Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 11 - Conic Sections

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Page No 202:

Question 1:

Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.

Answer:

Let us consider a circle that touches both the axes and has radius "a".
The centre of the circle is A(a, a) and radius is a units.
Now, the equation of the circle with centre (h, k) and radius r is:
${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$
Here, the equation of the circle is:
${(x - a)}^{2} + {(y - a)}^{2} = a^{2} \Rightarrow x^{2} + a^{2} - 2 a x + y^{2} + a^{2} - 2 a y = a^{2} \Rightarrow x^{2} + y^{2} - a^{2} - 2 a (x + y) = 0$

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Question 2:

Show that the point (x, y) given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of t such that –1 ≤ t ≤ 1 where a is any given real numbers.

Answer:

Given that, $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ .

$\Rightarrow x^{2} = \frac{4 a^{2} t^{2}}{{(1 + t^{2})}^{2}} and y^{2} = \frac{a^{2} {(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}}$

$\Rightarrow x^{2} + y^{2} = \frac{4 a^{2} t^{2} + a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}}$

$= \frac{4 a^{2} t^{2} + a^{2} + a^{2} t^{4} - 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} (1 + t^{4} + 2 t^{2})}{{(1 + t^{2})}^{2}} = \frac{a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} = a^{2}$
$\Rightarrow x^{2} + y^{2} = a^{2} . . . . . (1)$
Comparing (1) with equation of circle with centre (h, k) and radius r units, we get that centre is the origin i.e (0, 0) and
Radius = a units
Hence, proved.

Page No 202:

Question 3:

If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.

Answer:

Given that, the circle passes through the points (0, 0), (a, 0) and (0, b).
This means that these points satisfy the equation of the circle.
For (0, 0),
${(0 - h)}^{2} + {(0 - k)}^{2} = r^{2} \Rightarrow h^{2} + k^{2} = r^{2}$
For (a, 0),
${(a - h)}^{2} + {(0 - k)}^{2} = r^{2} \Rightarrow a^{2} + h^{2} - 2 a h + k^{2} = r^{2} . . . . . (1)$
For (0, b)
${(0 - h)}^{2} + {(b - k)}^{2} = r^{2} \Rightarrow h^{2} + b^{2} + k^{2} - 2 b k = r^{2} . . . . . (2) Solving (1) and (2), a (a - 2 h) = 0 \Rightarrow a = 0 or a = 2 h b (b - 2 k) = 0 \Rightarrow b = 0 or b = 2 k$
Since, the circle passes through the centre (0, 0), the coordinates are
$a = 2 h \Rightarrow h = \frac{a}{2} b = 2 k \Rightarrow k = \frac{b}{2}$
Therefore, the coordinates of the centre are $(\frac{a}{2}, \frac{b}{2}) .$

Page No 202:

Question 4:

Find the equation of the circle which touches x-axis and whose centre is (1, 2).

Answer:

Given that, the circle touches are x-axis and its centre is (1, 2).
Thus, r = 2.
Therefore, the equation of the circle is:

${(x - 1)}^{2} + {(y - 2)}^{2} = 2^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 4 \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$

Hence, the equation of the circle is $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ .

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Question 5:

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.

Answer:

Given that, 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle.
Now, $\frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2}; \frac{b_{1}}{b_{2}} = \frac{- 4}{- 8} = \frac{1}{2}; \frac{c_{1}}{c_{2}} = \frac{4}{- 7}$
$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Therefore, these lines are parallel to each other.
The distance between two parallel tangents of a circle is the diameter of the circle.
Distance between two parallel lines = $d = |\frac{c_{1} - c_{2}}{\sqrt{a^{2} + b^{2}}}|$
Where a = 3, b = – 4, c₁ = 4 and c₂ $= \frac{- 7}{2} .$
$\begin{array}{rcl} \Rightarrow & d = |\frac{4 + \frac{7}{2}}{\sqrt{9 + 16}}| \\ = & |\frac{\frac{15}{2}}{5}| \\ = & \frac{3}{2} units \end{array}$
Thus, the radius of the given circle is $\begin{array}{rcl} \frac{3}{2} units . \end{array}$

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Question 6:

Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.

Answer:

Given that, the circle lies in the third quadrant. Let a be the radius of the circle and A(–a, –a) be its centre.
The perpendicular distance between a point and a line $= |\frac{A x_{1} + B y_{1} + C_{1}}{\sqrt{A^{2} + B^{2}}}|$
Now, the distance of the given line from the centre of the circle is a units.

Here,
equation of line $\Rightarrow 3 x - 4 y + 8 = 0$
point = A(–a, –a)
and distance between them = a units
$\begin{array}{rcl} \Rightarrow & d = |\frac{3 (- a) - 4 (- a) + 8}{\sqrt{3^{2} + {(- 4)}^{2}}}| \\ = & |\frac{a + 8}{5}| \end{array}$

$\Rightarrow a = |\frac{a + 8}{5}| \Rightarrow \pm a = \frac{a + 8}{5} \Rightarrow 5 a = a + 8 or - 5 a = a + 8 \Rightarrow a = 2 or a = \frac{- 4}{3}$
Since a represents distance, it cannot be negative.
⇒ A(–2, –2) is the centre of the circle.
Therefore, equation of the circle:-

${(x + 2)}^{2} + {(y + 2)}^{2} = {(2)}^{2} \Rightarrow x^{2} + 4 + 4 x + y^{2} + 4 + 4 y = 4 \Rightarrow x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

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Question 7:

If one end of a diameter of the circle x²+ y²– 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.

Answer:

Given that, the equation of the circle is x²+ y²– 4x – 6y + 11 = 0 and one end of the diameter is (3, 4).
The diametrix form of a circle with centre = (–g, –f) is given as:
x²+ y²– 2gx – 2fy + c = 0
Here,
$x^{2} + y^{2} + 2 (- 2) x + 2 (- 3) y + 11 = 0 \Rightarrow g = - 2 and f = - 3$
Therefore, centre of the given circle is (2, 3).
Now, the centre of the circle is the midpoint of the diameter. Let (x₁, y₁) be the coordinates of the other end of the diameter.
Thus,
$2 = \frac{x_{1} + 3}{2} \Rightarrow x = 1 3 = \frac{y_{1} + 4}{2} \Rightarrow y_{1} = 2$

Hence, the coordinates of the other end of the diameter are (1, 2).

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Question 8:

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

Answer:

Given that, the centre of the circle is (1, –2) and it passes through the lines 3x + y = 14 and 2x + 5y = 18.
Solving the two equations,
3x + y = 14
⇒ $15 x + 5 y = 70 - 2 x - 5 y = - 18 13 x = 52$
$\Rightarrow x = 4 \Rightarrow y = 14 - 3 (4) = 2$
Therefore, the point of intersection of the two lines is (4, 2).
The circle also passes through this point. Now, the distance between the centre of the circle (1, –2) and (4, 2) is given as:
$r = \sqrt{{(4 - 1)}^{2} + {(2 + 2)}^{2}} = \sqrt{9 + 16} \Rightarrow r = 5 units$
Therefore, the equation of the circle is:
${(x - 1)}^{2} + {(y + 2)}^{2} = 5^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25 \Rightarrow x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

Page No 202:

Question 9:

If the line $y = \sqrt{3} x + k$ touches the circle x²+ y²= 16, then find the value of k.

Answer:

Given that, the line $y = \sqrt{3} x + k$ touches the circle x²+ y²= 16.
Thus, the centre of the circle is (0, 0).
Now, the equation of the given line is:
$\sqrt{3} x - y + k = 0$
The distance of the line from the centre of the circle is:
$d = |\frac{\sqrt{3} (0) + (- 1) (0) + k}{\sqrt{3 + 1}}| \Rightarrow r = \frac{\pm k}{2} \Rightarrow r = 4 = \frac{\pm k}{2} \Rightarrow k = \pm 8$

Hence, the value of k is $\pm 8 .$

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Question 10:

Find the equation of a circle concentric with the circle x²+ y²– 6x + 12y + 15 = 0 and has double of its area.

Answer:

Given that, the required circle is concentric with x²+ y²– 6x + 12y + 15 = 0 and has double the area.
Using the diametrix form of the equation of the circle with centre (–g, –f), we get
g = –3
f = 6
Therefore, centre of the required circle is (3, –6).
Also, the equation of the given circle can be written as:-
$x^{2} - 2 (3) x + y^{2} + 2 (6) y + 15 = 0 \Rightarrow x^{2} - 2 (3) x + 3^{2} + y^{2} + 2 (6) y + 6^{2} + 15 - 9 + 36 = 0 \Rightarrow {(x - 3)}^{2} + {(y - (- 6))}^{2} - 30 = 0 \Rightarrow {(x - 3)}^{2} + {(y - (- 6))}^{2} = {(\sqrt{30})}^{2}$
Thus, the radius of the given circle is $\sqrt{30}$ units.
Area of given circle $= π r^{2} = π \times (30) sq . units = 30 π sq . units$
Area of required circle = 60π sq. units
$\Rightarrow 60 π = π R^{2} \Rightarrow R^{2} = 60$
Therefore, the equation of the required circle is:-
${(x - 3)}^{2} + {(y + 6)}^{2} = 60 \Rightarrow x^{2} + 9 - 6 x + y^{2} + 36 + 12 y = 60 \Rightarrow x^{2} + y^{2} - 6 x + 12 y - 15 = 0$

Page No 202:

Question 11:

If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Answer:

Given that, the latus rectum of an ellipse is half of the minor axis.
Consider the ellipse: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Length of minor axis = 2b
Length of latus rectum $= \frac{2 b^{2}}{a}$
Now,
$\frac{2 b^{2}}{a} = \frac{1}{2} \times 2 b \Rightarrow 4 b^{2} = 2 a b \Rightarrow 2 b^{2} - a b = 0 \Rightarrow b = 0 or b = \frac{a}{2}$
Now,
$b^{2} = a^{2} (1 - e^{2}) \Rightarrow b = {(2 b)}^{2} (1 - e^{2}) = 4 b^{2} (1 - e^{2}) \Rightarrow 1 - e^{2} = \frac{1}{4} \Rightarrow e^{2} = \frac{3}{4} \Rightarrow e = \frac{\sqrt{3}}{2} (as e < 1 for an ellipse)$

Hence, the eccentricity of the ellipse is $\frac{\sqrt{3}}{2} .$

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Question 12:

Given the ellipse with equation 9x²+ 25y²= 225, find the eccentricity and foci.

Answer:

Given that, the equation of the ellipse is 9x²+ 25y²= 225.
$\Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
Now, b² = a²(1 – e²) and e < 1 for ellipse.
$\Rightarrow \frac{9}{25} = 1 - e^{2} \Rightarrow e^{2} = \frac{16}{25} \Rightarrow e = \frac{4}{5}$
Also, foci of the ellipse is given as $(\pm a e, 0)$
$\Rightarrow F (\pm 5 \times \frac{4}{5}, 0) = F (\pm 4, 0)$

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Question 13:

If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10, then find latus rectum of the ellipse.

Answer:

Given that, eccentricity of the ellipse is $\frac{5}{8}$ and the distance between the foci of the ellipse is 10.
Now,
2ae = 10
$\Rightarrow a e = 5 \Rightarrow a = \frac{5}{5} \times 8 = 8$
For an ellipse,
$b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 64 [1 - {(\frac{5}{8})}^{2}] = 64 [1 - \frac{25}{64}] \Rightarrow b^{2} = 39$
Now, length of latus rectum $= 2 b^{2} = \frac{2 \times 39}{8} = \frac{39}{4} units .$

Page No 203:

Question 14:

Find the equation of ellipse whose eccentricity is $\frac{2}{3}$ , latus rectum is 5 and the centre is (0, 0).

Answer:

Given that, the centre of the ellipse is (0, 0), eccentricity is $\frac{2}{3}$ and length of the latus rectum is 5 units.
Equation of an ellipse is:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, $\frac{2 b^{2}}{a} = 5$
$\Rightarrow b^{2} = \frac{5}{2} a$
Also, $b^{2} = a^{2} (1 - e^{2})$
$\Rightarrow \frac{5}{2} a = a^{2} (1 - e^{2})$
$\Rightarrow \frac{5}{2} a = a^{2} [1 - {(\frac{2}{3})}^{2}] = a^{2} [1 - \frac{4}{9}] = \frac{5 a^{2}}{9}$
$\Rightarrow \frac{5}{2} a = \frac{5}{9} a^{2} \Rightarrow \frac{5}{9} a^{2} - \frac{5}{2} a = 0 \Rightarrow a = 0 or a = \frac{9}{2}$
Now, a = 0 is not possible as the length of the latus rectum is 5 units.
$\Rightarrow b^{2} = \frac{5}{2} a \Rightarrow b^{2} = \frac{5}{2} \times \frac{9}{2} = \frac{45}{4}$
Therefore, equation of the elllipse is:-
$\frac{x^{2}}{{(\frac{9}{2})}^{2}} + \frac{y^{2}}{\frac{45}{4}} = 1 \Rightarrow \frac{4 x^{2}}{81} + \frac{4 y^{2}}{45} = 1$

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Question 15:

Find the distance between the directrices of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 .$

Answer:

Given that, the equation of ellipse is $\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 .$
$\Rightarrow \frac{x}{{(6)}^{2}} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1$
Now, b² = a²(1 – e²).
$\Rightarrow 20 = 36 (1 - e^{2}) \Rightarrow \frac{5}{9} = 1 - e^{2} \Rightarrow e^{2} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$
Distance between the directrices $= \frac{2 a}{e} = \frac{2 \times 6}{2} \times 3 = 18$
Hence, the distance between the two directrices is 18 units.

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Question 16:

Find the coordinates of a point on the parabola y²= 8x whose focal distance is 4.

Answer:

Given that, the equation of parabola is y² = 8x and its focal distance is 4 units.
Length of the latus rectum = 4a
Now,
4a = 8
⇒ a = 2
Focal length of an ellipse $= |a + x_{1}|$ where P $(x_{1}, y_{1})$ is any point on the parabola.
$\Rightarrow 4 = |a + x_{1}| \Rightarrow 4 = |2 + x_{1}| \Rightarrow \pm 4 = 2 + x_{1} \Rightarrow x_{1} = - 6 or 2 But y^{2} = 8 x Putting x = 2, y^{2} = 8 \times 2 = 16 \Rightarrow y = \pm 4$
Putting x = –6
⇒ y² = –48, which is not possible.
Therefore, the required points are (2, 4) and (2, –4).

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Question 17:

Find the length of the line-segment joining the vertex of the parabola y²= 4ax and a point on the parabola where the line-segment makes an angle θ to the x-axis.

Answer:

Given that, the equation of parabola is y²= 4ax.
Consider a point on the parabola $P (a t^{2}, 2 a t) .$
Now,
$\tan θ = \frac{2 a t}{a t^{2}} \Rightarrow \tan θ = \frac{2}{t} \Rightarrow t = \frac{2}{\tan θ} = 2 co t θ$
Now,
$\begin{array}{rcl} OP & = & \sqrt{{(a t^{2} - 0)}^{2} + {(2 a t - 0)}^{2}} \\ = & \sqrt{a^{2} t^{4} + 4 a^{2} t^{2}} \\ = & a t \sqrt{t^{2} + 4} \\ = & a (2 co t θ) \sqrt{4 c o t^{2} θ + 4} \\ = & 4 a c o t θ \cdot \cos e c θ \end{array}$

$\Rightarrow OP = 4 a \times \frac{\cos θ}{\sin θ} \times \frac{1}{\sin θ} = \frac{4 a \cos θ}{\sin^{2} θ}$

Hence, the required length of the line segment is $\frac{4 a \cos θ}{\sin^{2} θ} .$

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Question 18:

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Answer:

Given that, the vertex of the parabola is (0, 4) and focus of the parabola is (0, 2). Therefore, the equation of the directrix is y = 6.

Now,
FP = PM for a parabola.
Using distance formula,
$\begin{array}{rcl} FP & = & \sqrt{{(x - 0)}^{2} + {(y - 2)}^{2}} \\ = & \sqrt{x^{2} + y^{2} + 4 - 4 y} \end{array}$
Also, distance between y = 6 and P(x, y) is:
$PM = d = |\frac{x_{1} (0) + y - 6}{1}| \Rightarrow d = |y - 6| \Rightarrow \sqrt{x^{2} + y^{2} + 4 - 4 y} = |y - 6| \Rightarrow \sqrt{x^{2} + y^{2} + 4 - 4 y} = y - 6 (∵ value of square root is positive) \Rightarrow x^{2} + y^{2} + 4 - 4 y = y^{2} + 36 - 12 y \Rightarrow x^{2} + 8 y - 32 = 0$
is the required equation of the parabola.

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Question 19:

If the line y = mx + 1 is tangent to the parabola y²= 4x then find the value of m.

Answer:

Given that, the line y = mx + 1 is a tangent to the parabola y² = 4x.
Let their point of intersection be (x₁, y₁).
⇒ y₁= mx₁ + 1 and $y_{1}^{2} = 4 x_{1}$
$\Rightarrow {(m x_{1} + 1)}^{2} = 4 x_{1} \Rightarrow m^{2} x_{1}^{2} + 1 + 2 m x_{1} = 4 x_{1} \Rightarrow m^{2} x_{1}^{2} + (2 m - 4) x_{1} + 1 = 0$
According to the condition of tangency, D = 0.
Thus,
${(2 m - 4)}^{2} - 4 m^{2} (1) = 0 \Rightarrow 4 m^{2} + 16 - 16 m - 4 m^{2} = 0 \Rightarrow 16 - 16 m = 0 \Rightarrow m = 1$
Hence, the value of m is 1.

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Question 20:

If the distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$ , then obtain the equation of the hyperbola.

Answer:

Given that, the distance between the foci of the hyperbola is 16 and its eccentricity is $\sqrt{2}$ .
The equation of a hyperbola is given as:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Now,
$2 a e = 16 \Rightarrow a e = 8 \Rightarrow a (\sqrt{2}) = 8 \Rightarrow a = 4 \sqrt{2}$
For a hyperbola,
$b^{2} = a^{2} (e^{2} - 1) and e > 1 \Rightarrow b^{2} = 32 (2 - 1) = 32$
Thus, equation of hyperbola is:
$\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32$

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Question 21:

Find the eccentricity of the hyperbola 9y²– 4x²= 36.

Answer:

Given that, the equation of hyperbola is 9y²– 4x²= 36.
$\Rightarrow \frac{y^{2}}{4} - \frac{x^{2}}{9} = 1 \Rightarrow a = \pm 2 and b = \pm 3$
For hyperbola, b² = a²(e²– 1)
$\Rightarrow 9 = 4 (e^{2} - 1) \Rightarrow e^{2} = \frac{9}{4} + 1 = \frac{13}{4} \Rightarrow e = \frac{\sqrt{13}}{2}$
Hence, the eccentricity of the given hyperbola is $\frac{\sqrt{13}}{2}$ .

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Question 22:

Find the equation of the hyperbola with eccentricity $\frac{3}{2}$ and foci at (± 2, 0).

Answer:

Given that, the hyperbola has $e = \frac{3}{2}$ and foci at (±2, 0).
Now ae = 2
$\Rightarrow a = \frac{4}{3}$
For a hyperbola, b² = a² (e²– 1)
$\begin{array}{rcl} \Rightarrow & b^{2} = \frac{16}{9} (\frac{9}{4} - 1) \\ = & \frac{16}{9} \times \frac{5}{4} \\ \Rightarrow & b^{2} = \frac{20}{9} \end{array}$
Thus, the equation of the hyperbola is :
$\frac{x^{2}}{{(\frac{4}{3})}^{2}} - \frac{y^{2}}{\frac{20}{9}} = 1 \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}$

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Question 23:

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer:

Given that, the diameters of a circle of area 154 sq. units are given by the lines 2x – 3y = 5 and 3x – 4y = 7.
The intersection point of the two lines is the centre of the circle.
Solving the two equations,

$\begin{array}{ccllcc} 6 x & - & 9 y & = & 15 \\ - & 6 x & + & 8 y & = & - 14 \\ - y & = & 1 \\ \Rightarrow & y & = & - 1 \end{array} \begin{array}{l} \Rightarrow x = 1 \end{array}$
Therefore, the coordinates of the centre of the circle are (1, –1).
Also, the area of the circle is 154 sq. units.
⇒ 154 = πr²
$\begin{array}{rcl} \Rightarrow & r^{2} = \frac{154}{22} \times 7 \\ = & 49 \end{array}$
Thus, the equation of the circle with centre (1, –1) and r² = 49 is :
${(x - 1)}^{2} + {(y + 1)}^{2} = 49 \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 + 2 y = 49 \Rightarrow x^{2} + y^{2} - 2 x + 2 y - 47 = 0$

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Question 24:

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Answer:

Given that, the circle passes through the points (2, 3) and (4, 5). Also the centre lies on the line y – 4x + 3 = 0.
Now, the equation of a circle with centre (h, k) is:
(x – h)²+ (y – k)²= r²
Since, the circle passes through (2, 3) and (4, 5), (2 – h)² + (3 – k)² = r² .....(1)
${(4 - h)}^{2} + {(5 - k)}^{2} = r^{2} \Rightarrow {(2 - h)}^{2} + {(3 - k)}^{2} = {(4 - h)}^{2} + {(5 - k)}^{2} \Rightarrow {(4 - h)}^{2} - {(2 - h)}^{2} = {(3 - k)}^{2} - {(5 - k)}^{2} \Rightarrow 16 + h^{2} - 8 h - 4 - h^{2} + 4 h = 9 + k^{2} - 6 k - 25 - k^{2} + 10 k$
$\Rightarrow - 4 h + 12 = 4 k - 16 \Rightarrow 4 h + 4 k = 28 \Rightarrow h + k = 7$
Putting this is the equation of line on which the centre lies,
$k - 4 h + 3 = 0 \Rightarrow 7 - h - 4 h + 3 = 0 \Rightarrow - 5 h + 10 = 0 \Rightarrow h = 2 \Rightarrow k = 5$
Putting this in (1), (2 – 2)²+ (3 – 5)² = r²
$\Rightarrow r^{2} = 2^{2} = 4$
Hence, the equation of the circle is:
${(x - 2)}^{2} + {(y - 5)}^{2} = 4 \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 - 10 y = 4 \Rightarrow x^{2} + y^{2} - 4 x - 10 y + 25 = 0$

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Question 25:

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

Answer:

The perpendicular from the centre of the circle to the chord, bisects it.
⇒ AC = 3 units.
Now, the distance between O(3, –1) and the line 2x – 5y + 18 = 0 is:

$d = |\frac{2 (3) - 5 (- 1) + 18}{\sqrt{4 + 25}}| = |\frac{6 + 5 + 18}{\sqrt{29}}| \Rightarrow d = \sqrt{29} units$
Applying Pythagoras theorem is ΔAOC,

$\begin{array}{rcl} {AO}^{2} & = & {OC}^{2} + {AC}^{2} \\ = & 29 + 9 \\ = & 38 \\ \Rightarrow & r = \sqrt{38} units \end{array}$
Therefore, equation of the circle is:
${(x - 3)}^{2} + {(y + 1)}^{2} = 38 \Rightarrow x^{2} + 9 - 6 x + y^{2} + 1 + 2 y = 38 \Rightarrow x^{2} + y^{2} - 6 x + 2 y - 28 = 0$

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Question 26:

Find the equation of a circle of radius 5 which is touching another circle x²+ y²– 2x – 4y – 20 = 0 at (5, 5).

Answer:

The equation of the given circle is:-
x²+ y²– 2x – 4y – 20 = 0
⇒ g = –1 and f = –2
Therefore, centre of the given circle is (1, 2).

$\begin{array}{rcl} Radius of given circle & = & \sqrt{g^{2} + f^{2} - c} \\ = & \sqrt{25} \\ = & 5 units \end{array}$
Thus, point (5, 5) is the midpoint of the line segment joining the two centres.
$\Rightarrow 5 = \frac{1 + x}{2} \Rightarrow x = 9 and 5 = \frac{2 + y}{2} \Rightarrow y = 8$
Therefore, centre of the required circle is (9, 8).
Hence, the equation of the required circle is:
${(x - 9)}^{2} + {(y - 8)}^{2} = 25 \Rightarrow x^{2} - 18 x + 81 + y^{2} + 64 - 16 y = 25 \Rightarrow x^{2} + y^{2} - 18 x - 16 y + 120 = 0$

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Question 27:

Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.

Answer:

Given that, the circle has radius 3 units, passes through the point (7, 3) and the centre lies on the line y = x – 1.
The equation of the circle with centre (h, k) is:
(x – h)² + (y – k)² = r²
Since the point (7, 3) lies on the circle, (7 – h)² + (3 – k)² = 9 .....(1)
Now, the centre lies on the line y = x – 1.
⇒ k = h – 1 .....(2)
Putting (2) in (1),
${(7 - h)}^{2} + {(3 - h + 1)}^{2} = 9 \Rightarrow {(7 - h)}^{2} + {(4 - h)}^{2} = 9 \Rightarrow 49 + h^{2} - 14 h + 16 + h^{2} - 8 h = 9 \Rightarrow 2 h^{2} - 22 h + 56 = 0 \Rightarrow h^{2} - 11 h + 28 = 0 \Rightarrow h^{2} - 7 h - 4 h + 28 = 0 \Rightarrow h (h - 7) - 4 (h - 7) = 0 \Rightarrow h = 47 \Rightarrow k = 3 or 6$
Therefore, there are two required circles with centres (4, 3) and (7, 6).
Thus, the equations of the two circles are:
${(x - 4)}^{2} + {(y - 3)}^{2} = 9 \Rightarrow x^{2} + y^{2} + 16 + 9 - 8 x - 6 y = 9 \Rightarrow x^{2} + y^{2} - 8 x - 6 y + 16 = 0 and {(x - 7)}^{2} + {(y - 6)}^{2} = 9 \Rightarrow x^{2} + y^{2} + 49 + 36 - 14 x - 12 y = 9 \Rightarrow x^{2} + y^{2} - 14 x - 12 y + 76 = 0$

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Question 28:

Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0

Answer:

(a) For a parabola, FP = PM

$\begin{array}{rcl} \Rightarrow & PR = \sqrt{{(6 - x)}^{2} + {(0 - y)}^{2}} \\ = & \sqrt{36 + x^{2} - 12 x + y^{2}} . . . . . (1) \end{array}$
Now,
$\begin{array}{rcl} PM & = & \sqrt{{(x - 0)}^{2} + {(y - y)}^{2}} \\ = & \sqrt{x^{2}} \\ = & x . . . . . (2) \end{array}$
Equation (1) and(2),
$\Rightarrow x = \sqrt{36 + x^{2} - 2 x + y^{2}} \Rightarrow x^{2} = 35 + x^{2} - 12 x + y^{2} \Rightarrow y^{2} - 12 x + 36 = 0 \Rightarrow y^{2} = 36 - 12 x [is the required equation]$
(b) Consider a point P(x, y) on the parabola.

Now,
PF = PM.
$\begin{array}{rcl} \Rightarrow & \sqrt{{(x - 0)}^{2} + {(y - 2)}^{2}} = \sqrt{{(x - x)}^{2} + {(6 - y)}^{2}} \end{array} \Rightarrow \sqrt{x^{2} + {(y - 2)}^{2}} = 6 - y \Rightarrow x^{2} + y^{2} + 4 - 4 y = 36 + y^{2} - 12 y \Rightarrow x^{2} 6 + 8 y - 32 = 0$
$\Rightarrow x^{2} = 32 - 8 y$ is the required equation of the parabola.
(c) Since no clear information about the parabola is given, let us consider the following parabola and directrix:

For a parabola, PF = PM.
Now,
$PF = \sqrt{{(x_{1} + 1)}^{2} + {(y_{1} + 2)}^{2}} PM = d = |\frac{x_{1} (1) - 2 (y_{1}) + 3}{\sqrt{5}}| = |\frac{x_{1} - 12 y_{1} + 3}{\sqrt{5}}| \Rightarrow \sqrt{{(x_{1} + 1)}^{2} + {(y_{1} + 2)}^{2}} = \frac{x_{1} - 2 y_{1} + 3}{\sqrt{5}}$
Squaring both sides, ${(x_{1} + 1)}^{2} + {(y_{1} + 2)}^{2} = \frac{{(x_{1} - 2 y_{1} + 3)}^{2}}{\sqrt{5}}$
$\Rightarrow x_{1}^{2} + 2 x_{1} + 1 + y_{1}^{2} + 4 + 4 y_{1} = \frac{{(x_{1} - 2 y_{1} + 3)}^{2}}{5} \Rightarrow x_{1}^{2} + y_{1}^{2} + 2 x_{1} + 4 y_{1} + 5 = \frac{{(x_{1} - 2 y_{1} + 3)}^{2}}{5} \Rightarrow 5 x_{1}^{2} + 5 y_{1}^{2} + 10 x_{1} + 20 y_{1} + 25 = x_{1}^{2} + 4 y_{1}^{2} + 9 - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1}$
$\Rightarrow 4 x^{2} + y^{2} + 4 x + 32 y + 4 x y + 16 = 0$ is the required equation of the parabola.

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Question 29:

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Answer:

Consider a point P(x, y).

$\Rightarrow \sqrt{{(x - 3)}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - 9)}^{2} + {(y - 0)}^{2}} = 12 \Rightarrow \sqrt{x^{2} - 6 x + 9 + y^{2}} + \sqrt{x^{2} + y^{2} + 81 - 18 x} = 12$

Let x² – 6x + 9 + y²= Q.

$\Rightarrow \sqrt{Q} + \sqrt{Q - 12 x + 72} = 12 \Rightarrow \sqrt{Q - 12 x + 72} = 12 - \sqrt{Q}$

Squaring both sides,

$Q - 12 x + 72 = 144 + Q - 24 \sqrt{Q} \Rightarrow 24 \sqrt{Q} = 12 x + 72 \Rightarrow 2 \sqrt{Q} = x + 6 \Rightarrow 4 Q = x^{2} + 36 + 12 x \Rightarrow 4 x^{2} - 24 x + 36 + 4 y^{2} = x^{2} + 36 + 12 x$

$\Rightarrow 3 x^{2} - 36 x + 4 y^{2} = 0 is the required equation .$

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Question 30:

Find the equation of the set of all points whose distance from (0, 4) are $\frac{2}{3}$ of their distance from the line y = 9.

Answer:

Given, that,
$PM = \frac{2}{3} PN . . . . . (1)$
Now,

$\begin{array}{rcl} PM & = & \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} \\ = & \sqrt{x^{2} + {(y - 4)}^{2}} \end{array}$
And
$\begin{array}{rcl} PN & = & d = |\frac{0 (x) + 1 (y) - 9}{1}| \\ = & |y = 9| \end{array}$
$\Rightarrow \sqrt{x^{2} + {(y - 4)}^{2}} = \frac{2}{3} |y - 9| \Rightarrow x^{2} + {(y - 4)}^{2} = \frac{4}{9} {(y - 9)}^{2} \Rightarrow 9 x^{2} + 9 y^{2} + 144 - 72 y = 4 y^{2} + 324 - 72 y$
$\Rightarrow 9 x^{2} + 5 y^{2} - 180 = 0 is the required equation .$

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Question 31:

Show that the set of all points such that the difference of their distances from (4, 0) and (–4, 0) is always equal to 2 represent a hyperbola.

Answer:

Given that, PM = PN = 2

Now,
$\begin{array}{rcl} PM & = & \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2}} \\ = & \sqrt{{(x + 4)}^{2} + y^{2}} \\ PN & = & \sqrt{{(4 - x)}^{2} + {(0 - y)}^{2}} \\ = & \sqrt{{(4 - x)}^{2} + y^{2}} \end{array}$

Let x² + 16 + y²= k.

$\Rightarrow PM = \sqrt{k + 8 x} and PN = \sqrt{k - 8 x} \Rightarrow \sqrt{k + 8 x} - \sqrt{k - 8 x} = 2 \Rightarrow \sqrt{k + 8 x} = 2 + \sqrt{k - 8 x}$

Squaring both sides,
$k + 8 x = 4 + k - 8 x + 4 \sqrt{k - 8 x} \Rightarrow 16 x = 4 + 4 \sqrt{k - 8 x} \Rightarrow 4 x = 1 + \sqrt{k - 8 x} \Rightarrow \sqrt{k - 8 x} = 4 x - 1 \Rightarrow k - 8 x = 16 x^{2} - 8 x + 1 \Rightarrow k = 16 x^{2} + 1 \Rightarrow x^{2} + 16 + y^{2} = 16 x^{2} + 1 \Rightarrow 15 x^{2} - 15 - y^{2} = 0 \Rightarrow \frac{x^{2}}{1} - \frac{y^{2}}{15} = 1$
Thus, it is a hyperbola.

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Question 32:

Find the equation of the hyperbola with

(a) Vertices (±5, 0), foci (±7, 0)

(b) Vertices (0, ±7), $e = \frac{4}{3}$

(c) Foci $(0, \pm \sqrt{10}),$ passing through (2, 3)

Answer:

(a) Here, a = 5 and ae = 7

\Rightarrow e = \frac{7}{5}

For a hyperbola,

b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = 25 (e^{2} - 1) \Rightarrow b^{2} = 25 [\frac{49}{25} - 1] \Rightarrow b^{2} = 25 \times \frac{24}{25} \Rightarrow b^{2} = 24

Hence, the equation of the hyperbola is:

\frac{x^{2}}{25} - \frac{y^{2}}{24} = 1

(b) Here, a = 7 and

e = \frac{4}{3}

Again, b² = a²(e² – 1)

\begin{array}{rcl} \Rightarrow & b^{2} = 49 [\frac{16}{9} - 1] \\ = & 49 \times \frac{7}{9} = \frac{343}{9} \end{array}

Hence, the equation of the hyperbola is:

\frac{y^{2}}{49} - \frac{9 x^{2}}{343} = 1 \Rightarrow 7 y^{2} - 9 x^{2} = 343 \Rightarrow 9 x^{2} - 7 y^{2} + 343 = 0

a e = \sqrt{10}

and the equation of the hyperbola is given as:

\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

Now, b² = a²(e² – 1)

\begin{array}{rcl} \Rightarrow & b^{2} = a^{2} e^{2} - a^{2} \\ = & 10 - a^{2} \end{array}

Also, x = 2, y = 3 is a point on hyperbola.

\Rightarrow \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \Rightarrow 90 - 9 a^{2} - 4 a^{2} = (10 - a^{2}) a^{2} \Rightarrow 90 - 13 a^{2} = 10 a^{2} - a^{4} \Rightarrow 23 a^{2} - a^{4} - 90 = 0 \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0 \Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0 \Rightarrow a^{2} = 5 or 18

And b² = 10 – a² = 5 (a² = 18 not possible as b² will become negative)
Thus, the equation of the hyperbola is:

\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1 \Rightarrow y^{2} - x^{2} = 5 \Rightarrow x^{2} - y^{2} + 5 = 0

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Question 33:

State whether the following statement is True or False.
The line x + 3y = 0 is a diameter of the circle x²+ y²+ 6x + 2y = 0.

Answer:

Given that, the diameter of the circle is x + 3y = 0 and the equation of the circle is x² + y² + 6x + 2y = 0.
⇒ g = 3 and f = 1
Therefore, centre of the given circle is (–3, –1).
Now, this point should satisfy the equation of the diameter for the statement to be true.
⇒ (–3) + 3(–1) = –3 – 3 = 6 ≠ 0
Hence, the statement is false.

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Question 34:

State whether the following statement is True or False.
The shortest distance from the point (2, –7) to the circle x²+ y²– 14x – 10y – 151 = 0 is equal to 5.

Answer:

The centre of the given circle is:
g = –7 and f = –5
⇒ c(7, 5)
Now, the distance between the centre of the circle and point P(2, –7) is:
$\begin{array}{rcl} d & = & \sqrt{{(7 - 2)}^{2} + {(5 + 7)}^{2}} \\ = & \sqrt{25 + 144} \\ = & 13 units \end{array}$
$\begin{array}{rcl} Also, radius & = & \sqrt{g^{2} + f^{2} - c} \\ = & \sqrt{49 + 25 + 151} \\ = & \sqrt{225} \\ = & 15 units \end{array}$
Therefore, the distance between the circle and the point P(2, –7) is 2 units.
Hence, the statement is false.

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Question 35:

State whether the following statement is True or False.
If the line lx + my = 1 is a tangent to the circle x²+ y²= a², then the point (l, m) lies on a circle.

Answer:

If (l, m) lies on the circle, it must satisfy the equation of the circle.
⇒ l² + m² = a²
The centre of the circle is (0, 0) and its radius is a.
Now, distance between point and a line is:

$a = |\frac{l (0) + m (0) - 1}{\sqrt{l^{2} + m^{2}}}| = |\frac{- 1}{\sqrt{l^{2} + m^{2}}}| \Rightarrow a^{2} = \frac{1}{l^{2} + m^{2}} \Rightarrow l^{2} + m^{2} = {(\frac{1}{a})}^{2}$
Hence, the statement is true.

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Question 36:

State whether the following statement is True or False.
The point (1, 2) lies inside the circle x²+ y²– 2x + 6y + 1 = 0.

Answer:

Given that, the circle is given by:
x²+ y²– 2x + 6y + 1 = 0.
⇒ g = –1
f = 3
$\begin{array}{rcl} ∴ r & = & \sqrt{g^{2} + f^{2} - c} \\ = & \sqrt{1 + 9 - 1} \\ = & 3 units \end{array}$
Also, the centre of the circle is (1, –3).
Now, the distance between (1, 2) and the centre (1, –3) is:
$\begin{array}{rcl} d & = & \sqrt{(1 - 1) 1^{2} + {(2 + 3)}^{2}} \\ = & 5 units > r \end{array}$
Thus, the point lies outside the circle.
Hence, the statement is false.

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Question 37:

State whether the following statement is True or False.
The line lx + my + n = 0 will touch the parabola y²= 4ax if ln = am².

Answer:

Line ⇒ lx + my + n = 0
Parabola ⇒ y² = 4ax
Now, from the equation of line,
$y = \frac{- n - l x}{m} . . . . . (1)$
Putting in the equation of parabola, ${(\frac{- n - l x}{m})}^{2} = 4 a x$
$\Rightarrow \frac{n^{2} + l^{2} x^{2} + 2 n l x}{m^{2}} = 4 a x$
⇒ l²x² + n² + 2nlx –4am²x = 0.
⇒ l²x² + n² + 2x (nl – 2am²) = 0
As per the condition of tandency, D = 0.
⇒ D = (2nl – 4am²)² – 4l²n²
⇒ 4n²l² + 16a²m⁴ – 16anlm² – 4l²n² = 0
⇒ 16am² (am² – nl) = 0
⇒ am² – nl = 0
⇒ am² = nl
Hence, the statement is true.

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Question 38:

State whether the following statement is True or False.
If P is a point on the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ whose foci are S and S′, then PS + PS′ = 8.

Answer:

The sum of the distances of any point P on the ellipse from the two foci is equal to the length of major axis.
Here,
Length of major axis = 5 × 2 = 10
Hence, the statement is false.

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Question 39:

State whether the following statement is True or False.
The line 2x + 3y = 12 touches the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 2$ at the point (3, 2).

Answer:

If the point of intersection of the line and ellipse is (3, 2), then it satisfies the equations.
Now,
2x + 3y = 12
⇒ 2(3) + 3(2) = 6 + 6 = 12
And
$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 2 \Rightarrow \frac{3^{2}}{9} + \frac{2^{2}}{4} = \frac{9}{9} + \frac{4}{4} = 1 + 1 = 2$
Thus, (3, 2) satisfies both the equations.
Hence, the statement is true.

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Question 40:

State whether the following statement is True or False.
The locus of the point of intersection of lines $\sqrt{3} x - y - 4 \sqrt{3} k = 0$ and $\sqrt{3} k x + k y - 4 \sqrt{3} = 0$ for different value of k is a hyperbola whose eccentricity is 2.

Answer:

The two given lines are: $\sqrt{3} x - y - 4 \sqrt{3} k = 0$ and $\sqrt{3} k x + k y - 4 \sqrt{3} = 0 .$
$\Rightarrow k = \frac{- 4 + \sqrt{3} x}{4 \sqrt{3}} and k = \frac{4 \sqrt{3}}{\sqrt{3} x + y} \Rightarrow \frac{- 4 + \sqrt{3} x}{4 \sqrt{3}} = \frac{4 \sqrt{3}}{\sqrt{3} x + y} \Rightarrow 3 x^{2} - y^{2} = 48 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{48} = 1 which is the equation of a hyperbola .$
Now,
a² = 16 ⇒ a = 4
b² = 48
$e^{2} = 1 + \frac{48}{16} = 4 \Rightarrow e = 2$
Hence, the statement is true.

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Question 41:

Fill in the Blank.
The equation of the circle having centre at (3, –4) and touching the line 5x + 12y – 12 = 0 is ________________ .

Answer:

The distance between the centre (3, –4) and the line is given as:
$d = |\frac{5 (3) + 12 (- 4) - 12}{\sqrt{25 + 144}}| = \frac{45}{13} \Rightarrow r = \frac{45}{13} units$
Thus, the equation of the circle is ${(x - 3)}^{2} + {(y + 4)}^{2} = {(\frac{45}{13})}^{2}$

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Question 42:

Fill in the Blank.
The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is ________________ .

Answer:

The sides of the triangles are given by:
y = x + 2.....(1)
3y = 4x.....(2)
2y = 3k.....(3)
Solving (1), (2) and (3), the three points of the triangle are obtained as follows:
A (6, 8)
B (4, 6)
C (0, 0)
Let the equation of the circle circumscribing the given triangle be x² + y² + 2gx + 2fy + c = 0.
Putting A(6, 8) in the equation,
36 + 64 + 12g + 16f + c = 0
⇒ 12g + 16f + c = –100
Putting B(4, 6) in the equation,
16 + 36 + 8g + 12f + c = 0
⇒ 8g + 12f + c = –52
Putting C(0, 0) in the equation,
0 + 0 + 0 + 0 + c = 0
⇒ c = 0
Therefore,
3g + 4f = –25
2g + 3f = –13
Solving these equations,
g = –23
f = 11
Hence, the equation of the circle is x² + y² – 46x + 22y = 0.

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Question 43:

Fill in the Blank.
An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____________.

Answer:

Let the equation of ellipse be:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, a = 3 and b = 2.
For an ellipse,
b² = a² (1 – e²)
$\Rightarrow e^{2} = 1 - \frac{4}{9} = \frac{5}{9} \Rightarrow e = \frac{\sqrt{5}}{3}$
Consider a point x on the ellipse.
Then, sx + s'x = 2a
$\begin{array}{rcl} Thus, length of the string & = & s x + s' x + s s' \\ = & 2 a + 2 a e \\ = & 2 (3) + 2 (3) (\frac{\sqrt{5}}{3}) \\ = & 6 + 2 \sqrt{5} \end{array}$

Page No 205:

Question 44:

Fill in the Blank.
The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is ________________ .

Answer:

Given that, length of minor axis is 1.
$\Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2} Now, a^{2} = b^{2} (1 - e^{2}) \Rightarrow \frac{1}{4} = b^{2} (1 - e^{2}) Also, be = 1 . \Rightarrow b^{2} e^{2} = 1 \Rightarrow \frac{1}{4} = b^{2} - 1 \Rightarrow b^{2} = \frac{5}{4}$
Hence, the equation of ellipse is:
$\frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{5}{4}} = 1 \Rightarrow \frac{4 x^{2}}{1} + \frac{4 y^{2}}{5} = 1$

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Question 45:

Fill in the Blank.
The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ________________ .

Answer:

Given that, the focus of parabola is at (–1, –2) and the equation of its directrix is x – 2y + 3 = 0.
Consider a point P(x, y) on the parabola. Let the length of the perpendicular from S on the directrix be SP.
$\Rightarrow \frac{{(x - 2 y + 3)}^{2}}{5} = {(x + 1)}^{2} + {(y + 2)}^{2}$
⇒ x² + 4y²+ 9 – 4xy –12y + 6x = 5[x² + 2x + 1 + y² + 4 + 4y]
⇒ 4x² + y² + 4x + 32y + 4xy + 16 = 0.

Page No 205:

Question 46:

Fill in the Blank.
The equation of the hyperbola with vertices at (0, ± 6) and eccentricity $\frac{5}{3}$ is ______________ and its foci are ________________ .

Answer:

Given that, the hyperbola has vertices as $(0, \pm 6)$ and eccentricity $\frac{5}{3} .$
⇒ b = 6
Now, for a hyperbola,
a² = b²(e² – 1)
$\begin{array}{rcl} \Rightarrow & a^{2} = 6^{2} ({(\frac{5}{3})}^{2} - 1) \\ = & 36 [\frac{25}{3} - 1] \\ = & 36 [\frac{16}{9}] \\ \Rightarrow & a = \frac{6 \times 4}{3} = 8 \end{array}$
Thus, the equation of the hyperbola is:
$\frac{x^{2}}{64} - \frac{y^{2}}{36} = - 1 \Rightarrow \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1$
Hence, the foci is $(0, \pm 10)$ and the equation of the hyperbola is $\frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 .$

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Question 47:

Choose the correct answer out of the given four options.
The area of the circle centred at (1, 2) and passing through (4, 6) is
(A) 5π
(B) 10π
(C) 25π
(D) none of these

Answer:

Given that, the centre of the circle is (1, 2) and it passes through (4, 6).
$\begin{array}{rcl} r & = & \sqrt{{(4 - 1)}^{2} + (6 - 2) 2} \\ = & \sqrt{9 + 16} \\ = & 5 units \end{array}$
Thus, area of the circle = π × 5 × 5
= 25π sq. units
Hence, the correct answer is option C.

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Question 48:

Equation of a circle which passes through (3, 6) and touches the axes is
(A) x²+ y²+ 6x + 6y + 3 = 0
(B) x²+ y²– 6x – 6y – 9 = 0
(C) x²+ y²– 6x – 6y + 9 = 0
(D) none of these

Answer:

For a = 3, the equation of the circle is :
(x – 3)² + (y – 3)² = 9
⇒ x² + 9 – 6x + y² + 9 – 6y = 9
⇒ x² + y² – 6x – 6y + 9 = 0

Hence, the correct answer is option (c).

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Question 49:

Choose the correct answer out of the given four options.
49. Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
(A) x²+ y²+ 13y = 0
(B) 3x²+ 3y²+ 13x + 3 = 0
(C) 6x²+ 6y²– 13x = 0
(D) x²+ y²+ 13x + 3 = 0

Answer:

Let the equation of the circle be (x – h)² + (y – k)²= r²
Let the centre of the circle be (0, a).
Thus, the equation of the circle becomes:
(x – 0)² + (y – a)²= a²
⇒ x²+ y²+ a² – 2ay = a²
⇒ x²+ y² – 2ay = 0

Since the point (2, 3) lies on the circle,

$a = \sqrt{{(2 - 0)}^{2} {(3 - a)}^{2}} \Rightarrow a = \sqrt{4 + 9 + a^{2} - 6 a} \Rightarrow a = \sqrt{13 + a^{2} - 6 a} \Rightarrow a^{2} = 13 + a^{2} - 6 a = a = \frac{13}{6}$

Putting the value of a in the equation of circle,

$x^{2} + y^{2} + 2 (\frac{13}{6}) y = 0 \Rightarrow 3 x^{2} + 8 y^{2} - 13 y = 0$

Hence, the correct answer is option (a).

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Question 50:

Choose the correct answer out of the given four options.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(A) x²+ y²= 9a²
(B) x²+ y²= 16a²
(C) x²+ y²= 4a²
(D) x²+ y²= a²

Answer:

Consider an equilateral triangle ABC with median AD = 3a.

The centroid of the circle and its centre are at the origin i.e. (0, 0).

Figure

Now,

AG : GD = 2 : 1

$\begin{array}{rcl} \Rightarrow & AG = \frac{2}{3} AD \\ = & \frac{2}{3} \times 3 a \\ = & 2 a \end{array}$

Therefore, the equation of the circle is:

Now,

(x – 0)² + (y – 0)² = (2a)²

⇒ x² + y² = 4a²

Hence, the correct answer is option (c).

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Question 51:

Choose the correct answer out of the given four options.
If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
(A) x²= –12y
(B) x²= 12y
(C) y²= –12x
(D) y²= 12x

Answer:

For a parabola,

$\sqrt{{(x - 0)}^{2} + {(y + 3)}^{2}} = |\frac{y - 3}{\sqrt{0^{2} + 1^{2}}}| \Rightarrow \sqrt{x^{2} + y^{2} + 9 + 6 y} = |y - 3|$

Squaring both sides,

x² + y² + 9 + 6y = y² + 9 – 6y

⇒ x² + 9 + 6y = 9 – 6y

⇒ x² = –12y

Hence, the correct answer is option (a).

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Question 52:

Choose the correct answer out of the given four options.
If the parabola y²= 4ax passes through the point (3, 2), then the length of its latus rectum is

(A) $\frac{2}{3}$

(B) $\frac{4}{3}$

(C) $\frac{1}{3}$

(D) 4

Answer:

Given that, the equation of parabola is y²= 4ax.

Since the parabola passes through (3, 2), (2)² = 4a × 3

$\Rightarrow 4 = 12 a \Rightarrow a = \frac{1}{3}$

Now,

$\begin{array}{rcl} Length of Latus rectum & = & 4 a \\ = & 4 \times \frac{1}{3} \\ = & \frac{4}{3} \end{array}$

Hence, the correct answer is option (b).

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Question 53:

Choose the correct answer out of the given four options.
If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is
(A) y²= 8 (x + 3)
(B) x²= 8 (y + 3)
(C) y²= – 8 (x + 3)
(D) y²= 8 ( x + 5)

Answer:

Given that, the vertex is (–3, 0).

⇒ a = –3

Now, equation of directrix is: x + 5 = 0

Figure

For a parabola,

AF = AD

$\Rightarrow - 3 = \frac{x_{1} - 5}{2} \Rightarrow x_{1} = - 1 and 0 = \frac{0 + y_{1}}{2} \Rightarrow y_{1} = 0$

Thus,

F (–1, 0) is the focus of the parabola

Now, $\sqrt{{(x + 1)}^{2} + {(y - 0)}^{2}} = |\frac{x + 5}{\sqrt{1^{2} + 0^{2}}}|$
Squaring both sides,

${(x + 1)}^{2} + y^{2} = {(x + 5)}^{2} \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 25 + 10 x \Rightarrow y^{2} = 10 x - 2 x + 24 \Rightarrow y^{2} = 8 x + 24 \Rightarrow y^{2} = 8 (x + 3)$

Hence, the correct answer is option (a).

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Question 54:

Choose the correct answer out of the given four options.
The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity $\frac{1}{2}$ is
(A) 7x²+ 2xy + 7y²– 10x + 10y + 7 = 0
(B) 7x²+ 2xy + 7y²+ 7 = 0
(C) 7x²+ 2xy + 7y²+ 10x – 10y – 7 = 0
(D) none

Answer:

The focus of the ellipse is (1, –1) and equation of its directrix is x – y – 3 = 0.
Consider P(x, y) on the parabola.
Now, $e = \frac{1}{2}$
$\Rightarrow \frac{PF}{Distance of P from directrix} = e \Rightarrow \frac{{(x - 1)}^{2} + {(y + 1)}^{2}}{|\frac{x - y - 3}{\sqrt{1^{2} + {(- 1)}^{2}}}|} = \frac{1}{2} \Rightarrow 2 \sqrt{x^{2} + 1 - 2 x + y^{2} + 1 + 2 y} = |\frac{x - y - 3}{\sqrt{2}}| \Rightarrow 4 (x^{2} + y^{2} - 2 x + 2 y + 2) = \frac{x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x}{2} \Rightarrow 8 x^{2} + 8 y^{2} - 16 x + 16 y + 16 = x^{2} + y^{2} - 2 x y + 6 y - 6 x + 9 \Rightarrow 7 x^{2} + 7 y^{2} + 2 x y - 10 x + 10 y + 7 = 0$

Hence, the correct answer is option (A).

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Question 55:

Choose the correct answer out of the given four options.
The length of the latus rectum of the ellipse 3x²+ y²= 12 is

(A) 4

(B) 3

(C) 8

(D) $\frac{4}{\sqrt{3}}$

Answer:

Given that, the equation of ellipse is:
3x²+ y²= 12
$\Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{12} = 1 \Rightarrow a^{2} = 4 b^{2} = 12 \Rightarrow a = 2 b = 2 \sqrt{3}$
Now,
$\begin{array}{rcl} Length of latus rectum & = & \frac{2 a^{2}}{b} \\ = & \frac{2 \times 4}{2 \sqrt{3}} \\ = & \frac{4}{\sqrt{3}} \end{array}$

Hence, the correct answer is option (D).

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Question 56:

Choose the correct answer out of the given four options.
If e is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b),$ then
(A) b²= a²(1 – e²)
(B) a²= b²(1 – e²)
(C) a²= b²(e²– 1)
(D) b²= a²(e²– 1)

Answer:

Given that, the equation of ellipse is:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a < b$
Now, for an ellipse,
$e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e^{2} = 1 - \frac{a^{2}}{b^{2}} \Rightarrow \frac{a^{2}}{b^{2}} = 1 - e^{2} \Rightarrow a^{2} = b^{2} (1 - e^{2})$

Hence, the correct answer is option (B).

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Question 57:

Choose the correct answer out of the given four options.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is

(A) $\frac{4}{3}$

(B) $\frac{4}{\sqrt{3}}$

(C) $\frac{2}{\sqrt{3}}$

(D) none of these

Answer:

Length of latus rectum $= \frac{2 b^{2}}{a} = 8$
⇒ b² = 4a
Now, distance between the foci = 2ae
Length of major axis = 2a
Length of minor axis = 2b
$\Rightarrow \frac{1}{2} (2 a e) = 2 b \Rightarrow b = \frac{a e}{2} \Rightarrow b^{2} = \frac{a^{2} e^{2}}{4} \Rightarrow 4 a = \frac{a^{2} e^{2}}{4} \Rightarrow a = \frac{16}{e^{2}}$
Now,
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow 4 a = a^{2} (e^{2} - 1) \Rightarrow \frac{4}{a} = e^{2} - 1 \Rightarrow \frac{4}{\frac{16}{e^{2}}} = e^{2} - 1 \Rightarrow \frac{e^{2}}{4} = e^{2} - 1 \Rightarrow e^{2} = \frac{4}{3} \Rightarrow e = \frac{2}{\sqrt{3}}$

Hence, the correct answer is option (C).

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Question 58:

Choose the correct answer out of the given four options.
The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2} .$ Its equation is

(A) x²– y²= 32

(B) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

(C) 2x – 3y²= 7

(D) none of these

Answer:

Distance between the foci = 2ae
Now, 2ae = 16
⇒ ae = 8
It is given that $e = \sqrt{2}$
$\Rightarrow \sqrt{2} a = 8 \Rightarrow a = 4 \sqrt{2}$
Now,
$b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = 32 (2 - 1) \Rightarrow b^{2} = 32$
Therefore, the equation of hyperbola is:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32$
Hence, the correct answer is option (A).

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Question 59:

Choose the correct answer out of the given four options.
Equation of the hyperbola with eccentricty $\frac{3}{2}$ and foci at (± 2, 0) is

(A) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}$

(B) $\frac{x^{2}}{9} - \frac{y^{2}}{9} = \frac{4}{9}$

(C) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

(D) none of these

Answer:

Given that, the eccentricity of hyperbola is $\frac{3}{2}$ and foci at (±2, 0).
Thus, ae = 2
$\Rightarrow a = \frac{4}{3}$
For a hyperbola.

$b^{2} = a^{2} (e^{2} - 1) \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) \Rightarrow b^{2} = \frac{20}{9}$

Therefore, equation of the hyperbola is:
$\frac{x^{2}}{{(\frac{4}{3})}^{2}} - \frac{y^{2}}{\frac{20}{9}} = 1 \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{20} = \frac{1}{9} \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9}$
Hence, the correct answer is option (A).

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