b^{2} \sin α}{b^{2} \sin α} = p \Rightarrow y (- a^{2} \cos^{2} α + b^{2} \sin α) = p b^{2} \sin α \Rightarrow y = \frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)} So, from (iii), we get x = \frac{- y a^{2} \cos α}{b^{2} \sin α} = \frac{- y a^{2} \cos α}{b^{2} \sin α} = \frac{- a^{2} \cos α}{b^{2} \sin α} \times \frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)} \Rightarrow x = \frac{- p a^{2} \cos α}{(b^{2} \sin α - a^{2} \cos^{2} α)} Substituting the values x and y in (ii), we get \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{1}{a^{2}} \times {(\frac{- p a^{2} \cos α}{(b^{2} \sin α - a^{2} \cos^{2} α)})}^{2} - \frac{1}{b^{2}} \times {(\frac{p b^{2} \sin α}{(b^{2} \sin α - a^{2} \cos^{2} α)})}^{2} = 1 \Rightarrow \frac{p^{2} a^{2} \cos^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} - \frac{p^{2} b^{2} \sin^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} a^{2} \cos^{2} α - p^{2} b^{2} \sin^{2} α}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} (a^{2} \cos^{2} α - b^{2} \sin^{2} α)}{{(b^{2} \sin α - a^{2} \cos^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2} (a^{2} \cos^{2} α - b^{2} \sin^{2} α)}{{(a^{2} \cos^{2} α - b^{2} \sin^{2} α)}^{2}} = 1 \Rightarrow \frac{p^{2}}{(a^{2} \cos^{2} α - b^{2} \sin^{2} α)} = 1 ∴ p^{2} = a^{2} \cos^{2} α - b^{2} \sin^{2} α$

Page No 15.41:

Question 1:

The equation to the normal to the curve y = sin x at (0, 0) is

(a) x = 0
(b) y = 0
(c) x + y = 0
(d) x − y = 0

Answer:

(c) x + y = 0

$Given : y = \sin x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = \cos x Slope of the tangent = {(\frac{d y}{d x})}_{(0, 0)} = cos 0 =1 Slope of the normal, m = \frac{- 1}{1} =-1 Given : (x_{1}, y_{1}) = (0, 0) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = - 1 (x - 0) \Rightarrow y = - x \Rightarrow x + y = 0$

Page No 15.41:

Question 2:

The equation of the normal to the curve y = x + sin x cos x at x = π/2 is

(a) x = 2
(b) x = π
(c) x + π = 0
(d) 2x = π

Answer:

(d) 2x = π

$Given : y = x + \sin x \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = 1 + \cos^{2} x - \sin^{2} x Slope of the tangent= {(\frac{d y}{d x})}_{x = \frac{π}{2}} {=1+cos}^{2} \frac{π}{2} {-sin}^{2} \frac{π}{2} =1-1=0 Slope of the normal, m = \frac{- 1}{0} When x = \frac{π}{2}, y= \frac{π}{2} +cos \frac{π}{2} sin \frac{π}{2} = \frac{π}{2} Now, (x_{1}, y_{1}) = (\frac{π}{2}, \frac{π}{2}) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{π}{2} = \frac{- 1}{0} (x - \frac{π}{2}) \Rightarrow x = \frac{π}{2} \Rightarrow 2 x = π$

Page No 15.41:

Question 3:

The equation of the normal to the curve y = x(2 − x) at the point (2, 0) is

(a) x − 2y = 2
(b) x − 2y + 2 = 0
(c) 2x + y = 4
(d) 2x + y − 4 = 0

Answer:

(a) x − 2y = 2

$Here, y = x (2 - x) = 2 x - x^{2} \Rightarrow \frac{d y}{d x} = 2 - 2 x Slope of the tangent= {(\frac{d y}{d x})}_{(2, 0)} = 2 - 4 = - 2 S lope of the normal, m = \frac{- 1}{- 2} = \frac{1}{2} Given : (x_{1}, y_{1}) = (2, 0) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = \frac{1}{2} (x - 2) \Rightarrow 2 y = x - 2 \Rightarrow x - 2 y = 2$

Page No 15.41:

Question 4:

The point on the curve y² = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.41:

Question 5:

If the tangent to the curve x = a t², y = 2 at is perpendicular to x-axis, then its point of contact is

(a) (a, a)
(b) (0, a)
(c) (0, 0)
(d) (a, 0)

Answer:

(c) (0, 0)

Let the required point be (x₁, y₁).

$Since, the point lies on the curve . Hence, x_{1} = a t^{2} and y_{1} = 2 a t Now, x = a t^{2} and y = 2 a t \Rightarrow \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t} = \frac{2 a}{y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{2 a}{y_{1}} It is given that the tangent is perpendicular to the y -axis. It means that it is parallel to the x -axis. ∴ Slope of the tangent = Slope of the x -axis \frac{2 a}{y_{1}} = 0 \Rightarrow a = 0 Now, x_{1} = a t^{2} = 0 and y_{1} = 2 a t = 0 ∴ (x_{1}, y_{1}) = (0, 0)$

Page No 15.41:

Question 6:

The point on the curve y = x² − 3x + 2 where tangent is perpendicular to y = x is

(a) (0, 2)
(b) (1, 0)
(c) (−1, 6)
(d) (2, −2)

Answer:

(b) (1, 0)

y = x
$\Rightarrow \frac{d y}{d x} = 1$

$Let (x_{1}, y_{1}) be the required point. Since, the point lies on the curve, Hence, y_{1} = {x_{1}}^{2} - 3 x_{1} + 2 Now, y = x^{2} - 3 x + 2 ∴ \frac{d y}{d x} = 2 x - 3 Slope of the tangent= {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} - 3$
The tangent is perpendicular to this line.
∴Slope of the tangent = $\frac{- 1}{Slope of the line} = \frac{- 1}{1} = - 1$

Now,
$2 x_{1} - 3 = - 1 \Rightarrow 2 x_{1} = 2 \Rightarrow x_{1} = 1 and y_{1} = {x_{1}}^{2} - 3 x_{1} + 2 = 1 - 3 + 2 = 0 ∴ (x_{1}, y_{1}) = (1, 0)$

Page No 15.41:

Question 7:

The point on the curve y² = x where tangent makes 45° angle with x-axis is

(a) (1/2, 1/4)
(b) (1/4, 1/2)
(c) (4, 2)
(d) (1, 1)

Answer:

(b) (1/4, 1/2)

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.41:

Question 8:

The point at the curve y = 12x − x² where the slope of the tangent is zero will be

(a) (0, 0)
(b) (2, 16)
(c) (3, 9)
(d) none of these

Answer:

(d) none of these

$Let (x_{1}, y_{1}) be the required point. Since, the point lies on the curve, Hence, y_{1} = 12 x_{1} - {x_{1}}^{2} Now, y = 12 x - x^{2} \Rightarrow \frac{d y}{d x} = 12 - 2 x S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 12 - 2 x_{1} Given: Slope of the tangent=0 12 - 2 x_{1} = 0 \Rightarrow x_{1} = 6 Now, y_{1} = 12 x_{1} - {x_{1}}^{2} = 72 - 36 = 36 ∴ (x_{1}, y_{1}) = (6, 36)$

Page No 15.41:

Question 9:

The angle between the curves y² = x and x²= y at (1, 1) is

(a) $\tan^{- 1} \frac{4}{3}$
(b) $\tan^{- 1} \frac{3}{4}$
(c) 90°
(d) 45°

Answer:

(b) $\tan^{- 1} \frac{3}{4}$

$Given : y^{2} = x . . . (1) x^{2} = y . . . (2) Point = (1, 1) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} \Rightarrow m_{1} = \frac{1}{2} On d ifferentiating (2) w.r.t. x, we get 2 x = \frac{d y}{d x} \Rightarrow m_{2} = 2 (1) = 2 Now, \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2}| = \frac{3}{4} \Rightarrow θ = \tan^{- 1} (\frac{3}{4})$

Page No 15.42:

Question 10:

The equation of the normal to the curve 3x² − y² = 8 which is parallel to x + 3y = 8 is

(a) x + 3y = 8
(b) x + 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0

Answer:

(c) x + 3y ± 8 = 0

The slope of line x + 3y = 8 is $\frac{- 1}{3}$ .

$Since the normal is parallel to the given line, the equation of normal will be of the given form. x + 3 y = k 3 x^{2} - y^{2} = 8 Let (x_{1}, y_{1}) be the point of intersection of the two curves . Then, x_{1} + 3 y_{1} = k . . . (1) 3 {x_{1}}^{2} - {y_{1}}^{2} = 8 . . . (2) Now, 3 x^{2} - y^{2} = 8 On d ifferentiating both sides w.r.t. x, we get 6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{6 x}{2 y} = \frac{3 x}{y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{3 x_{1}}{y_{1}} Slope of the normal, m = \frac{- 1}{(\frac{3 x}{y})} = \frac{- y_{1}}{3 x_{1}} Given: Slope of the normal = Slope of the given line \Rightarrow \frac{- y_{1}}{3 x_{1}} = \frac{- 1}{3} \Rightarrow y_{1} = x_{1} . . . (3) From (2), we get 3 {x_{1}}^{2} - {x_{1}}^{2} = 8 \Rightarrow 2 {x_{1}}^{2} = 8 \Rightarrow {x_{1}}^{2} = 4 \Rightarrow x_{1} = \pm 2 Case 1: When x_{1} =2 From (3), we get y_{1} = x_{1} = 2 ∴ (x_{1}, y) = (2, 2) From (1), we get 2 + 3 (2) = k \Rightarrow 2 + 6 = k \Rightarrow k = 8 ∴ Equation of the normal from (1) \Rightarrow x + 3 y = 8 \Rightarrow x + 3 y - 8 = 0 Case 2: When x_{1} =-2 From (3), we get y_{1} = x_{1} = - 2, ∴ (x_{1}, y) = (- 2, - 2) From (1), we get - 2 + 3 (- 2) = k \Rightarrow - 2 - 6 = k \Rightarrow k = - 8 ∴ Equation of the normal from (1) \Rightarrow x + 3 y = - 8 \Rightarrow x + 3 y + 8 = 0 From both the cases, we get the equation of the normal as: x + 3 y \pm 8 = 0$

Page No 15.42:

Question 11:

The equations of tangent at those points where the curve y = x² − 3x + 2 meets x-axis are

(a) x − y + 2 = 0 = x − y − 1
(b) x + y − 1 = 0 = x − y − 2
(c) x − y − 1 = 0 = x − y
(d) x − y = 0 = x + y

Answer:

(b) x + y − 1 = 0 = x − y − 2

Let the tangent meet the x-axis at point (x, 0).
Now,
$y = x^{2} - 3 x + 2 \Rightarrow \frac{d y}{d x} = 2 x - 3 The tangent passes through point (x, 0). ∴ 0 = x^{2} - 3 x + 2 \Rightarrow (x - 2) (x - 1) = 0 \Rightarrow x = 2 or x = 1 Case 1: When x =2: Slope of the tangent, m = {(\frac{d y}{d x})}_{(2, 0)} =4-3=1 ∴ (x_{1}, y_{1}) = (2, 0) Equation of the tangent: y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = 1 (x - 2) \Rightarrow x - y - 2 = 0 Case 2: When x =1: Slope of the tangent, m = {(\frac{d y}{d x})}_{(2, 0)} =2-3=-1 ∴ (x_{1}, y_{1}) = (1, 0) Equation of the tangent: y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = - 1 (x - 1) \Rightarrow x + y - 1 = 0$

Page No 15.42:

Question 12:

The slope of the tangent to the curve x = t² + 3 t − 8, y = 2t² − 2t − 5 at point (2, −1) is

(a) 22/7
(b) 6/7
(c) −6
(d) none of these

Answer:

(b) 6/7

$x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} The g iven point is (2, -1). ∴ x =2 and y =-1 Now, t^{2} + 3 t - 8 = 2 and 2 t^{2} - 2 t - 5 = - 1 Let us solve one of these to get the value of t. t^{2} + 3 t - 10 = 0 and 2 t^{2} - 2 t - 4 = 0 \Rightarrow (t + 5) (t - 2) = 0 and (2 t + 2) (t - 2) = 0 \Rightarrow t = - 5 or t= 2 and t =-1 or t =2 These two have t = 2 as a common solution . ∴ Slope of the tangent = {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$

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Question 13:

At what point the slope of the tangent to the curve x² + y² − 2x − 3 = 0 is zero

(a) (3, 0), (−1, 0)
(b) (3, 0), (1, 2)
(c) (−1, 0), (1, 2)
(d) (1, 2), (1, −2)

Answer:

(d) (1, 2), (1, −2)

Let (x₁, y₁) be the required point.

$Since, the point lie on the curve . Hence, {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 . . . (1) Now, x^{2} + y^{2} - 2 x - 3 = 0 \Rightarrow 2 x + 2 y \frac{d y}{d x} - 2 = 0 ∴ \frac{d y}{d x} = \frac{2 - 2 x}{2 y} = \frac{1 - x}{y} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} S lope of the tangent=0 (Given) ∴ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1 From (1), we get {x_{1}}^{2} + {y_{1}}^{2} - 2 x_{1} - 3 = 0 \Rightarrow 1 + {y_{1}}^{2} - 2 - 3 = 0 \Rightarrow {y_{1}}^{2} - 4 = 0 \Rightarrow y_{1} = \pm 2 So, the points are (1, 2) and (1, - 2) .$

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Question 14:

The angle of intersection of the curves xy = a² and x² − y² = 2a²is

(a) 0°
(b) 45°
(c) 90°
(d) none of these

Answer:

(c) 90°

$Given : x y = a^{2} . . . (1) x^{2} - y^{2} = 2 a^{2} . . . (2) Let (x_{1}, y_{1}) be the point of intersection. On d ifferentiating (1) w.r.t. x, we get x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- y_{1}}{x_{1}} On d ifferentiating (2) w.r.t. x, we get 2 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{y} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{x_{1}}{y_{1}} Now, m_{1} \times m_{2} = \frac{- y_{1}}{x_{1}} \times \frac{x_{1}}{y_{1}} \Rightarrow m_{1} \times m_{2} = - 1 ∵ m_{1} \times m = - 1 So, the angle between the curves is 90°.$

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Question 15:

If the curve ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then a is equal to

(a) 1
(b) −6
(c) 6
(d) 0

Answer:

(c) 6

$Given : a y + x^{2} = 7 . . . (1) x^{3} = y . . . (2) Point= (1, 1) On d ifferentiating (1) w.r.t. x, we get a \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{a} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 2}{a} Again, on d ifferentiating (2) w.r.t. x, we get 3 x^{2} = \frac{d y}{d x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 1)} = 3 It is given that the curves are orthogonal at the given point. ∴ m_{1} \times m_{2} = - 1 \Rightarrow \frac{- 2}{a} \times 3 = - 1 \Rightarrow a = 6$

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Question 16:

If the line y = x touches the curve y = x² + bx + c at a point (1, 1) then

(a) b = 1, c = 2
(b) b = −1, c = 1
(c) b = 2, c = 1
(d) b = −2, c = 1

Answer:

(b) b = −1, c = 1

We can find the slope of the line by differentiating w.r.t. x.
Slope of the given line = 1
Now,
$y = x^{2} + b x + c . . . (1) \Rightarrow \frac{d y}{d x} = 2 x + b Slope of the tangent= {(\frac{d y}{d x})}_{(1, 1)} =2+ b Given: Slope of the tangent = 1 \Rightarrow 2 + b = 1 \Rightarrow b = - 1 On s ubstituting b = - 1, x =1 and y =1 in (1), we get \Rightarrow 1 = 1 - 1 + c \Rightarrow c = 1$

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Question 17:

The slope of the tangent to the curve x = 3t² + 1, y = t³ −1 at x = 1 is

(a) 1/2
(b) 0
(c) −2
(d) ∞

Answer:

(b) 0

$Given : x = 3 t^{2} + 1 y = t^{3} - 1 x = 1 Now, 3 t^{2} + 1 = 1 \Rightarrow 3 t^{2} = 0 \Rightarrow t = 0 \frac{d x}{d t} = 6 t and \frac{d y}{d t} = 3 t^{2} ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{3 t^{2}}{6 t} = \frac{t}{2} Thus, we get Slope of the tangent= {(\frac{d y}{d x})}_{t = 0} = \frac{0}{2} =0$

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Question 18:

The curves y = ae^x and y = be^−x cut orthogonally, if

(a) a = b
(b) a = −b
(c) ab = 1
(d) ab = 2

Answer:

(c) ab = 1

$Given : y = a e^{x} . . . (1) y = b e^{- x} . . . (2) Let the point of intersection of these two curves be (x_{1}, y_{1}) . Now, On d ifferentiating (1) w.r.t. x, we get \frac{d y}{d x} = a e^{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = a e^{x_{1}} Again, on d ifferentiating (2) w.r.t. x, we get \frac{d y}{d x} = - b e^{- x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = - b e^{- x_{1}} It is given that the curves cut orthogonally. ∴ m_{1} \times m_{2} = - 1 \Rightarrow a e^{x_{1}} \times (- b e^{- x_{1}}) = - 1 \Rightarrow a b = 1$

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Question 19:

The equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at the point θ = π/4 is

(a) x = 0
(b) y = 0
(c) c = y
(d) x + y = a

Answer:

(c) x = y

$Here, x = a \cos^{3} θ and y = a \sin^{3} θ \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ and \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} = - \tan θ Now, Slope of the tangent = {(\frac{d y}{d x})}_{θ = \frac{π}{4}} =-tan \frac{π}{4} =-1 (x_{1}, y_{1}) = (a \cos^{3} \frac{π}{4}, a \sin^{3} \frac{π}{4}) = (\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}) ∴ Equation of the normal = y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{a}{2 \sqrt{2}} = 1 (x - \frac{a}{2 \sqrt{2}}) \Rightarrow x = y$

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Question 20:

If the curves y = 2 e^x and y = ae^−x intersect orthogonally, then a =

(a) 1/2
(b) −1/2
(c) 2
(d) 2e²

Answer:

(a) 1/2

$Given : y = 2 e^{x} . . . (1) y = a e^{- x} . . . (2) Let the point of intersection of these curves be (x_{1}, y_{1}) . On d ifferentiating (1) w.r.t. x, we get \frac{d y}{d x} = 2 e^{x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 e^{x_{1}} Now, on d ifferentiating (2) w.r.t. x, we get \frac{d y}{d x} = - a e^{- x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = - a e^{- x_{1}} It is given that the curves are orthogonal. ∴ m_{1} \times m_{2} = - 1 \Rightarrow 2 e^{x_{1}} \times (- a e^{- x_{1}}) = - 1 \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2}$

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Question 21:

The point on the curve y = 6x − x² at which the tangent to the curve is inclined at π/4 to the line x + y = 0 is

(a) (−3, −27)
(b) (3, 9)
(c) (7/2, 35/4)
(d) (0, 0)

Answer:

(b) (3, 9)

Let (x₁, y₁) be the point where the given curve intersect the given line at the given angle.
$Since, the point lie on the curve . Hence, y_{1} = 6 x_{1} - {x_{1}}^{2} Now, y = 6 x - x^{2} \Rightarrow \frac{d y}{d x} = 6 - 2 x \Rightarrow m_{1} = 6 - 2 x_{1} and x + y = 0 \Rightarrow 1 + \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - 1 \Rightarrow m_{2} = - 1 It is given that the angle between them is \frac{π}{4} . ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| \Rightarrow \tan \frac{π}{4} = |\frac{6 - 2 x_{1} + 1}{1 - 6 + 2 x_{1}}| \Rightarrow 1 = |\frac{7 - 2 x_{1}}{2 x_{1} - 5}| \Rightarrow \frac{7 - 2 x_{1}}{2 x_{1} - 5} = \pm 1 \Rightarrow \frac{7 - 2 x_{1}}{2 x_{1} - 5} = 1 or \frac{7 - 2 x_{1}}{2 x_{1} - 5} =-1 \Rightarrow 7 - 2 x_{1} = 2 x_{1} - 5 or 7 - 2 x_{1} = - 2 x_{1} + 5 \Rightarrow 4 x_{1} = 12 or 2 = 0 (It is not true.) \Rightarrow x_{1} = 3 and y_{1} = 6 x_{1} - {x_{1}}^{2} = 18 - 9 = 9 ∴ (x_{1}, y_{1}) = (3, 9)$

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Question 22:

The angle of intersection of the parabolas y² = 4 ax and x² = 4ay at the origin is

(a) π/6
(b) π/3
(c) π/2
(d) π/4

Answer:

(c) π/2

$Given : y^{2} = 4 a x . . . (1) x^{2} = 4 a y . . . (2) Point = (0, 0) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} \Rightarrow m_{1} = \infty Now, on d ifferentiating (2) w.r.t. x, we get 2 x = 4 a \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x}{2 a} = 0 ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\infty}{1 + 0}| = \infty \Rightarrow θ = \tan^{- 1} \infty = \frac{π}{2}$

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Question 23:

The angle of intersection of the curves y = 2 sin²x and y = cos 2 x at $x = \frac{π}{6}$ is

(a) π/4
(b) π/2
(c) π/3
(d) none of these

Answer:

(c) π/3

$Given: x = \frac{π}{6} Now, y = 2 \sin^{2} x \Rightarrow \frac{d y}{d x} = 4 \sin x \cos x \Rightarrow \frac{d y}{d x} = 2 \sin 2 x \Rightarrow m_{1} = {(\frac{d y}{d x})}_{x = \frac{π}{6}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} Also, y = \cos 2 x \Rightarrow \frac{d y}{d x} = - 2 \sin 2 x \Rightarrow m_{2} = {(\frac{d y}{d x})}_{x = \frac{π}{6}} = - 2 \times \frac{\sqrt{3}}{2} = - \sqrt{3} ∴ \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\sqrt{3} + \sqrt{3}}{1 - \sqrt{3} \sqrt{3}}| = |\frac{2 \sqrt{3}}{- 2}| = \sqrt{3} \Rightarrow θ = \tan^{- 1} (\sqrt{3}) = \frac{π}{3}$

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Question 24:

Any tangent to the curve y = 2x⁷ + 3x + 5

(a) is parallel to x-axis
(b) is parallel to y-axis
(c) makes an acute angle with x-axis
(d) makes an obtuse angle with x-axis

Answer:

(c) makes an acute angle with x-axis

We have, y = 2x⁷ + 3x + 5
$\frac{d y}{d x} = 14 x^{6} + 3 \Rightarrow \frac{d y}{d x} > 3 (∵ x^{6} is always positive for any real value of x) \Rightarrow \frac{d y}{d x} > 0 S o, \tan θ > 0 Hence, θ lies in first quadrant . Thus, the tangent to the curve makes an acute angle with x - axis$

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Question 25:

The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is

(a) $(4, \frac{8}{3})$
(b) $(- 4, \frac{8}{3})$
(c) $(4, - \frac{8}{3})$
(d) none of these

Answer:

(a) $(4, \frac{8}{3})$ and (c) $(4, - \frac{8}{3})$

Let (x₁, y₁) be the required point.
$Since, (x_{1}, y_{1}) lies on the given curve ∴ 9 {y_{1}}^{2} = {x_{1}}^{3} . . . (1) Now, 9 y^{2} = x^{3} 18 y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{18 y} = \frac{x^{2}}{6 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{{x_{1}}^{2}}{6 y_{1}} Slope of the normal = \frac{- 1}{\frac{{x_{1}}^{2}}{6 y_{1}}} = \frac{- 6 y_{1}}{{x_{1}}^{2}} It is given that the normal makes equal intercepts with the axes. ∴ Slope of the normal = ±1 Now, \frac{- 6 y_{1}}{{x_{1}}^{2}} = \pm 1 \Rightarrow \frac{- 6 y_{1}}{{x_{1}}^{2}} = 1 or \frac{- 6 y_{1}}{{x_{1}}^{2}} =-1 \Rightarrow y_{1} = \frac{- {x_{1}}^{2}}{6} or y_{1} = \frac{{x_{1}}^{2}}{6} . . . (2) Case 1: When y_{1} = \frac{- {x_{1}}^{2}}{6} From (1), we have 9 (\frac{{x_{1}}^{4}}{36}) = {x_{1}}^{3} \Rightarrow {x_{1}}^{4} = 4 {x_{1}}^{3} \Rightarrow {x_{1}}^{4} - 4 {x_{1}}^{3} = 0 \Rightarrow {x_{1}}^{3} (x_{1} - 4) = 0 \Rightarrow x_{1} = 0, 4 Putting x_{1} = 0 in (1), we get, 9 {y_{1}}^{2} = 0 \Rightarrow y_{1} = 0 Putting x_{1} = 4 in (1), we get, 9 {y_{1}}^{2} = 4^{3} \Rightarrow y_{1} = \pm \frac{8}{3} But, the line making the equal intercepts with the coordinate axes can not pass through the origin . So, the points are (4, \pm \frac{8}{3})$

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Question 26:

The slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at the point (2, −1) is

(a) $\frac{22}{7}$
(b) $\frac{6}{7}$
(c) $\frac{7}{6}$
(d) $- \frac{6}{7}$

Answer:

(b) $\frac{6}{7}$

Given:
x = t² + 3t − 8 and y = 2t² − 2t − 5

$\Rightarrow \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} The given point is (2, -1). ∴ x = 2 and y = -1 Now, t^{2} + 3 t - 8 = 2 and 2 t^{2} - 2 t - 5 = - 1 \Rightarrow t^{2} + 3 t - 10 = 0 and t^{2} - t - 2 = 0 \Rightarrow (t + 5) (t - 2) = 0 and (t + 1) (t - 2) = 0 \Rightarrow t = - 5, 2 and t = - 1, 2 We can see that t = 2 satisfies both of these. ∴ Slope of the tangent = {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$
Thus, the correct option is (b).

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Question 27:

The line y = mx + 1 is a tangent to the curve y² = 4x, if the value of m is

(a) 1
(b) 2
(c) 3
(d) $\frac{1}{2}$

Answer:

(a) 1
Let (x₁, y₁) be the required point.
The slope of the given line is m.
We have
$y^{2} = 4 x \Rightarrow 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{4}{2 y} = \frac{2}{y} Slope of the tangent = (\frac{d y}{d x})_{(x_{1}, y_{1})} = \frac{2}{y_{1}} Given: Slope of the tangent = m Now, \frac{2}{y_{1}} = m . . . (1)$

Because the given line is a tangent to the given curve at point (x₁, y₁), this point lies on both the line and the curve.

$∴ y_{1} = m x_{1} + 1 and {y_{1}}^{2} = 4 x_{1} \Rightarrow x_{1} = \frac{y_{1} - 1}{m} and x_{1} = \frac{{y_{1}}^{2}}{4} So, \frac{y_{1} - 1}{m} = \frac{{y_{1}}^{2}}{4} \Rightarrow \frac{y_{1} - 1}{(\frac{2}{y_{1}})} = \frac{{y_{1}}^{2}}{4} [From (1)] \Rightarrow \frac{y_{1} (y_{1} - 1)}{2} = \frac{{y_{1}}^{2}}{4} \Rightarrow 2 {y_{1}}^{2} - 2 y_{1} = {y_{1}}^{2} \Rightarrow {y_{1}}^{2} - 2 y_{1} = 0 \Rightarrow {y_{1}}^{2} - 2 y_{1} = 0 \Rightarrow y_{1} (y_{1} - 2) = 0 \Rightarrow y_{1} = 0, 2 So, For y_{1} =0, m = \frac{2}{0} = \infty For y_{1} =2, m = \frac{2}{2} = 1$

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Question 28:

The normal at the point (1, 1) on the curve 2y + x² = 3 is

(a) x + y = 0
(b) x − y = 0
(c) x + y + 1 = 0
(d) x − y = 1

Answer:

(b) x − y = 0

$Given : 2 y + x^{2} = 3 \Rightarrow 2 \frac{d y}{d x} + 2 x = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{2} = - x Slope of the tangent = {(\frac{d y}{d x})}_{(1, 1)} =-1 Slope of the normal, m = \frac{- 1}{Slope of the tangent} = \frac{- 1}{- 1} =1 Now, (x_{1}, y_{1}) = (1, 1) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = 1 (x - 1) \Rightarrow x - y = 0$

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Question 29:

The normal to the curve x² = 4y passing through (1, 2) is

(a) x + y = 3
(b) x − y = 3
(c) x + y = 1
(d) x − y = 1

Answer:

Disclaimer: None of the given options is correct.

$Given : x^{2} = 4 y \Rightarrow 2 x = 4 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{2 x}{4} = \frac{x}{2} Slope of the tangent = {(\frac{d y}{d x})}_{(1, 2)} = \frac{1}{2} Slope of the normal, m = \frac{- 1}{Slope of the tangent} = \frac{- 1}{\frac{1}{2}} =-2 Also, (x_{1}, y_{1}) = (1, 2) ∴ Equation of the normal = y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = - 2 (x - 1) \Rightarrow y - 2 = - 2 x + 2 \Rightarrow 2 x + y = 4$

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Question 30:

The abscissa of the point on the curve 3y = 6x - 5x³, the normal at which passes through the origin is
(a) 1 (b) $\frac{1}{3}$ (c) 2 (d) $\frac{1}{2}$

Answer:

Let (h, k) be the point on the curve 3y = 6x − 5x³, the normal at which passes through the origin.

∴ 3k = 6h − 5h³ .....(1)

3y = 6x − 5x³

Differentiating both sides with respect to x, we get

$3 \frac{d y}{d x} = 6 - 15 x^{2}$

$\Rightarrow \frac{d y}{d x} = 2 - 5 x^{2}$

∴ Slope of tangent to the given curve at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 2 - 5 h^{2}$

Slope of normal to the given curve at (h, k) = $- \frac{1}{{(\frac{d y}{d x})}_{(h, k)}} = - \frac{1}{(2 - 5 h^{2})}$

Equation of the normal to the given curve at (h, k) is

$y - k = - \frac{1}{(2 - 5 h^{2})} (x - h)$

This equation passes through the origin.

$∴ 0 - k = - \frac{1}{(2 - 5 h^{2})} (0 - h)$

$\Rightarrow k = - \frac{h}{2 - 5 h^{2}}$ .....(2)

From (1) and (2), we have

$- \frac{h}{2 - 5 h^{2}} = \frac{6 h - 5 h^{3}}{3}$

$\Rightarrow \frac{1}{5 h^{2} - 2} = \frac{6 - 5 h^{2}}{3}$

$\Rightarrow 30 h^{2} - 25 h^{4} - 12 + 10 h^{2} = 3$

$\Rightarrow 25 h^{4} - 40 h^{2} + 15 = 0$

$\Rightarrow 5 h^{4} - 8 h^{2} + 3 = 0$

$\Rightarrow 5 h^{4} - 5 h^{2} - 3 h^{2} + 3 = 0$

$\Rightarrow 5 h^{2} (h^{2} - 1) - 3 (h^{2} - 1) = 0$

$\Rightarrow (h^{2} - 1) (5 h^{2} - 3) = 0$

⇒ h² − 1 = 0 or 5h² − 3 = 0

⇒ h² = 1 or $h^{2} = \frac{3}{5}$

⇒ h = ±1 or $h = \pm \sqrt{\frac{3}{5}}$

Thus, the abscissa of the points on the given curve, the normal at which passes through the origin are $- \sqrt{\frac{3}{5}}$ , −1, 1 and $\sqrt{\frac{3}{5}}$ .

Hence, the correct answer is option (a).

Page No 15.43:

Question 31:

Two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2
(a) touch each other (b) cut at right angle
(c) cut an angle $\frac{π}{3}$ (d) cut at an angle $\frac{π}{4}$

Answer:

The given curves are x³ − 3xy² + 2 = 0 and 3x²y − y³ = 2.

We know that the angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.

Let (x₁, y₁) be the point of intersection of two curves,

Consider the first curve C₁≡ x³ − 3xy² + 2 = 0.

x³ − 3xy² + 2 = 0

Differentiating both sides with respect to x, we get

$3 x^{2} - 3 (x \times 2 y \frac{d y}{d x} + y^{2} \times 1) + 0 = 0$

$\Rightarrow 3 x^{2} - 6 x y \frac{d y}{d x} - 3 y^{2} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{3 x^{2} - 3 y^{2}}{6 x y}$

∴ Slope of tangent to the first curve at (x₁, y₁) = ${(\frac{d y}{d x})}_{C_{1}} = \frac{x_{1}^{2} - y_{1}^{2}}{2 x_{1} y_{1}}$

Now, consider the second curve C₂≡ 3x²y − y³ = 2.

3x²y − y³ = 2

Differentiating both sides with respect to x, we get

$3 (x^{2} \times \frac{d y}{d x} + y \times 2 x) - 3 y^{2} \frac{d y}{d x} = 0$

$\Rightarrow (3 x^{2} - 3 y^{2}) \frac{d y}{d x} = - 6 x y$

$\Rightarrow \frac{d y}{d x} = - \frac{6 x y}{3 x^{2} - 3 y^{2}}$

∴ Slope of tangent to the second curve at (x₁, y₁) = ${(\frac{d y}{d x})}_{C_{2}} = - \frac{2 x_{1} y_{1}}{x_{1}^{2} - y_{1}^{2}}$

Now, ${(\frac{d y}{d x})}_{C_{1}} \times {(\frac{d y}{d x})}_{C_{2}} = (\frac{x_{1}^{2} - y_{1}^{2}}{2 x_{1} y_{1}}) \times (- \frac{2 x_{1} y_{1}}{x_{1}^{2} - y_{1}^{2}}) = - 1$

The tangents to the two curves are perpendicular to each other.

Thus, the two curves cut at right angle.

Hence, the correct answer is option (b).

Page No 15.43:

Question 32:

The tangent to the curve x = e^t cost, y = e^t sin t at t = $\frac{π}{4}$ makes with x-axis an angle
(a) 0 (b) $\frac{π}{4} (c) \frac{π}{3} (d) \frac{π}{2}$

Answer:

The given curve is x = e^t cost, y = e^t sint.

x = e^t cost

Differentiating both sides with respect to t, we get

$\frac{d x}{d t} = e^{t} \times (- \sin t) + \cos t \times e^{t}$

$\Rightarrow \frac{d x}{d t} = e^{t} (\cos t - \sin t)$

y = e^t sint

Differentiating both sides with respect to t, we get

$\frac{d y}{d t} = e^{t} \times \cos t + \sin t \times e^{t}$

$\Rightarrow \frac{d y}{d t} = e^{t} (\cos t + \sin t)$

Now,

Slope of tangent to the given curve $= \frac{d y}{d x}$ $= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $= \frac{e^{t} (\cos t + \sin t)}{e^{t} (\cos t - \sin t)}$ $= \frac{\cos t + \sin t}{\cos t - \sin t}$

∴ Slope of tangent to the given curve at $t = \frac{π}{4}$ $= {(\frac{d y}{d x})}_{t = \frac{π}{4}}$ $= \frac{\cos \frac{π}{4} + \sin \frac{π}{4}}{\cos \frac{π}{4} - \sin \frac{π}{4}}$ $= \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}$ $= \frac{\sqrt{2}}{0} = \infty$

Let the angle made by the tangent to the given curve at t = $\frac{π}{4}$ with the x-axis be θ.

∴ tanθ = Slope of tangent to the given curve at $t = \frac{π}{4}$ = $\infty$

$\Rightarrow \tan θ = \tan \frac{π}{2}$

$\Rightarrow θ = \frac{π}{2}$

Thus, the tangent to the curve x = e^t cost, y = e^t sint at t = $\frac{π}{4}$ makes with x-axis an angle $\frac{π}{2}$ .

Hence, the correct answer is option (d).

Page No 15.43:

Question 33:

The tangent to the curve y = e^2x at the point (0,1) meets x-axis at :
(a) (0, 1) (b) $(- \frac{1}{2}, 0)$ (c) (2, 0) (d) (0, 2)

Answer:

The given curve is y = e^2x.

y = e^2x

$\Rightarrow \frac{d y}{d x} = 2 e^{2 x}$

$\Rightarrow {(\frac{d y}{d x})}_{(0, 1)} = 2 e^{2 \times 0} = 2 \times 1 = 2$         .....(1)

So, the equation of tangent at (0, 1) is

$y - 1 = {(\frac{d y}{d x})}_{(0, 1)} (x - 0)$

$\Rightarrow y - 1 = 2 x$               [Using (1)]

$\Rightarrow 2 x - y + 1 = 0$    .....(2)

This tangent meet the x-axis where y = 0.

Putting y = 0 in (2), we get

$2 x - 0 + 1 = 0$

$\Rightarrow x = - \frac{1}{2}$

Thus, the tangent to the given curve y = e^2x at the point (0,1) meets x-axis at $(- \frac{1}{2}, 0)$ .

Hence, the correct answer is option (b).

Page No 15.43:

Question 34:

The equation of tangent to the curve y(1 + x²) = 2 - x, where it crosses x-axis is
(a) x + 5y = 2 (b) x - 5y = 2 (c) 5x - y = 2 (d) 5x + y = 2

Answer:

The equation of given curve is

y(1 + x²) = 2 − x .....(1)

This cuts the x-axis at the point, where y = 0.

Putting y = 0 in (1), we get

0 × (1 + x²) = 2 − x

⇒ 2 − x = 0

⇒ x = 2

So, the given curve intersects the x-axis at (2, 0).

y(1 + x²) = 2 − x

Differentiating both sides with respect to x, we get

$y \times 2 x + (1 + x^{2}) \times \frac{d y}{d x} = - 1$

$\Rightarrow (1 + x^{2}) \frac{d y}{d x} = - 1 - 2 x y$

$\Rightarrow \frac{d y}{d x} = - \frac{1 + 2 x y}{1 + x^{2}}$

∴ Slope of tangent to the given curve at (2, 0) = ${(\frac{d y}{d x})}_{(2, 0)} = - \frac{1 + 2 \times 2 \times 0}{1 + {(2)}^{2}} = - \frac{1}{5}$

So, the equation of tangent at (2, 0) is

$y - 0 = {(\frac{d y}{d x})}_{(2, 0)} (x - 2)$

$\Rightarrow y = - \frac{1}{5} (x - 2)$

$\Rightarrow 5 y = - x + 2$

$\Rightarrow x + 5 y = 2$

Thus, the equation of tangent to the given curve where it crosses x-axis is x + 5y = 2.

Hence, the correct answer is option (a).

Page No 15.43:

Question 35:

The points at which the tangents to the curve y = x³ - 12x + 18 are parallel to x-axis are
(a) (2, -2)(-2, -34) (b) (2, 34)(-2, 0)
(c) (0, 34)(-2, 0) (d) (2, 2)(-2, 34)

Answer:

Let (x₁, y₁) be the point of contact of tangent with the given curve y = x³ − 12x + 18.

∴ $y_{1} = x_{1}^{3} - 12 x + 18$ .....(1)

y = x³ − 12x + 18

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 3 x^{2} - 12$

∴ Slope of tangent at (x₁, y₁) = ${(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 x_{1}^{2} - 12$

It is given that, the tangent is parallel to the x-axis.

$∴ {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 0$

$\Rightarrow 3 x_{1}^{2} - 12 = 0$

$\Rightarrow x_{1}^{2} = 4$

⇒ x₁ = −2 or x₁ = 2

Putting x₁ = −2 in (1), we get

$y_{1} = {(- 2)}^{3} - 12 \times (- 2) + 18 = - 8 + 24 + 18 = 34$

Putting x₁ = 2 in (1), we get

$y_{1} = {(2)}^{3} - 12 \times 2 + 18 = 8 - 24 + 18 = 2$

So, the coordinates of the point of contact are (−2, 34) and (2, 2).

Thus, the points at which the tangents to the given curve are parallel to x-axis are (−2, 34) and (2, 2).

Hence, the correct answer is option (d).

Page No 15.43:

Question 36:

The curve y = x^1/5 has at (0, 0)
(a) a vertical tangent (b) a horizontal tangent
(c) an oblique tangent (d) no tangent

Answer:

The given curve is

$y = x^{\frac{1}{5}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{1}{5} x^{- \frac{4}{5}}$

At (0, 0), we have

${(\frac{d y}{d x})}_{(0, 0)} = \frac{1}{5} \times {(0)}^{- \frac{4}{5}} = \frac{1}{5} \times \infty = \infty$

We know that, if $\frac{d y}{d x} = \infty$ , then the tangent is parallel to the y-axis.

Now, ${(\frac{d y}{d x})}_{(0, 0)} = \infty$

So, the tangent to the given curve at (0, 0) is parallel to the y-axis.

Thus, the curve $y = x^{\frac{1}{5}}$ has at (0, 0) a vertical tangent.

Hence, the correct answer is option (a).

Page No 15.44:

Question 1:

The equation of the normal to the curve y = tan x at (0, 0) is ______________.

Answer:

The given curve is

y = tanx

Differentiating both sides with respect to x, we have

$\frac{d y}{d x} = \sec^{2} x$

∴ Slope of tangent at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)} = \sec^{2} 0 = {(1)}^{2} = 1$

⇒ Slope of normal at (0, 0) = $- \frac{1}{{(\frac{d y}{d x})}_{(0, 0)}} = - \frac{1}{1} = - 1$

So, the equation of normal at (0, 0) is

$y - 0 = - \frac{1}{{(\frac{d y}{d x})}_{(0, 0)}} (x - 0)$

$\Rightarrow y = - 1 \times x$

$\Rightarrow x + y = 0$

Thus, the equation of the normal to the given curve at (0, 0) is x + y = 0.

The equation of the normal to the curve y = tanx at (0, 0) is ___x + y = 0___.

Page No 15.44:

Question 2:

The value of 'a' for which y = x² + ax + 25 touches the axis of x are ______________.

Answer:

The given curve is y = x² + ax + 25.

Suppose this curve touches the x-axis at (h, 0).

∴ 0 = h² + ah + 25 .....(1)

y = x² + ax + 25

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 x + a$

∴ Slope of tangent at (h, 0) = ${(\frac{d y}{d x})}_{(h, 0)} = 2 h + a$

The given curve touches the x-axis at (h, 0). Therefore,

${(\frac{d y}{d x})}_{(h, 0)} = 0$ (Slope of the x-axis = 0)

$\Rightarrow 2 h + a = 0$

$\Rightarrow h = - \frac{a}{2}$

Putting $h = - \frac{a}{2}$ in (1), we get

$0 = {(- \frac{a}{2})}^{2} + a \times (- \frac{a}{2}) + 25$

$\Rightarrow \frac{a^{2}}{4} - \frac{a^{2}}{2} + 25 = 0$

$\Rightarrow \frac{a^{2}}{4} = 25$

$\Rightarrow a^{2} = 100$

$\Rightarrow a = \pm 10$

So, the values of 'a' are −10 and 10.

Thus, the values of 'a' for which y = x² + ax + 25 touches the x-axis are −10 and 10.

The value of 'a' for which y = x² + ax + 25 touches the axis of x are __−10 and 10__.

Page No 15.44:

Question 3:

The points on the curve y = 12x - x³ at which the gradient is zero are _______________.

Answer:

Let (h, k) be the point on the curve y = 12x − x³ at which the gradient (or slope) of the tangent is zero.

∴ k = 12h − h³ .....(1)

y = 12x − x³

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 12 - 3 x^{2}$

$∴ {(\frac{d y}{d x})}_{(h, k)} = 12 - 3 h^{2} = 0$ (Given)

$\Rightarrow 12 - 3 h^{2} = 0$

$\Rightarrow h^{2} = 4$

⇒ h = −2 or h = 2

Putting h = −2 in (1), we get

k = 12 × (−2) − (−2)³ = −24 + 8 = −16

Putting h = 2 in (1), we get

k = 12 × 2 − (2)³ = 24 − 8 = 16

So, the coordinates of the required point are (−2, −16) and (2, 16).

Thus, the points on the curve y = 12x − x³ at which the gradient is zero are (−2, −16) and (2, 16).

The points on the curve y = 12x − x³ at which the gradient is zero are ___(−2, −16) and (2, 16)___.

Page No 15.44:

Question 4:

The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are_______________.

Answer:

Let (h, k) be the point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3.

y = x log_ex

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = x \times \frac{1}{x} + \log_{e} x \times 1$

$\Rightarrow \frac{d y}{d x} = 1 + \log_{e} x$

It is given that, the normal is parallel to the line 2x − 2y = 3.

∴ Slope of normal at (h, k) = Slope of the given line

$\Rightarrow - \frac{1}{{(\frac{d y}{d x})}_{(h, k)}} = - \frac{2}{(- 2)}$

$\Rightarrow - \frac{1}{1 + \log_{e} h} = 1$

$\Rightarrow 1 + \log_{e} h = - 1$

$\Rightarrow \log_{e} h = - 2$

$\Rightarrow h = e^{- 2} = \frac{1}{e^{2}}$

Now, (h, k) lies on the given curve.

$∴ k = h \log_{e} h$ .....(1)

Putting $h = \frac{1}{e^{2}}$ in (1), we get

$k = \frac{1}{e^{2}} \times \log_{e} \frac{1}{e^{2}}$

$\Rightarrow k = \frac{1}{e^{2}} \times \log_{e} e^{- 2}$

$\Rightarrow k = - \frac{2}{e^{2}} (\log a^{b} = b \log a)$

So, the coordinates of the required points are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

Thus, the coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $(\frac{1}{e^{2}}, - \frac{2}{e^{2}})$ .

Page No 15.44:

Question 5:

The coordinates of the point on the curve y = 2 + $\sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are _________________.

Answer:

Let (h, k) be the point on the curve $y = 2 + \sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ .

$y = 2 + \sqrt{4 x + 1}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 0 + \frac{1}{2 \sqrt{4 x + 1}} \times 4$

$\Rightarrow \frac{d y}{d x} = \frac{2}{\sqrt{4 x + 1}}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = \frac{2}{5}$ (Given)

$\Rightarrow \frac{2}{\sqrt{4 h + 1}} = \frac{2}{5}$

$\Rightarrow \sqrt{4 h + 1} = 5$

$\Rightarrow 4 h + 1 = 25$

$\Rightarrow 4 h = 24$

$\Rightarrow h = 6$

Now, (h, k) lies on the given curve $y = 2 + \sqrt{4 x + 1}$ .

$∴ k = 2 + \sqrt{4 h + 1}$ .....(1)

Putting h = 6 in (1), we get

$k = 2 + \sqrt{4 \times 6 + 1}$

$\Rightarrow k = 2 + \sqrt{25}$

$\Rightarrow k = 2 + 5 = 7$

So, the coordinates of the required point are (6, 7).

Thus, the coordinates of the point on the curve $y = 2 + \sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are (6, 7).

The coordinates of the point on the curve y = 2 + $\sqrt{4 x + 1}$ where tangent has slope $\frac{2}{5}$ are ___(6, 7)___.

Page No 15.44:

Question 6:

The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ________________.

Answer:

The given curve is x = 3t² + 1, y = t³ − 1.

When x = 1, we have

3t² + 1 = 1

⇒ 3t² = 0

⇒ t = 0

Now,

x = 3t² + 1

Differentiating both sides with respect to t, we get

$\frac{d x}{d t} = 6 t$        .....(1)

y = t³ − 1

Differentiating both sides with respect to t, we get

$\frac{d y}{d t} = 3 t^{2}$       .....(2)

$∴ \frac{d y}{d x}$ $= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $= \frac{3 t^{2}}{6 t}$ $= \frac{t}{2}$       [Using (1) and (2)]

At t = 0, slope of the tangent = ${(\frac{d y}{d x})}_{t = 0}$ = $\frac{0}{2}$ = 0 $(\frac{d y}{d x} = \frac{t}{2})$

So, the slope of the tangent to the given curve at t = 0 (or x = 1) is 0.

Thus, the slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is 0.

The slope of the tangent to the curve x = 3t² + 1, y = t³ − 1 at x = 1 is ___0___.

Page No 15.44:

Question 7:

The angle of intersection of the curves y = x² and x = y² at (0, 0) is __________________.

Answer:

The given curves are y = x² and x = y².

Let C₁ represents the curve y = x² and C₂ represents the curve x = y².

y = x²

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 x$

∴ Slope of the tangent to the curve C₁ at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)}$ = 2 × 0 = 0

So, the the tangent to the curve y = x² at (0, 0) is parallel to the x-axis.

x = y²

Differentiating both sides with respect to x, we get

$1 = 2 y \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 y}$

∴ Slope of the tangent to the curve C₂ at (0, 0) = ${(\frac{d y}{d x})}_{(0, 0)}$ $= \frac{1}{2 \times 0} = \frac{1}{0} = \infty$

So, the tangent to the curve x = y² at (0, 0) is parallel to the y-axis.

Now, the tangent to the curve y = x² at (0, 0) is parallel to the x-axis and tangent to the curve x = y² at (0, 0) is parallel to the y-axis. So, the angle between the tangents to the given curves at (0, 0) is $\frac{π}{2}$ .

Thus, the angle of intersection of the given curves = x² and x = y² at (0, 0) is $\frac{π}{2}$ .

The angle of intersection of the curves y = x² and x = y² at (0, 0) is $\frac{π}{2}$ .

Page No 15.44:

Question 8:

The slope of the tangent to the curve y = be^−x/a where it crosses y-axis is ______________.

Answer:

The equation of given curve is

$y = b e^{- \frac{x}{a}}$ .....(1)

It crosses the y-axis at the point, where x = 0.

Putting x = 0 in (1), we get

$y = b e^{- \frac{0}{a}}$

$\Rightarrow y = b e^{0}$

$\Rightarrow y = b$

So, the coordinates of the point where the given curve crosses y-axis is (0, b).

$y = b e^{- \frac{x}{a}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = b e^{- \frac{x}{a}} \times (- \frac{1}{a})$

$\Rightarrow \frac{d y}{d x} = - \frac{b}{a} e^{- \frac{x}{a}}$

∴ Slope of the tangent at (0, b) $= {(\frac{d y}{d x})}_{(0, b)}$ $= - \frac{b}{a} e^{- \frac{0}{a}}$ $= - \frac{b}{a} e^{0}$ $= - \frac{b}{a}$

Thus, the slope of tangent to the given curve $y = b e^{- \frac{x}{a}}$ where it crosses y-axis i.e. (0, b) is $- \frac{b}{a}$ .

The slope of the tangent to the curve $y = b e^{- \frac{x}{a}}$ where it crosses y-axis is $- \frac{b}{a}$ .

Page No 15.44:

Question 9:

The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point __________________.

Answer:

The given curve is

y = e^2x

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 2 e^{2 x}$

∴ Slope of tangent at (0, 1) = ${(\frac{d y}{d x})}_{(0, 1)} = 2 e^{2 \times 0} = 2 \times 1 = 2$         .....(1)

So, the equation of tangent at (0, 1) is

$y - 1 = {(\frac{d y}{d x})}_{(0, 1)} (x - 0)$

$\Rightarrow y - 1 = 2 x$               [Using (1)]

$\Rightarrow 2 x - y + 1 = 0$    .....(2)

This tangent cuts the x-axis where y = 0.

Putting y = 0 in (2), we get

$2 x - 0 + 1 = 0$

$\Rightarrow x = - \frac{1}{2}$

So, the coordinates of the required point are $(- \frac{1}{2}, 0)$ .

Thus, the tangent to the given curve y = e^2x at (0, 1) cuts the x-axis at $(- \frac{1}{2}, 0)$ .

The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point $(- \frac{1}{2}, 0)$ .

Page No 15.44:

Question 10:

The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to _________________.

Answer:

The equation of given curve is

y³ − xy − 8 = 0

Differentiating both sides with respect to x, we get

$3 y^{2} \frac{d y}{d x} - (x \frac{d y}{d x} + y) - 0 = 0$

$\Rightarrow (3 y^{2} - x) \frac{d y}{d x} = y$

$\Rightarrow \frac{d y}{d x} = \frac{y}{3 y^{2} - x}$

Now, slope of tangent at (0, 2) = ${(\frac{d y}{d x})}_{(0, 2)} = \frac{2}{3 \times {(2)}^{2} - 0} = \frac{2}{12} = \frac{1}{6}$ .....(1)

∴ Slope of the normal at (0, 2)

$= - \frac{1}{{(\frac{d y}{d x})}_{(0, 2)}}$

$= - \frac{1}{(\frac{1}{6})}$ [From (1)]

$= - 6$

Thus, the slope of normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is −6.

The slope of the normal to the curve y³ − xy − 8 = 0 at the point (0, 2) is equal to ___−6___.

Page No 15.44:

Question 11:

If the normal to the curve y² = 5x -1, at the point(1, -2) is of the form ax - 5y + b = 0, then a + b = _________________.

Answer:

The equation of given curve is

y² = 5x −1

Differentiating both sides with respect to x, we get

$2 y \frac{d y}{d x} = 5$

$\Rightarrow \frac{d y}{d x} = \frac{5}{2 y}$

∴ Slope of normal at (1, −2)

$= - \frac{1}{{(\frac{d y}{d x})}_{(1, - 2)}}$

$= - \frac{1}{\frac{5}{2 \times (- 2)}}$

$= \frac{4}{5}$

So, the equation of normal at (1, −2) is

$y - (- 2) = - \frac{1}{{(\frac{d y}{d x})}_{(1, - 2)}} (x - 1)$

$\Rightarrow y - (- 2) = \frac{4}{5} (x - 1)$

$\Rightarrow 5 y + 10 = 4 x - 4$

$\Rightarrow 4 x - 5 y - 14 = 0$

Comparing with the given equation of normal ax − 5y + b = 0, we get

a = 4 and b = −14

∴ a + b = 4 + (−14) = −10

Thus, the value of a + b is −10.

If the normal to the curve y² = 5x −1, at the point(1, −2) is of the form ax − 5y + b = 0, then a + b = ___−10___.

Page No 15.44:

Question 12:

If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is _____________.

Answer:

The equation of given curve is xy = 1.

Let (x₁, y₁) be the point of contact of normal with the curve.

$∴ x_{1} y_{1} = 1$ .....(1)

Now,

xy = 1

Differentiating both sides with respect to x, we get

$x \frac{d y}{d x} + y = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{y}{x}$

∴ Slope of the normal at (x₁, y₁)

$= - \frac{1}{{(\frac{d y}{d x})}_{(x_{1}, y_{1})}}$

$= - \frac{1}{(- \frac{y_{1}}{x_{1}})}$

$= \frac{x_{1}}{y_{1}}$

$= x_{1}^{2}$ [Using (1)]

The given equation of normal to the curve is ax + by + c = 0.

Slope of this line = $- \frac{a}{b}$

$∴ x_{1}^{2} = - \frac{a}{b}$

$\Rightarrow \frac{a}{b} = - x_{1}^{2}$

Now, $x_{1}^{2}$ is always positive.

∴ $\frac{a}{b}$ < 0 ∀ x ∈ R

Thus, the set of values of $\frac{a}{b}$ is the set of all negative real numbers i.e. (−∞, 0).

If the line ax + by + c = 0 is normal to the curve xy = 1, then the set of values of $\frac{a}{b}$ , is __(−∞, 0)__.

Page No 15.44:

Question 13:

If the normal to the curve y = f(x) at (3, 4) makes an angle $\frac{3 π}{4}$ with positive x-axis, then f '(3) is equal to ________________.

Answer:

It is given that, the normal to the curve y = f(x) at (3,4) makes an angle $\frac{3 π}{4}$ with positive x-axis.

∴ Slope of the normal = $\tan \frac{3 π}{4} = \tan (π - \frac{π}{4}) = - \tan \frac{π}{4}$ = −1 .....(1)

Now, y = f(x)

$\Rightarrow \frac{d y}{d x} = f' (x)$

∴ Slope of the normal at (3, 4) $= - \frac{1}{{(\frac{d y}{d x})}_{(3, 4)}} = - \frac{1}{f' (3)}$ .....(2)

From (1) and (2), we have

$- \frac{1}{f' (3)} = - 1$

$\Rightarrow f' (3) = 1$

Thus, the value of $f' (3)$ is 1.

If the normal to the curve y = f(x) at (3, 4) makes an angle $\frac{3 π}{4}$ with positive x-axis, then f '(3) is equal to ___1___.

Page No 15.44:

Question 14:

The equation of the tangent to the curve y = x + $\frac{4}{x^{2}}$ , that is parallel to x-axis, is ________________.

Answer:

The equation of the given curve is $y = x + \frac{4}{x^{2}}$ .

Let (h, k) be the point of contact of tangent with the curve.

$∴ k = h + \frac{4}{h^{2}}$ .....(1)

$y = x + \frac{4}{x^{2}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 1 - \frac{8}{x^{3}}$

∴ Slope of tangent at (h, k) $= {(\frac{d y}{d x})}_{(h, k)} = 1 - \frac{8}{h^{3}}$

It is given that, the tangent to the given curve is parallel to the x-axis.

∴ Slope of tangent = 0

$\Rightarrow {(\frac{d y}{d x})}_{(h, k)} = 0$

$\Rightarrow 1 - \frac{8}{h^{3}} = 0$

$\Rightarrow h^{3} = 8$

$\Rightarrow h = 2$

Putting h = 2 in (1), we get

$k = 2 + \frac{4}{{(2)}^{2}} = 2 + 1 = 3$

Thus, the point of contact of tangent with the curve is (2, 3).

So, the equation of the tangent at (2, 3) is

$y - 3 = {(\frac{d y}{d x})}_{(2, 3)} (x - 2)$

$\Rightarrow y - 3 = 0 [{(\frac{d y}{d x})}_{(2, 3)} = 0]$

$\Rightarrow y = 3$

Thus, the equation of tangent to the given curve $y = x + \frac{4}{x^{2}}$ , that is parallel to x-axis, is y = 3.

The equation of the tangent to the curve y = x + $\frac{4}{x^{2}}$ , that is parallel to x-axis, is ___y = 3___.

Page No 15.44:

Question 16:

If slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the point is _________________.

Answer:

The equation of given curve is y = x³.

Let (h, k) be the point on the curve at which slope of tangent is equal to ordinate of the point.

∴ k = h³        .....(1)

Now,

y = x³

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 3 x^{2}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 3 h^{2}$

It is given that,

Slope of tangent at (h, k) = Ordinate of (h, k)

⇒ 3h² = k       .....(2)

From (1) and (2), we get

$3 h^{2} = h^{3}$

$\Rightarrow h^{3} - 3 h^{2} = 0$

$\Rightarrow h^{2} (h - 3) = 0$

⇒ h = 0 or h − 3 = 0

⇒ h = 0 or h = 3

When h = 0,

k = (0)³ = 0           [From (1)]

When h = 3,

k = (3)³ = 27           [From (1)]

So, the coordinates of the required points are (0, 0) and (3, 27).

Thus, if slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the points are (0, 0) and (3, 27).

If slope of tangent to curve y = x³ at a point is equal to ordinate of point, then the point is ___(0, 0) and (3, 27)___.

Page No 15.44:

Question 17:

The slope of the normal to the curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is _________________.

Answer:

The equation of given curve is

x² + y² − 2x + 4y − 5 = 0

Differentiating both sides with respect to x, we get

$2 x + 2 y \frac{d y}{d x} - 2 + 4 \frac{d y}{d x} - 0 = 0$

$\Rightarrow (2 y + 4) \frac{d y}{d x} = 2 - 2 x$

$\Rightarrow \frac{d y}{d x} = \frac{1 - x}{y + 2}$

∴ Slope of normal at (2, 1)

$= - \frac{1}{{(\frac{d y}{d x})}_{(2, 1)}}$

$= - \frac{1}{(\frac{1 - 2}{1 + 2})}$

$= - \frac{1}{(- \frac{1}{3})}$

= 3

Thus, the slope of normal to the given curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is 3.

The slope of the normal to the curve x² + y² − 2x + 4y − 5 = 0 at (2, 1) is ___3___.

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Question 18:

The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is ____________________.

Answer:

Let (h, k) be the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis.

y² = x

Differentiating both sides with respect to x, we get

$2 y \frac{d y}{d x} = 1$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 y}$

∴ Slope of tangent at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = \frac{1}{2 k}$      .....(1)

It is given that, the tangent makes an angle of 45º with x-axis.

∴ Slope of the tangent = tan45º = 1        .....(2)

From (1) and (2), we have

$\frac{1}{2 k} = 1$

$\Rightarrow k = \frac{1}{2}$

Now, (h, k) lies on the curve.

∴ k² = h     .....(3)

Putting $k = \frac{1}{2}$ in (3), we get

$h = {(\frac{1}{2})}^{2} = \frac{1}{4}$

So, the coordinates of the required point are $(\frac{1}{4}, \frac{1}{2})$ .

Thus, the point on the curve y² = x, the tangent at which makes an angle of 45º with x-axis is $(\frac{1}{4}, \frac{1}{2})$ .

The point on the curve y² = x, the tangent at which makes an angle of 45^o with x-axis is $(\frac{1}{4}, \frac{1}{2})$ .

Page No 15.44:

Question 19:

The curve y = 4x² + 2x − 8 and y = x³ − x + 13 touch each other at the point _________________.

Answer:

Disclaimer: The solution has been provided for the following question.

The curve y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point _________________.

Solution:

Let the given curves each other at the point (h, k).

∴ k = 4h² + 2h − 8 .....(1)

k = h³ − h + 10 .....(2)

Consider the first curve C₁ ≡ y = 4x² + 2x − 8.

y = 4x² + 2x − 8

$\Rightarrow \frac{d y}{d x} = 8 x + 2$

∴ Slope of tangent to the curve C₁ at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 8 h + 2$

Consider the second curve C₂ ≡ y = x³ − x + 10.

y = x³ − x + 10

$\Rightarrow \frac{d y}{d x} = 3 x^{2} - 1$

∴ Slope of tangent to the curve C₂ at (h, k) = ${(\frac{d y}{d x})}_{(h, k)} = 3 h^{2} - 1$

It is given that, the two curves touch each other at (h, k).

∴ Slope of tangent to the curve C₁ at (h, k) = Slope of tangent to the curve C₂ at (h, k)

$\Rightarrow 8 h + 2 = 3 h^{2} - 1$

$\Rightarrow 3 h^{2} - 8 h - 3 = 0$

$\Rightarrow 3 h^{2} - 9 h + h - 3 = 0$

$\Rightarrow 3 h (h - 3) + 1 (h - 3) = 0$

$\Rightarrow (3 h + 1) (h - 3) = 0$

⇒ 3h + 1 = 0 or h − 3 = 0

⇒ h = $- \frac{1}{3}$ or h = 3

Putting h = $- \frac{1}{3}$ in (1), we get

$k = 4 {(- \frac{1}{3})}^{2} + 2 \times (- \frac{1}{3}) - 8 = \frac{4}{9} - \frac{2}{3} - 8 = - \frac{74}{9}$

Putting h = $- \frac{1}{3}$ in (2), we get

$k = {(- \frac{1}{3})}^{3} - (- \frac{1}{3}) + 10 = - \frac{1}{27} + \frac{1}{3} + 10 = \frac{278}{27}$

Here, the values of k are not same for the two curves for h = $- \frac{1}{3}$ .

So, the two curves do not touch each other when h = $- \frac{1}{3}$ . However, the tangents are parallel to each other at h = $- \frac{1}{3}$ .

Putting h = 3 in (1), we get

$k = 4 \times {(3)}^{2} + 2 \times 3 - 8 = 36 + 6 - 8 = 34$

Putting h = 3 in (2), we get

$k = {(3)}^{3} - 3 + 10 = 27 - 3 + 10 = 34$

Here, the values of k are same for the two curves when h = 3.

So, the two curves touch each other at (3, 34).

Thus, the curves y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point (3, 34).

The curve y = 4x² + 2x − 8 and y = x³ − x + 10 touch each other at the point ___(3, 34)___.

Page No 15.45:

Question 1:

Find the point on the curve y = x² − 2x + 3, where the tangent is parallel to x-axis.

Answer:

The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
Here,
$Since, the point lies on the curve . Hence, y_{1} = {x_{1}}^{2} - 2 x_{1} + 3 . . . (1) Now, y = x^{2} - 2 x + 3 \frac{d y}{d x} = 2 x - 2 Slope of the tangent at (x, y) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 2 x_{1} - 2 Given: Slope of the tangent at (x_{1}, y_{1}) =Slope of the x -axis = 2 x_{1} - 2 = 0 \Rightarrow x_{1} = 1 and y_{1} = 1 - 2 + 3 = 2 [From (1)] ∴ Required point= (x_{1}, y_{1}) = (1, 2)$

Page No 15.45:

Question 2:

Find the slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at t = 2.

Answer:

$Here, x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} Now, Slope of the tangent= {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$

Page No 15.45:

Question 3:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to x-axis, then write the value of $\frac{d y}{d x}$ .

Answer:

The slope of the x-axis is 0.
Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the x-axis.
∴ Slope of the tangent $(\frac{d y}{d x})$ = Slope of the x-axis = 0

Page No 15.45:

Question 4:

Write the value of $\frac{d y}{d x}$ , if the normal to the curve y = f(x) at (x, y) is parallel to y-axis.

Answer:

The slope of the y-axis is $\infty$ .
Also, the normal at (x, y) on the curve y = f(x) is parallel to the y-axis.
∴ Slope of the normal = Slope of the y-axis = $\infty$
$⇒ \frac{d y}{d x} =Slope of the tangent= \frac{- 1}{Slope of the normal} = \frac{- 1}{\infty} =0$

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Question 5:

If the tangent to a curve at a point (x, y) is equally inclined to the coordinates axes then write the value of $\frac{d y}{d x}$ .

Answer:

Because the tangent to the curve at (x, y) is equally inclined to the coordinate axes, the angle made by the tangent with the axes can be $\pm 45 °$ .
$∴ \frac{d y}{d x} =Slope of the tangent=tan (\pm 45) =±1$

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Question 6:

If the tangent line at a point (x, y) on the curve y = f(x) is parallel to y-axis, find the value of $\frac{d x}{d y}$ .

Answer:

Slope of the y-axis is $\infty$ .
Also, the tangent at (x, y) on the curve y = f(x) is parallel to the y-axis,
∴ Slope of the tangent, $\frac{d y}{d x}$ = Slope of the y-axis = $\infty$
$\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})} = \frac{1}{\infty} = 0$

Page No 15.45:

Question 7:

Find the slope of the normal at the point 't' on the curve $x = \frac{1}{t}, y = t$ .

Answer:

$Here, x = \frac{1}{t} and y = t \frac{d x}{d t} = \frac{- 1}{t^{2}} and \frac{d y}{d t} = 1 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{1}{(\frac{- 1}{t^{2}})} = - t^{2} Now, Slope of the tangent= {(\frac{d y}{d x})}_{} = - t^{2} Slope of the normal= \frac{- 1}{Slope of the tangent} = \frac{- 1}{- t^{2}} = \frac{1}{t^{2}}$

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Question 8:

Write the coordinates of the point on the curve y² = x where the tangent line makes an angle $\frac{π}{4}$ with x-axis.

Answer:

Let the required point be (x₁, y₁).
The tangent makes an angle of 45^o with the x-axis.
∴ Slope of the tangent = tan 45^o = 1

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = x_{1} Now, y^{2} = x \Rightarrow 2 y \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} Slope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1}{2 y_{1}} Given: \frac{1}{2 y_{1}} = 1 \Rightarrow 2 y_{1} = 1 \Rightarrow y_{1} = \frac{1}{2} Now, x_{1} = {y_{1}}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} ∴ (x_{1}, y_{1}) = (\frac{1}{4}, \frac{1}{2})$

Page No 15.45:

Question 9:

Write the angle made by the tangent to the curve x = e^t cos t, y = e^t sin t at $t = \frac{π}{4}$ with the x-axis.

Answer:

$Here, x = e^{t} \cos t and y = e^{t} \sin t \frac{d x}{d t} = e^{t} c os t - e^{t} \sin t and \frac{d y}{d t} = e^{t} \sin t + e^{t} \cos t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{e^{t} \sin t + e^{t} \cos t}{e^{t} c os t - e^{t} \sin t} = \frac{\sin t + \cos t}{\cos t - \sin t} Now, Slope of the tangent = {(\frac{d y}{d x})}_{t = \frac{π}{4}} = \frac{\sin \frac{π}{4} + \cos \frac{π}{4}}{\cos \frac{π}{4} - \sin \frac{π}{4}} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = \frac{\frac{2}{\sqrt{2}}}{0} =∞ Let θ be the angle made by the tangent with the x -axis. ∴ tan θ =∞ \Rightarrow θ = \frac{π}{2}$

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Question 10:

Write the equation of the normal to the curve y = x + sin x cos x at $x = \frac{π}{2}$ .

Answer:

$Here, y = x + \sin x \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = 1 + \cos^{2} x - \sin^{2} x Now, Slope of the tangent = {(\frac{d y}{d x})}_{x = \frac{π}{2}} {= 1+cos}^{2} \frac{π}{2} {-sin}^{2} \frac{π}{2} = 1-1=0 When x = \frac{π}{2}, y = \frac{π}{2} +sin \frac{π}{2} cos \frac{π}{2} = \frac{π}{2} ∴ (x_{1}, y_{1}) = (\frac{π}{2}, \frac{π}{2}) Equation of the normal = y - y_{1} = \frac{- 1}{Slope of the tangent} (x - x_{1}) \Rightarrow y - \frac{π}{2} = \frac{- 1}{0} (x - \frac{π}{2}) \Rightarrow x = \frac{π}{2} \Rightarrow 2 x = π$

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Question 11:

Find the coordinates of the point on the curve y² = 3 − 4x where tangent is parallel to the line 2x + y − 2 = 0.

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = $-$ 2

$Since, the point lies on the curve . Hence, {y_{1}}^{2} = 3 - 4 x_{1} . . . (1) Now, y^{2} = 3 - 4 x \Rightarrow 2 y \frac{d y}{d x} = - 4 ∴ \frac{d y}{d x} = \frac{- 4}{2 y} = \frac{- 2}{y} S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- 2}{y_{1}} Given: Slope of the tangent = Slope of the line \Rightarrow \frac{- 2}{y_{1}} = - 2 \Rightarrow y_{1} = 1 From (1), we get 1 = 3 - 4 x_{1} \Rightarrow - 2 = - 4 x_{1} \Rightarrow x_{1} = \frac{1}{2} ∴ (x_{1}, y_{1}) = (\frac{1}{2}, 1)$

Page No 15.45:

Question 12:

Write the equation on the tangent to the curve y = x² − x + 2 at the point where it crosses the y-axis.

Answer:

When the curve crosses the y-axis, the point on the curve is of the form (0, y).
Here,
$y = x^{2} - x + 2 \Rightarrow y = 0 - 0 + 2 = 2 So, the point where the curve crosses the y -axis is (0, 2). Now, y = x^{2} - x + 2 \Rightarrow \frac{d y}{d x} = 2 x - 1 Slope of the tangent, m = {(\frac{d y}{d x})}_{(0, 2)} =2 (0) -1=-1 ∴ (x_{1}, y_{1}) = (0, 2) and Equation of tangent = y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 = - 1 (x - 0) \Rightarrow y - 2 = - x \Rightarrow x + y - 2 = 0$

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Question 13:

Write the angle between the curves y² = 4x and x² = 2y − 3 at the point (1, 2).

Answer:

$Given : y^{2} = 4 x . . . (1) x^{2} = 2 y - 3 . . . (2) On d ifferentiating (1) w.r.t. x, we get 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(1, 2)} = \frac{2}{2} = 1 On d ifferentiating (2) w.r.t. x, we get 2 x = 2 \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = x \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(1, 2)} = 1 Thus, we get \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| \Rightarrow \tan θ = |\frac{1 - 1}{1 + 1}| \Rightarrow \tan θ = 0 \Rightarrow θ = 0^{o}$

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Question 14:

Write the angle between the curves y = e^−x and y = e^x at their point of intersections.

Answer:

$Given : y = e^{- x} . . . (1) y = e^{x} . . . (2) On s ubstituting the value of y in (1), we get e^{- x} = e^{x} \Rightarrow x = 0 and y = 1 [From (2)] On d ifferentiating (1) w.r.t. x,we get \frac{d y}{d x} = - e^{- x} \Rightarrow m_{1} = {(\frac{d y}{d x})}_{(0, 1)} = - 1 On d ifferentiating (2) w.r.t . x, we get \frac{d y}{d x} = e^{x} \Rightarrow m_{2} = {(\frac{d y}{d x})}_{(0, 1)} = 1 ∵ m_{1} \times m_{2} = - 1 Since the multiplication of the slopes is - 1 so the slopes are perpendicular to each other . ∴ Required angle = \frac{π}{2}$

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Question 15:

Write the slope of the normal to the curve $y = \frac{1}{x}$ at the point $(3, \frac{1}{3})$ .

Answer:

$Given : y = \frac{1}{x} On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = \frac{- 1}{x^{2}} Now, Slope of the tangent = {(\frac{d y}{d x})}_{(3, \frac{1}{3})} = \frac{- 1}{9} Slope of the normal = \frac{- 1}{Slope of tangent} = \frac{- 1}{(\frac{- 1}{9})} = 9$

Page No 15.45:

Question 16:

Write the coordinates of the point at which the tangent to the curve y = 2x² − x + 1 is parallel to the line y = 3x + 9.

Answer:

Let (x₁, y₁) be the required point.
Slope of the given line = 3
$Since, the point lies on the curve . Hence, y_{1} = 2 {x_{1}}^{2} - x_{1} + 1 . . . (1) Now, y = 2 x^{2} - x + 1 ∴ \frac{d y}{d x} = 4 x - 1 Now, S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 4 x_{1} - 1 Given: Slope of the tangent = Slope of line ∴ 4 x_{1} - 1 = 3 \Rightarrow x_{1} = 1 From (1), we get y_{1} = 2 - 1 + 1 = 2 ∴ (x_{1}, y_{1}) = (1, 2)$

Page No 15.46:

Question 17:

Write the equation of the normal to the curve y = cos x at (0, 1).

Answer:

$Given : y = \cos x On d ifferentiating both sides w.r.t. x, we get \frac{d y}{d x} = - \sin x Now, Slope of the tangent= {(\frac{d y}{d x})}_{(0, 1)} =-sin 0=0 and Equation of the normal = y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = \frac{- 1}{0} (x - 0) \Rightarrow x = 0$

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Question 18:

Write the equation of the tangent drawn to the curve $y = \sin x$ at the point (0,0).

Answer:

We have,

$y = \sin x$

$\Rightarrow \frac{d y}{d x} = \cos x$

Slope at (0, 0) = m = ${[\frac{d y}{d x}]}_{x = 0} = \cos 0 = 1$

So, the equation of the tangent at (0,0) is given by,

y = mx

Putting m = 1, we get

The equation of the tangent is y = x.

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