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Question 1:

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

(i)
(ii)
(iii) y = x3x 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)

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.

Question 3:

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

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).

Let (x1, y1) be the required point.

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.

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

Question 6:

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

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

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.

Let the required point be (x1, y1).
Slope of the tangent at this point = tan 45$°$ = 1
Given:

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.

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

Question 9:

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

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

Question 10:

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

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

Question 11:

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

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 $±$45$°$.
∴ Slope of the tangent = tan ($±$45) = $±$ 1  ...(1)

Question 12:

At what points on the curve y = 2x2x + 1 is the tangent parallel to the line y = 3x + 4?

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

Question 13:

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

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

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.

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

Question 15:

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

Let (x1, y1) represent the required points.
The slope of the x-axis is 0.
Here,

Question 16:

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

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

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.

(i) The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.

(ii) The slope of the y-axis is $\infty$.
Now, let (x1, y1) be the required point.

Question 18:

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

Let (x1, y1) be the required point.

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.

(i) The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.

(ii) The slope of the y-axis is $\infty$.
Let (x1, y1) be the required point.
Given:

Question 20:

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

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

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.

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

Question 1:

Find the equation of the tangent to the curve $\sqrt{x}+\sqrt{y}=a$, at the point .

Question 2:

Find the equation of the normal to y = 2x3x2 + 3 at (1, 4).

Question 3:

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

(i) x4 − bx3 + 13x2 − 10x + 5 at (0, 5)                  [NCERT]
(ii) x4 − 6x3 + 13x2 − 10x + 5 at x = 1                  [NCERT, CBSE 2011]
(iii) x2 at (0, 0)                  [NCERT]
(iv) y = 2x2 − 3x − 1 at (1, −2)
(v)
(vi) y = x2 + 4x + 1 at x = 3                       [CBSE 2004]
(vii)
(viii)
(ix) y2 = 4ax at $\left(\frac{a}{{m}^{2}},\frac{2a}{m}\right)$
(x)
(ix) xy = c2 at $\left(ct,\frac{c}{t}\right)$
(xii)
(xiii)              [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)           [CBSE 2014]

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(xvii) Equation of tangent:

(xviii)

(xix)

Question 4:

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

Question 5:

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

(i) x = θ + sinθy = 1 + cosθ at θ = $\frac{\mathrm{\pi }}{2}$
(ii)
(iii) x = at2y = 2at at t = 1
(iv) x = asecty = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi)
x = 3cosθ − cos3θy = 3sinθ − sin3θ                    [NCERT EXEMPLAR]

(i)

(ii) Equation of tangent:

(iii)

Equation of normal:

(iv)

(v)

(vi)

Question 6:

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

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

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

Question 7:

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

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.

The slope of the given line y = 4x − 5 is 4

Question 9:

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

Let (x0, y0) be the point of intersection of both the curve and the tangent.

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.

Let (x1, y1) be a point on the curve where we need to find the normal.
Slope of the given line = $\frac{-1}{14}$

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.

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 = $\frac{-1}{\left(\frac{-1}{9}\right)}=9$

Question 12:

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

Slope of the given line is 1

Question 13:

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

(i) Slope of the given line is 2

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

Question 14:

Find the equations of all lines having slope 2 and that are tangent to the curve .

Slope of given tangent = 2

So, there does not exist any such tangent.

Question 15:

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

Slope of the given tangent is 0.

Question 16:

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

Slope of the given line is 2

Question 17:

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

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

Question 18:

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

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

Question 19:

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

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.

Slope of x - axis is 0
Let (x1, y1) be the required point.

Question 21:

Find the equation of  the tangents to the curve 3x2y2 = 8, which passes through the point (4/3, 0).

We have,

3x2 – y2 = 8                   ...(i)

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

$6x-2y\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}⇒2y\frac{dy}{dx}=6x\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{6x}{2y}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{3x}{y}$

Let tangent at (h, k) pass through .

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

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

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

Since, the tangent passess through .

Using (ii), we get

$12-{k}^{2}=8\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=4\phantom{\rule{0ex}{0ex}}⇒k=±2$

So, the points on curve (i) at which tangents pass through  are .

Now, from (iii), the equation of tangents are

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 yx2                      [NCERT EXEMPLAR]

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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 3x2yy3 = 2
(iii) x2 + 4y2 = 8 and x2 − 2y2 = 4

So, the given curves intersect orthogonally.

So, the given curves intersect orthogonally.

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 2x2y2 = 10 at

Hence,  the given curves intersect orthogonally at the given point.

Hence,  the given curves intersect orthogonally at the given point.

Hence,  the given curves intersect orthogonally at the given point.

Question 4:

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

Question 5:

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

Question 6:

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

Question 7:

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

Question 8:

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

(i)
(ii)

(ii) The condition for the curves to intersect orthogonally is given below:

Question 9:

Show that the curves intersect at right angles.

Question 10:

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

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) xy = 0

(c) x + y = 0

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

(d) 2x = π

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) 2xy = 4
(d) 2x + y − 4 = 0

(a) x − 2y = 2

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)

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

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

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)

(c) (0, 0)

Let the required point be (x1, y1).

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)

(b) (1, 0)

y = x
$⇒\frac{dy}{dx}=1$

The tangent is perpendicular to this line.
∴Slope of the tangent =

Now,

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)

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

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

Question 8:

The point at the curve y = 12xx2 where the slope of the tangent is zero will be

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

(d) none of these

Question 9:

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

(a)
(b)
(c) 90°
(d) 45°

(b)

Question 10:

The equation of the normal to the curve 3x2y2 = 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

(c) x + 3y ± 8 = 0

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

Question 11:

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

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

(b) x + y − 1 = 0 = xy − 2

Let the tangent meet the x-axis at point (x, 0).
Now,

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

(b) 6/7

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)

(d) (1, 2), (1, −2)

Let (x1, y1) be the required point.

Question 14:

The angle of intersection of the curves xy = a2 and x2y2 = 2a2 is

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

(c) 90°

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

(c) 6

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

(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,

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) ∞

(b) 0

Question 18:

The curves y = aex and y = bex cut orthogonally, if

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

(c) ab = 1

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

(c) x = y

Question 20:

If the curves y = 2 ex and y = aex intersect orthogonally, then a =

(a) 1/2
(b) −1/2
(c) 2
(d) 2e2

(a) 1/2

Question 21:

The point on the curve y = 6xx2 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)

(b) (3, 9)

Let (x1, y1) be the point where the given curve intersect the given line at the given angle.

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

(c) π/2

Question 23:

The angle of intersection of the curves y = 2 sin2 x and y = cos 2 x at $x=\frac{\mathrm{\pi }}{6}$ is

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

(c) π/3

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

(c) makes an acute angle with x-axis

We have, y = 2x7 + 3x + 5

Question 25:

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

(a)
(b)
(c)
(d) none of these

(a) and (c)

Let (x1, y1) be the required point.

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) $-\frac{6}{7}$

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

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

Thus, the correct option is (b).

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) $\frac{1}{2}$

(a) 1
Let (x1, y1) be the required point.
The slope of the given line is m.
We have

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.

Question 28:

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

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

(b) xy = 0

Question 29:

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

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

Disclaimer: None of the given options is correct.

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)         (c) 2                (d) $\frac{1}{2}$

Let (hk) 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

$3\frac{dy}{dx}=6-15{x}^{2}$

$⇒\frac{dy}{dx}=2-5{x}^{2}$

∴ Slope of tangent to the given curve at (hk) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=2-5{h}^{2}$

Slope of normal to the given curve at (hk) = $-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}}=-\frac{1}{\left(2-5{h}^{2}\right)}$

Equation of the normal to the given curve at (hk) is

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

This equation passes through the origin.

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

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

From (1) and (2), we have

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

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

$⇒30{h}^{2}-25{h}^{4}-12+10{h}^{2}=3$

$⇒25{h}^{4}-40{h}^{2}+15=0$

$⇒5{h}^{4}-8{h}^{2}+3=0$

$⇒5{h}^{4}-5{h}^{2}-3{h}^{2}+3=0$

$⇒5{h}^{2}\left({h}^{2}-1\right)-3\left({h}^{2}-1\right)=0$

$⇒\left({h}^{2}-1\right)\left(5{h}^{2}-3\right)=0$

h2 − 1 = 0 or 5h2 − 3 = 0

h2 = 1 or ${h}^{2}=\frac{3}{5}$

h = ±1 or $h=±\sqrt{\frac{3}{5}}$

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).

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  $\frac{\mathrm{\pi }}{3}$                       (d) cut at an angle $\frac{\mathrm{\pi }}{4}$

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 C≡ x3 − 3xy2 + 2 = 0.

x3 − 3xy2 + 2 = 0

Differentiating both sides with respect to x, we get

$⇒3{x}^{2}-6xy\frac{dy}{dx}-3{y}^{2}=0$

$⇒\frac{dy}{dx}=\frac{3{x}^{2}-3{y}^{2}}{6xy}$

∴ Slope of tangent to the first curve at (x1, y1) = ${\left(\frac{dy}{dx}\right)}_{{C}_{1}}=\frac{{x}_{1}^{2}-{y}_{1}^{2}}{2{x}_{1}{y}_{1}}$

Now, consider the second curve C≡ 3x2 y3 = 2.

3x2 y3 = 2

Differentiating both sides with respect to x, we get

$3\left({x}^{2}×\frac{dy}{dx}+y×2x\right)-3{y}^{2}\frac{dy}{dx}=0$

$⇒\left(3{x}^{2}-3{y}^{2}\right)\frac{dy}{dx}=-6xy$

$⇒\frac{dy}{dx}=-\frac{6xy}{3{x}^{2}-3{y}^{2}}$

∴ Slope of tangent to the second curve at (x1, y1) = ${\left(\frac{dy}{dx}\right)}_{{C}_{2}}=-\frac{2{x}_{1}{y}_{1}}{{x}_{1}^{2}-{y}_{1}^{2}}$

Now, ${\left(\frac{dy}{dx}\right)}_{{C}_{1}}×{\left(\frac{dy}{dx}\right)}_{{C}_{2}}=\left(\frac{{x}_{1}^{2}-{y}_{1}^{2}}{2{x}_{1}{y}_{1}}\right)×\left(-\frac{2{x}_{1}{y}_{1}}{{x}_{1}^{2}-{y}_{1}^{2}}\right)=-1$

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).

Question 32:

The tangent to the curve x = et cost, y = et sin t at t$\frac{\mathrm{\pi }}{4}$ makes with x-axis an angle
(a) 0              (b)

The given curve is x = et costy = et sint.

x = et cost

Differentiating both sides with respect to t, we get

$\frac{dx}{dt}={e}^{t}×\left(-\mathrm{sin}t\right)+\mathrm{cos}t×{e}^{t}$

$⇒\frac{dx}{dt}={e}^{t}\left(\mathrm{cos}t-\mathrm{sin}t\right)$

y = et sint

Differentiating both sides with respect to t, we get

$\frac{dy}{dt}={e}^{t}×\mathrm{cos}t+\mathrm{sin}t×{e}^{t}$

$⇒\frac{dy}{dt}={e}^{t}\left(\mathrm{cos}t+\mathrm{sin}t\right)$

Now,

Slope of tangent to the given curve $=\frac{dy}{dx}$$=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$=\frac{{e}^{t}\left(\mathrm{cos}t+\mathrm{sin}t\right)}{{e}^{t}\left(\mathrm{cos}t-\mathrm{sin}t\right)}$$=\frac{\mathrm{cos}t+\mathrm{sin}t}{\mathrm{cos}t-\mathrm{sin}t}$

∴ Slope of tangent to the given curve at $t=\frac{\mathrm{\pi }}{4}$ $={\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{4}}$ $=\frac{\mathrm{cos}\frac{\mathrm{\pi }}{4}+\mathrm{sin}\frac{\mathrm{\pi }}{4}}{\mathrm{cos}\frac{\mathrm{\pi }}{4}-\mathrm{sin}\frac{\mathrm{\pi }}{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 = $\frac{\mathrm{\pi }}{4}$ with the x-axis be θ.

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

$⇒\mathrm{tan}\theta =\mathrm{tan}\frac{\mathrm{\pi }}{2}$

$⇒\theta =\frac{\mathrm{\pi }}{2}$

Thus, the tangent to the curve x = et costy = et sint at t = $\frac{\mathrm{\pi }}{4}$ makes with x-axis an angle $\frac{\mathrm{\pi }}{2}$.

Hence, the correct answer is option (d).

Question 33:

The tangent to the curve y = e2x at the point (0,1) meets x-axis at :
(a) (0, 1)              (b) $\left(-\frac{1}{2},0\right)$           (c) (2, 0)            (d) (0, 2)

The given curve is y = e2x.

y = e2x

$⇒\frac{dy}{dx}=2{e}^{2x}$

$⇒{\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}=2{e}^{2×0}=2×1=2$        .....(1)

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

$y-1={\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}\left(x-0\right)$

$⇒y-1=2x$              [Using (1)]

$⇒2x-y+1=0$        .....(2)

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

Putting y = 0 in (2), we get

$2x-0+1=0$

$⇒x=-\frac{1}{2}$

Thus, the tangent to the given curve y = e2x at the point (0,1) meets x-axis at $\left(-\frac{1}{2},0\right)$.

Hence, the correct answer is option (b).

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

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

$y×2x+\left(1+{x}^{2}\right)×\frac{dy}{dx}=-1$

$⇒\left(1+{x}^{2}\right)\frac{dy}{dx}=-1-2xy$

$⇒\frac{dy}{dx}=-\frac{1+2xy}{1+{x}^{2}}$

∴ Slope of tangent to the given curve at (2, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}=-\frac{1+2×2×0}{1+{\left(2\right)}^{2}}=-\frac{1}{5}$

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

$y-0={\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}\left(x-2\right)$

$⇒y=-\frac{1}{5}\left(x-2\right)$

$⇒5y=-x+2$

$⇒x+5y=2$

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).

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)

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

${y}_{1}={x}_{1}^{3}-12x+18$       .....(1)

y = x3 − 12x + 18

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=3{x}^{2}-12$

∴ Slope of tangent at (x1, y1) = ${\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}=3{x}_{1}^{2}-12$

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

$\therefore {\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}=0$

$⇒3{x}_{1}^{2}-12=0$

$⇒{x}_{1}^{2}=4$

x1 = −2 or x1 = 2

Putting x1 = −2 in (1), we get

${y}_{1}={\left(-2\right)}^{3}-12×\left(-2\right)+18=-8+24+18=34$

Putting x1 = 2 in (1), we get

${y}_{1}={\left(2\right)}^{3}-12×2+18=8-24+18=2$

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).

Question 36:

The curve y = x1/5 has at (0, 0)
(a) a vertical tangent                      (b) a horizontal tangent
(c) an oblique tangent                   (d) no tangent

The given curve is

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

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{1}{5}{x}^{-\frac{4}{5}}$

At (0, 0), we have

${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}=\frac{1}{5}×{\left(0\right)}^{-\frac{4}{5}}=\frac{1}{5}×\infty =\infty$

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

Now, ${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}=\infty$

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).

Question 1:

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

The given curve is

y = tanx

Differentiating both sides with respect to x, we have

$\frac{dy}{dx}={\mathrm{sec}}^{2}x$

∴ Slope of tangent at (0, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}={\mathrm{sec}}^{2}0={\left(1\right)}^{2}=1$

⇒ Slope of normal at (0, 0) = $-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}}=-\frac{1}{1}=-1$

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

$y-0=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}}\left(x-0\right)$

$⇒y=-1×x$

$⇒x+y=0$

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___.

Question 2:

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

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

$\frac{dy}{dx}=2x+a$

∴ Slope of tangent at (h, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(h,0\right)}=2h+a$

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

${\left(\frac{dy}{dx}\right)}_{\left(h,0\right)}=0$            (Slope of the x-axis = 0)

$⇒2h+a=0$

$⇒h=-\frac{a}{2}$

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

$0={\left(-\frac{a}{2}\right)}^{2}+a×\left(-\frac{a}{2}\right)+25$

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

$⇒\frac{{a}^{2}}{4}=25$

$⇒{a}^{2}=100$

$⇒a=±10$

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__.

Question 3:

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

Let (hk) 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

$\frac{dy}{dx}=12-3{x}^{2}$

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=12-3{h}^{2}=0$           (Given)

$⇒12-3{h}^{2}=0$

$⇒{h}^{2}=4$

⇒ 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)___.

Question 4:

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

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

y = x logex

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=x×\frac{1}{x}+{\mathrm{log}}_{e}x×1$

$⇒\frac{dy}{dx}=1+{\mathrm{log}}_{e}x$

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

$⇒-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}}=-\frac{2}{\left(-2\right)}$

$⇒-\frac{1}{1+{\mathrm{log}}_{e}h}=1$

$⇒1+{\mathrm{log}}_{e}h=-1$

$⇒{\mathrm{log}}_{e}h=-2$

$⇒h={e}^{-2}=\frac{1}{{e}^{2}}$

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

$\therefore k=h{\mathrm{log}}_{e}h$       .....(1)

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

$k=\frac{1}{{e}^{2}}×{\mathrm{log}}_{e}\frac{1}{{e}^{2}}$

$⇒k=\frac{1}{{e}^{2}}×{\mathrm{log}}_{e}{e}^{-2}$

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

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

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

Question 5:

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

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

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

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=0+\frac{1}{2\sqrt{4x+1}}×4$

$⇒\frac{dy}{dx}=\frac{2}{\sqrt{4x+1}}$

∴ Slope of tangent at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=\frac{2}{5}$            (Given)

$⇒\frac{2}{\sqrt{4h+1}}=\frac{2}{5}$

$⇒\sqrt{4h+1}=5$

$⇒4h+1=25$

$⇒4h=24$

$⇒h=6$

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

$\therefore k=2+\sqrt{4h+1}$     .....(1)

Putting h = 6 in (1), we get

$k=2+\sqrt{4×6+1}$

$⇒k=2+\sqrt{25}$

$⇒k=2+5=7$

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

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

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

Question 6:

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

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

$\frac{dx}{dt}=6t$       .....(1)

y = t3 − 1

Differentiating both sides with respect to t, we get

$\frac{dy}{dt}=3{t}^{2}$      .....(2)

$\therefore \frac{dy}{dx}$$=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $=\frac{3{t}^{2}}{6t}$ $=\frac{t}{2}$      [Using (1) and (2)]

At t = 0, slope of the tangent = ${\left(\frac{dy}{dx}\right)}_{t=0}$  = $\frac{0}{2}$ = 0            $\left(\frac{dy}{dx}=\frac{t}{2}\right)$

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___.

Question 7:

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

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

$\frac{dy}{dx}=2x$

∴ Slope of the tangent to the curve C1 at (0, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}$  = 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

$1=2y\frac{dy}{dx}$

$⇒\frac{dy}{dx}=\frac{1}{2y}$

∴ Slope of the tangent to the curve C2 at (0, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}$  $=\frac{1}{2×0}=\frac{1}{0}=\infty$

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 $\frac{\mathrm{\pi }}{2}$.

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

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

Question 8:

The slope of the tangent to the curve y = bex/a where it crosses y-axis is ______________.

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

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

$⇒y=b{e}^{0}$

$⇒y=b$

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

$\frac{dy}{dx}=b{e}^{-\frac{x}{a}}×\left(-\frac{1}{a}\right)$

$⇒\frac{dy}{dx}=-\frac{b}{a}{e}^{-\frac{x}{a}}$

∴ Slope of the tangent at (0, b) $={\left(\frac{dy}{dx}\right)}_{\left(0,b\right)}$ $=-\frac{b}{a}{e}^{-\frac{0}{a}}$ $=-\frac{b}{a}{e}^{0}$ $=-\frac{b}{a}$

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 .

Question 9:

The tangent to the curve y = e2x at (0, 1) cuts x-axis at the point __________________.

The given curve is

y = e2x

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=2{e}^{2x}$

∴ Slope of tangent at (0, 1) = ${\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}=2{e}^{2×0}=2×1=2$        .....(1)

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

$y-1={\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}\left(x-0\right)$

$⇒y-1=2x$              [Using (1)]

$⇒2x-y+1=0$        .....(2)

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

Putting y = 0 in (2), we get

$2x-0+1=0$

$⇒x=-\frac{1}{2}$

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

Thus, the tangent to the given curve y = e2x at (0, 1) cuts the x-axis at $\left(-\frac{1}{2},0\right)$.

The tangent to the curve y = e2x at (0, 1) cuts x-axis at the point .

Question 10:

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

The equation of given curve is

y3 xy − 8 = 0

Differentiating both sides with respect to x, we get

$3{y}^{2}\frac{dy}{dx}-\left(x\frac{dy}{dx}+y\right)-0=0$

$⇒\left(3{y}^{2}-x\right)\frac{dy}{dx}=y$

$⇒\frac{dy}{dx}=\frac{y}{3{y}^{2}-x}$

Now, slope of tangent at (0, 2) = ${\left(\frac{dy}{dx}\right)}_{\left(0,2\right)}=\frac{2}{3×{\left(2\right)}^{2}-0}=\frac{2}{12}=\frac{1}{6}$        .....(1)

∴ Slope of the normal at (0, 2)

$=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(0,2\right)}}$

$=-\frac{1}{\left(\frac{1}{6}\right)}$           [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___.

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 = _________________.

The equation of given curve is

y2 = 5x −1

Differentiating both sides with respect to x, we get

$2y\frac{dy}{dx}=5$

$⇒\frac{dy}{dx}=\frac{5}{2y}$

∴ Slope of normal at (1, −2)

$=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}}$

$=-\frac{1}{\frac{5}{2×\left(-2\right)}}$

$=\frac{4}{5}$

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

$y-\left(-2\right)=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}}\left(x-1\right)$

$⇒y-\left(-2\right)=\frac{4}{5}\left(x-1\right)$

$⇒5y+10=4x-4$

$⇒4x-5y-14=0$

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___.

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 _____________.

The equation of given curve is xy = 1.

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

$\therefore {x}_{1}{y}_{1}=1$          .....(1)

Now,

xy = 1

Differentiating both sides with respect to x, we get

$x\frac{dy}{dx}+y=0$

$⇒\frac{dy}{dx}=-\frac{y}{x}$

∴ Slope of the normal at (x1, y1)

$=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}}$

$=-\frac{1}{\left(-\frac{{y}_{1}}{{x}_{1}}\right)}$

$=\frac{{x}_{1}}{{y}_{1}}$

$={x}_{1}^{2}$           [Using (1)]

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

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

$\therefore {x}_{1}^{2}=-\frac{a}{b}$

$⇒\frac{a}{b}=-{x}_{1}^{2}$

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)__.

Question 13:

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

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

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

Now, y = f(x)

$⇒\frac{dy}{dx}=f\text{'}\left(x\right)$

∴ Slope of the normal at (3, 4) $=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(3,4\right)}}=-\frac{1}{f\text{'}\left(3\right)}$        .....(2)

From (1) and (2), we have

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

$⇒f\text{'}\left(3\right)=1$

Thus, the value of $f\text{'}\left(3\right)$ is 1.

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

Question 14:

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

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.

$\therefore k=h+\frac{4}{{h}^{2}}$       .....(1)

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

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=1-\frac{8}{{x}^{3}}$

∴ Slope of tangent at (h, k) $={\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=1-\frac{8}{{h}^{3}}$

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

∴ Slope of tangent = 0

$⇒{\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=0$

$⇒1-\frac{8}{{h}^{3}}=0$

$⇒{h}^{3}=8$

$⇒h=2$

Putting h = 2 in (1), we get

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

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

$y-3={\left(\frac{dy}{dx}\right)}_{\left(2,3\right)}\left(x-2\right)$

$⇒y=3$

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___.

Question 16:

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

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

$\frac{dy}{dx}=3{x}^{2}$

∴ Slope of tangent at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=3{h}^{2}$

It is given that,

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

⇒ 3h2 = k       .....(2)

From (1) and (2), we get

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

$⇒{h}^{3}-3{h}^{2}=0$

$⇒{h}^{2}\left(h-3\right)=0$

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)___.

Question 17:

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

The equation of given curve is

x2 + y2 − 2x + 4y − 5 = 0

Differentiating both sides with respect to x, we get

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

$⇒\left(2y+4\right)\frac{dy}{dx}=2-2x$

$⇒\frac{dy}{dx}=\frac{1-x}{y+2}$

∴ Slope of normal at (2, 1)

$=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{\left(2,1\right)}}$

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

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

= 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___.

Question 18:

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

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

$2y\frac{dy}{dx}=1$

$⇒\frac{dy}{dx}=\frac{1}{2y}$

∴ Slope of tangent at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=\frac{1}{2k}$     .....(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

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

$⇒k=\frac{1}{2}$

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

k2 = h     .....(3)

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

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

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

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

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

Question 19:

The curve y = 4x2 + 2x − 8 and y = x3x + 13 touch each other at the point _________________.

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

The curve y = 4x2 + 2x − 8 and y = x3x + 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 = h3h + 10            .....(2)

Consider the first curve C1y = 4x2 + 2x − 8.

y = 4x2 + 2x − 8

$⇒\frac{dy}{dx}=8x+2$

∴ Slope of tangent to the curve C1 at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=8h+2$

Consider the second curve C2y = x3x + 10.

y = x3x + 10

$⇒\frac{dy}{dx}=3{x}^{2}-1$

∴ Slope of tangent to the curve C2 at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=3{h}^{2}-1$

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)

$⇒8h+2=3{h}^{2}-1$

$⇒3{h}^{2}-8h-3=0$

$⇒3{h}^{2}-9h+h-3=0$

$⇒3h\left(h-3\right)+1\left(h-3\right)=0$

$⇒\left(3h+1\right)\left(h-3\right)=0$

⇒ 3h + 1 = 0 or h − 3 = 0

h = $-\frac{1}{3}$ or h = 3

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

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

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

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

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

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

Putting h = 3 in (2), we get

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

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 = x3x + 10 touch each other at the point (3, 34).

The curve y = 4x2 + 2x − 8 and y = x3x + 10 touch each other at the point ___(3, 34)___.

Question 1:

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

The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
Here,

Question 2:

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

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 $\frac{dy}{dx}$.

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 $\left(\frac{dy}{dx}\right)$= Slope of the x-axis = 0

Question 4:

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

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 = $\infty$

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 $\frac{dy}{dx}$.

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 $±45°$.

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 $\frac{dx}{dy}$.

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, $\frac{dy}{dx}$ = Slope of the y-axis = $\infty$
$\frac{dx}{dy}=\frac{1}{\left(\frac{dy}{dx}\right)}=\frac{1}{\infty }=0$

Question 7:

Find the slope of the normal at the point 't' on the curve .

Question 8:

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

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

Question 9:

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

Question 10:

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

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.

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

Question 12:

Write the equation on the tangent to the curve y = x2x + 2 at the point where it crosses the y-axis.

When the curve crosses the y-axis, the point on the curve is of the form (0, y).
Here,

Question 13:

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

Question 14:

Write the angle between the curves y = ex and y = ex at their point of intersections.

Question 15:

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

Question 16:

Write the coordinates of the point at which the tangent to the curve y = 2x2x + 1 is parallel to the line y = 3x + 9.

Let (x1, y1) be the required point.
Slope of the given line = 3

Question 17:

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

Question 18:

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

We have,

$y=\mathrm{sin}x$

$⇒\frac{dy}{dx}=\mathrm{cos}x$

Slope at (0, 0) = m${\left[\frac{dy}{dx}\right]}_{x=0}=\mathrm{cos}0=1$

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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