JEE Advanced Maths PYQ – Conic Sections (Hyperbola) | Previous Year Questions with Solutions

Q 1 :

The locus of the orthocentre of the triangle formed by the lines

$(1 + p) x - p y + p (1 + p) = 0,$

$(1 + q) x - q y + q (1 + q) = 0,$

and $y = 0,$ where $p \neq q$ , is [2009]

a hyperbola
a parabola
an ellipse
a straight line

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(4)

$The triangle is formed by the lines$

$A B : (1 + p) x - p y + p (1 + p) = 0$

$A C : (1 + q) x - q y + q (1 + q) = 0$

$B C : y = 0$

$So that the vertices of ∆ A B C are$

$A (p q, (p + 1) (q + 1)), B (- p, 0) and C (- q, 0)$

$Let H (h, k) be the orthocentre of ∆ A B C . Then as$ $A H ⊥ B C$ $and passes through A (p q, (p + 1) (q + 1))$

$∴$ $Equation of A H is x = p q$

$∴$ $h = p q$ ...(i)

$∵$ $B H is perpendicular to A C$

$∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \frac{k - 0}{h + p} \times \frac{1 + q}{q} = - 1$

$\Rightarrow \frac{k}{p q + p} \times \frac{1 + q}{q} = - 1 [using (i)]$

$∴$ $k = - p q$ ...(ii)

$From (i) and (ii), we observe that$ $h + k = 0$

$∴$ $Locus of (h, k) is x + y = 0,$ $which is a straight line.$

Q 2 :

Consider a branch of the hyperbola $x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$ with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is [2008]

$1 - \sqrt{\frac{2}{3}}$
$\sqrt{\frac{3}{2}} - 1$
$1 + \sqrt{\frac{2}{3}}$
$\sqrt{\frac{3}{2}} + 1$

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(2)

$Given hyperbola is$

$x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$

$\Rightarrow (x^{2} - 2 \sqrt{2} x + 2) - 2 (y^{2} + 2 \sqrt{2} y + 2) = 6 + 2 - 4$

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

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

$∴$ $a = 2, b = \sqrt{2} \Rightarrow e = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$

$Clearly ∆ A B C is a right triangle.$

$∴$ $Area (∆ A B C) = \frac{1}{2} \times A C \times B C$

$= \frac{1}{2} (a e - a) \times \frac{b^{2}}{a}$

$= \frac{1}{2} (e - 1) \times b^{2}$

$= \frac{1}{2} (\sqrt{\frac{3}{2}} - 1) \times 2$

$= \sqrt{\frac{3}{2}} - 1$

Q 3 :

Let $a$ and $b$ be non-zero real numbers. Then, the equation $(a x^{2} + b y^{2} + c) (x^{2} - 5 x y + 6 y^{2}) = 0$ represents [2008]

four straight lines, when $c = 0$ and $a, b$ are of the same sign.
two straight lines and a circle, when $a = b$ , and $c$ is of sign opposite to that of $a$ .
two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$ .
a circle and an ellipse, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$ .

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(2)

$∵$ $x^{2} - 5 x y + 6 y^{2} = 0$

$\Rightarrow (x - 3 y) (x - 2 y) = 0$

$∴$ $x^{2} - 5 x y + 6 y^{2} = 0 represents a pair of straight lines given by$ $x - 3 y = 0 and x - 2 y = 0 .$

$Also a x^{2} + b y^{2} + c = 0 will represent a circle if a = b$ $and c is of sign opposite to that of a .$

Q 4 :

A hyperbola, having the transverse axis of length $2 \sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12 .$ Then its equation is [2007]

$x^{2} c o s e c^{2} θ - y^{2} \sec^{2} θ = 1$
$x^{2} \sec^{2} θ - y^{2} c o s e c^{2} θ = 1$
$x^{2} \sin^{2} θ - y^{2} \cos^{2} θ = 1$
$x^{2} \cos^{2} θ - y^{2} \sin^{2} θ = 1$

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(1)

$The length of transverse axis = 2 \sin θ = 2 a$

$\Rightarrow a = \sin θ$

$Also the given ellipse is 3 x^{2} + 4 y^{2} = 12$

$\Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1 \Rightarrow a^{2} = 4, b^{2} = 3$

$∴$ $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

$∴$ $Focus of ellipse = (2 \times \frac{1}{2}, 0) \Rightarrow (1, 0)$

$Since hyperbola is confocal with ellipse, therefore focus of hyperbola$

$= (1, 0) \Rightarrow a e = 1 \Rightarrow \sin θ \times e = 1$

$\Rightarrow e = c o s e c θ$

$∴$ $b^{2} = a^{2} (e^{2} - 1)$ $= \sin^{2} θ (\cos e c^{2} θ - 1) = \cos^{2} θ$

$∴$ $Equation of hyperbola is$

$\frac{x^{2}}{\sin^{2} θ} - \frac{y^{2}}{\cos^{2} θ} = 1$

$\Rightarrow x^{2} c o s e c^{2} θ - y^{2} \sec^{2} θ = 1$

Q 5 :

If the line $2 x + \sqrt{6} y = 2$ touches the hyperbola $x^{2} - 2 y^{2} = 4,$ then the point of contact is [2004]

$(- 2, \sqrt{6})$
$(- 5, 2 \sqrt{6})$
$(\frac{1}{2}, \frac{1}{\sqrt{6}})$
$(4, - \sqrt{6})$

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(4)

$Equation of tangent to hyperbola x^{2} - 2 y^{2} = 4 at any point (x_{1}, y_{1}) is$

$x x_{1} - 2 y y_{1} = 4$ ....(i)

$But the given tangent is 2 x + \sqrt{6} y = 2$

$On comparing equation (i) with 2 x + \sqrt{6} y = 2$

$i.e., 4 x + 2 \sqrt{6} y = 4, we get$

$x_{1} = 4 and - 2 y_{1} = 2 \sqrt{6}$ $\Rightarrow (4, - \sqrt{6}) is the required point of contact.$

Q 6 :

For hyperbola $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = 1,$ which of the following remains constant with change in $' α'$ [2003]

abscissae of vertices
abscissae of foci
eccentricity
directrix

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(2)

$Given equation of hyperbola is$

$\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = 1$ $\Rightarrow a = \cos α, b = \sin α$

$\Rightarrow e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \tan^{2} α} = \sec α$

$\Rightarrow a e = \cos α \cdot \sec α = 1$ $∴$ $Foci (\pm 1, 0)$

$Hence, foci remain constant with respect to α .$

Q 7 :

The equation of the common tangent to the curves $y^{2} = 8 x$ and $x y = - 1$ is [2002]

$3 y = 9 x + 2$
$y = 2 x + 1$
$2 y = x + 8$
$y = x + 2$

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(4)

$Given equation of curves are$

$y^{2} = 8 x ...(i)$

and $x y = - 1 ...(ii)$

If $m$ is the slope of tangent to (i), then equation of tangent is $y = m x + \frac{2}{m}$

If this tangent is also a tangent to (ii), then putting value of $y$ in curve (ii)

$x (m x + \frac{2}{m}) = - 1$ $\Rightarrow m x^{2} + \frac{2}{m} x + 1 = 0$ $\Rightarrow m^{2} x^{2} + 2 x + m = 0$

We should get repeated roots for the equation (condition of tangency)

$\Rightarrow D = 0$ $∴$ ${(2)}^{2} - 4 m^{2} \cdot m = 0$ $\Rightarrow m^{3} = 1$ $\Rightarrow m = 1$

Hence equation of required tangent is $y = x + 2$

Q 8 :

Consider the hyperbola $\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1$ with foci at $S$ and $S_{1}$ , where $S$ lies on the positive $x$ -axis. Let $P$ be a point on the hyperbola, in the first quadrant. Let $∠ S P S_{1} = α,$ with $α < \frac{π}{2}$ . The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S_{1} P$ at $P_{1}$ . Let $δ$ be the distance of $P$ from the straight line $S P_{1}$ and $β = S_{1} P .$ Then the greatest integer less than or equal to $\frac{β δ}{9} \sin \frac{α}{2}$ is ________. [2022]

(7)

In $∆ S_{1} Q P,$ $\sin \frac{α}{2} = \frac{S_{1} Q}{β}$

$\Rightarrow S_{1} Q = β \sin \frac{α}{2}$

Product of distances of any tangent from two foci $= b^{2}$

$δ \cdot S_{1} Q = δ \times β \sin \frac{α}{2} = b^{2}$

So, $\frac{β δ \sin \frac{α}{2}}{9} = \frac{b^{2}}{9} = \frac{64}{9}$

$∴$ $[\frac{64}{9}] = 7_{s s}$

Q 9 :

The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ If this line passes through the point of intersection of the nearest directrix and the $x$ -axis, then the eccentricity of the hyperbola is ____. [2010]

(2)

Intersection point of nearest directrix $x = \frac{a}{e}$ and $x$ -axis is $(\frac{a}{e}, 0)$

Since $2 x + y = 1$ passes through $(\frac{a}{e}, 0)$ ,

$∴$ $\frac{2 a}{e} = 1$ $\Rightarrow a = \frac{e}{2}$

Also $y = - 2 x + 1$ is a tangent to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$∴$ $1 = a^{2} {(- 2)}^{2} - b^{2}$ $\Rightarrow 4 a^{2} - b^{2} = 1$

$\Rightarrow 4 a^{2} - a^{2} (e^{2} - 1) = 1$ $\Rightarrow 4 \times \frac{e^{2}}{4} - \frac{e^{2}}{4} (e^{2} - 1) = 1$

$\Rightarrow e^{2} = 4$ as $e > 1$ for hyperbola, $\Rightarrow e = 2$

Q 10 :

Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a .$ Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ Suppose the tangent to the hyperbola at $P$ passes through the point (1, 0) and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $Δ$ denote the area of the triangle formed by the tangent at $P$ , the normal at $P$ and the $x$ -axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE? [2020]

$1 < e < \sqrt{2}$
$\sqrt{2} < e < 2$
$Δ = a^{4}$
$Δ = b^{4}$

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Select one or more options

(1, 4)

Let $P (a \sec θ, b \tan θ)$

Equation of tangent at $P$

$\frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1$

$∵$ $(1, 0) lies on the tangent$

so $a = \sec θ$

Equation of normal at $P$

$\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$

since normal at $P$ makes equal intercept on co-ordinate axes,

$∴$ $Slope of normal is - 1$

so $- \frac{a}{b} \sin θ = - 1$

$\Rightarrow b = \tan θ,$ hence, $a^{2} - b^{2} = 1 ...(i)$

$∵$ $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ $= \sqrt{1 + \frac{a^{2} - 1}{a^{2}}}$ $= \sqrt{2 - \frac{1}{a^{2}}} from (i)$

Since $a > 1,$ so $e \in (1, \sqrt{2})$

Hence, option (1) is true.

Area of $∆ P A B$ $= \frac{1}{2} A P \cdot P B$

$= \frac{1}{2} \sqrt{{(a^{2} - 1)}^{2} + {(b^{2})}^{2} \times \sqrt{2 b}}$ ⁴

$= \frac{1}{2} \sqrt{2 b^{4}} \sqrt{2 b^{4}}$ $= b^{4} from (i)$

Hence, option (4) is true.

Q 11 :

If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{16} = 1,$ then which of the following cannot be sides of a right angled triangle? [2017]

$a, 4, 1$
$a, 4, 2$
$2 a, 8, 1$
$2 a, 4, 1$

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Select one or more options

(1, 2, 3)

Given $2 x - y + 1 = 0$ i.e. $y = 2 x + 1$ is a tangent to hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{16} = 1$

$∴$ $c^{2} = a^{2} m^{2} - b^{2}$

$\Rightarrow 1^{2} = a^{2} \times 2^{2} - 16$

$\Rightarrow a^{2} = \frac{17}{4}$ $\Rightarrow a = \frac{\sqrt{17}}{2}$

$∴$ $a, 4, 1; a, 4, 2; 2 a, 8, 1$ i.e. $\frac{\sqrt{17}}{2}, 4, 1; \frac{\sqrt{17}}{2}, 4, 2; \sqrt{17}, 8, 1$ cannot be the sides of a right triangle.

Q 12 :

Consider the hyperbola $H : x^{2} - y^{2} = 1$ and a circle $S$ with center $N (x_{2}, 0)$ . Suppose that $H$ and $S$ touch each other at a point $P (x_{1}, y_{1})$ with $x_{1} > 1$ and $y_{1} > 0$ . The common tangent to $H$ and $S$ at $P$ intersects the $x$ -axis at point $M$ . If $(l, m)$ is the centroid of the triangle PMN, then the correct expression(s) is/are [2015]

$\frac{d l}{d x_{1}} = 1 - \frac{1}{3 x_{1}^{2}} for x_{1} > 1$
$\frac{d m}{d x_{1}} = \frac{x_{1}}{3 (\sqrt{x_{1}^{2} - 1})} for x_{1} > 1$
$\frac{d l}{d x_{1}} = 1 + \frac{1}{3 x_{1}^{2}} for x_{1} > 1$
$\frac{d m}{d y_{1}} = \frac{1}{3} for y_{1} > 0$

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Select one or more options

(1, 2, 4)

$H : x^{2} - y^{2} = 1$ is a hyperbola and $S : Circle with centre N (x_{2}, 0) .$ Common tangent to $H$ and $S$ at $P (x_{1}, y_{1})$ is

$x x_{1} - y y_{1} = 1$ $\Rightarrow m_{1} = \frac{x_{1}}{y_{1}}$

Now, radius of circle $S$ with centre $N (x_{2}, 0)$ through the point of contact $(x_{1}, y_{1})$ is perpendicular to the tangent.

$∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \frac{x_{1}}{y_{1}} \times \frac{0 - y_{1}}{x_{2} - x_{1}} = - 1$

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

$∵$ $M$ is the point of intersection of tangent at $P$ and $x$ -axis.

$∴$ $M (\frac{1}{x_{1}}, 0)$ $∵$ $Centroid of ∆ P M N is (ℓ, m)$

$∴$ $x_{1} + \frac{1}{x_{1}} + x_{2} = 3 ℓ$ and $y_{1} = 3 m$

$\Rightarrow \frac{1}{3} (3 x_{1} + \frac{1}{x_{1}}) = ℓ$ and $\frac{y_{1}}{3} = m [∵ x_{2} = 2 x_{1}]$

$∴$ $\frac{d ℓ}{d x_{1}} = 1 - \frac{1}{3 x_{1}^{2}}, \frac{d m}{d y_{1}} = \frac{1}{3}$

Also $(x_{1}, y_{1})$ lies on $H$ , $∴$ $x_{1}^{2} - y_{1}^{2} = 1$ $\Rightarrow y_{1} = \sqrt{x_{1}^{2} - 1}$

$∴$ $m = \frac{1}{3} \sqrt{x_{1}^{2} - 1}$ $∴$ $\frac{d m}{d x_{1}} = \frac{x_{1}}{3 \sqrt{x_{1}^{2} - 1}}$

Q 13 :

Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1,$ parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are [2012]

$(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}})$
$(- \frac{9}{2 \sqrt{2}}, - \frac{1}{\sqrt{2}})$
$(3 \sqrt{3}, - 2 \sqrt{2})$
$(- 3 \sqrt{3}, 2 \sqrt{2})$

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Select one or more options

(1, 2)

If slope of tangents to hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $m$ , then equation of tangent to the hyperbola is

$y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$ with the points of contact $(\frac{\pm a^{2} m}{\sqrt{a^{2} m^{2} - b^{2}}}, \frac{\pm b^{2}}{\sqrt{a^{2} m^{2} - b^{2}}})$

$∵$ $Tangent to hyperbola \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 is parallel to 2 x - y = 1$

$∴$ $Slope of tangent = 2$

$∴$ $Points of contact are$ $(\frac{\pm 9 \times 2}{\sqrt{9 \times 4 - 4}}, \frac{\pm 4}{\sqrt{9 \times 4 - 4}})$

i.e. $(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}})$ and $(\frac{- 9}{2 \sqrt{2}}, \frac{- 1}{\sqrt{2}})$

Q 14 :

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4 .$ If the hyperbola passes through a focus of the ellipse, then [2011]

the equation of the hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$
a focus of the hyperbola is $(2, 0)$
the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
the equation of the hyperbola is $x^{2} - 3 y^{2} = 3$

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Select one or more options

(2, 4)

Given ellipse $x^{2} + 4 y^{2} = 4$ $\Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$

Its focus is $(\pm \sqrt{3}, 0)$ and eccentricity, $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

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

Its eccentricity is $\sqrt{1 + \frac{b^{2}}{a^{2}}}$

According to the question, $\sqrt{1 + \frac{b^{2}}{a^{2}}} = \frac{2}{\sqrt{3}}$ $\Rightarrow \frac{b}{a} = \frac{1}{\sqrt{3}}$

As hyperbola passes through the eccentricity of the ellipse $(\pm \sqrt{3}, 0)$

$∴$ $\frac{3}{a^{2}} = 1$ or $a = \sqrt{3}$ $∴$ $b = 1$ and focus of hyperbola $(\pm 2, 0)$

$∴$ Equation of hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{1} = 1$ $\Rightarrow x^{2} - 3 y^{2} = 3$

Q 15 :

An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

equation of ellipse is $x^{2} + 2 y^{2} = 2$
the foci of ellipse are $(\pm 1, 0)$
equation of ellipse is $x^{2} + 2 y^{2} = 4$
the foci of ellipse are $(\pm \sqrt{2}, 0)$

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Select one or more options

(1, 2)

The given hyperbola is

$x^{2} - y^{2} = \frac{1}{2} ...(i)$

which is a rectangular hyperbola $(∵ a = b)$ $∴$ $e = \sqrt{2}$

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Its eccentricity $= \frac{1}{\sqrt{2}}$

$∴$ $b^{2} = a^{2} (1 - \frac{1}{2})$ $\Rightarrow b^{2} = \frac{a^{2}}{2}$

Hence, the equation of ellipse becomes

$x^{2} + 2 y^{2} = a^{2} ...(ii)$

Let the hyperbola (i) and ellipse (ii) intersect each other at $P (x_{1}, y_{1})$

Then slope of hyperbola (i) at $P$ is given by

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

and that of ellipse (ii) at $P$ is

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

As the two curves intersect orthogonally,

$∴$ $m_{1} m_{2} = - 1$

$\Rightarrow \frac{x_{1}}{y_{1}} (- \frac{x_{1}}{2 y_{1}}) = - 1$ $\Rightarrow x_{1}^{2} = 2 y_{1}^{2} ...(iii)$

Also $P (x_{1}, y_{1})$ lies on $x^{2} - y^{2} = \frac{1}{2}$

$∴$ $x_{1}^{2} - y_{1}^{2} = \frac{1}{2} ...(iv)$

On solving (iii) and (iv), we get $y_{1}^{2} = \frac{1}{2}$ and $x_{1}^{2} = 1$

Also $P (x_{1}, y_{1})$ lies on ellipse $x^{2} + 2 y^{2} = a^{2}$

$∴$ $x_{1}^{2} + 2 y_{1}^{2} = a^{2}$ $\Rightarrow 1 + 1 = a^{2}$ or $a^{2} = 2$

$∴$ Equation of required ellipse is $x^{2} + 2 y^{2} = 2,$ whose foci are $(\pm a e, 0) = (\pm \sqrt{2} \times \frac{1}{\sqrt{2}}, 0) = (\pm 1, 0)$

Q 16 :

Let a hyperbola passes through the focus of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 .$ The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then [2006]

the equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
the equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1$
focus of hyperbola is $(5, 0)$
vertex of hyperbola is $(5 \sqrt{3}, 0)$

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Select one or more options

(1, 3)

For the given ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ $\Rightarrow e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

$\Rightarrow$ Eccentricity of hyperbola $= \frac{5}{3}$

Let the hyperbola be $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ then

$B^{2} = A^{2} (\frac{25}{9} - 1)$ $= \frac{16}{9} A^{2}$ $∴$ $\frac{x^{2}}{A^{2}} - \frac{9 y^{2}}{16 A^{2}} = 1$

As it passes through focus of ellipse i.e. $(3, 0)$

$∴$ $we get A^{2} = 9$ $\Rightarrow B^{2} = 16$

$∴$ Equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Its focus is (5, 0) and vertex is (3, 0)

Q 17 :

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ where $a > b > 0$ , be a hyperbola in the $x y$ -plane whose conjugate axis $L M$ subtends an angle of $60 °$ at one of its vertices $N$ . Let the area of the triangle $L M N$ be $4 \sqrt{3}$ . [2018]

List I List II

P. The length of the conjugate axis of $H$ is 1. $8$

Q. The eccentricity of $H$ is 2. $\frac{4}{\sqrt{3}}$

R. The distance between the foci of $H$ is 3. $\frac{2}{\sqrt{3}}$

S. The length of the latus rectum of $H$ is 4. $4$

The correct option is:

P → 4; Q → 2; R → 1; S → 3
P → 4; Q → 3; R → 1; S → 2
P → 4; Q → 1; R → 3; S → 2
P → 3; Q → 4; R → 2; S → 1

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(2)

$Area of ∆ L M N = 4 \sqrt{3} (given)$

$\Rightarrow \frac{1}{2} \times L M \times O N = 4 \sqrt{3}$ $\Rightarrow \frac{1}{2} (2 b) (\sqrt{3} b) = 4 \sqrt{3}$

$∴$ $b^{2} = 4$ $\Rightarrow b = 2$

So, length of the conjugate axis of hyperbola $= 2 b = 4$

Now, $\tan 30 ° = \frac{O L}{O N} = \frac{b}{a}$ $\Rightarrow a = \sqrt{3} b$ $\Rightarrow a = 2 \sqrt{3}$

$∴$ $b^{2} = a^{2} (e^{2} - 1)$ $\Rightarrow 4 = 12 (e^{2} - 1)$ $\Rightarrow e^{2} = 1 + \frac{1}{3} = \frac{4}{3}$

$∴$ $The eccentricity of hyperbola$ $e = \frac{2}{\sqrt{3}}$

$The distance between the foci of hyperbola$ $= 2 a e$

$= 2 \times 2 \sqrt{3} \times \frac{2}{\sqrt{3}}$ $= 8$

$And length of latus rectum of hyperbola$

$= \frac{2 b^{2}}{a}$ $= \frac{2 \times 4}{2 \sqrt{3}}$ $= \frac{4}{\sqrt{3}}$

Q 18 :

By appropriately matching the information given in the three columns of the following table. Column 1, 2, and 3 contain conics, equations of tangents to the conics and points of contact, respectively. [2017]

Column 1 Column 2 Column 3

(I) $x^{2} + y^{2} = a^{2}$ (i) $m y = m^{2} x + a$ (P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$

(II) $x^{2} + a^{2} y^{2} = a^{2}$ (ii) $y = m x + a \sqrt{m^{2} + 1}$ (Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$

(III) $y^{2} = 4 a x$ (iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$ (R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$

(IV) $x^{2} - a^{2} y^{2} = a^{2}$ (iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$ (S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

Q. For $a = \sqrt{2}$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact $(- 1, 1)$ , then which of the following options is the only correct combination for obtaining its equation?

(I)(i)(P)
(I)(ii)(Q)
(II)(ii)(Q)
(III)(i)(P)

1

2

3

4

(2)

For $a = \sqrt{2}$ and point of contact $(- 1, 1)$

Equation of circle is satisfied

$x^{2} + y^{2} = 2$

then equation of tangent is

$- x + y = 2$ $\Rightarrow m = 1$ and point of contact

$(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$ $= (\frac{- \sqrt{2}}{\sqrt{2}}, \frac{\sqrt{2}}{\sqrt{2}})$ $= (- 1, 1)$

$∴$ $(I), (i i), (Q) is the correct combination.$

Q 19 :

By appropriately matching the information given in the three columns of the following table. Column 1, 2, and 3 contain conics, equations of tangents to the conics and points of contact, respectively. [2017]

Column 1 Column 2 Column 3

(I) $x^{2} + y^{2} = a^{2}$ (i) $m y = m^{2} x + a$ (P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$

(II) $x^{2} + a^{2} y^{2} = a^{2}$ (ii) $y = m x + a \sqrt{m^{2} + 1}$ (Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$

(III) $y^{2} = 4 a x$ (iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$ (R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$

(IV) $x^{2} - a^{2} y^{2} = a^{2}$ (iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$ (S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

Q. If a tangent to a suitable conic (column 1) is found to be $y = x + 8$ and its point of contact is (8, 16), then which of the following options is the only correct combination?

(I)(ii)(Q)
(II)(iv)(R)
(III)(i)(P)
(III)(ii)(Q)

1

2

3

4

(3)

Tangent $y = x + 8$ $\Rightarrow m = 1$ Point $(8, 16)$

$∴$ Both the coordinates as well as $m$ , are positive. The only possibility of point is $(\frac{a}{m^{2}}, \frac{2 a}{m}) = (8, 16)$ $∴$ $a = 8$

Also it satisfies the equation of curve $y^{2} = 4 a x$ for the point $(8, 16)$

And equation of tangent $m y = m^{2} x + a$ is satisfied by $m = 1$ and $a = 8$

$∴$ $(I I I), (i), (P) is the correct combination.$

Q 20 :

By appropriately matching the information given in the three columns of the following table. Column 1, 2, and 3 contain conics, equations of tangents to the conics and points of contact, respectively. [2017]

Column 1 Column 2 Column 3

(I) $x^{2} + y^{2} = a^{2}$ (i) $m y = m^{2} x + a$ (P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$

(II) $x^{2} + a^{2} y^{2} = a^{2}$ (ii) $y = m x + a \sqrt{m^{2} + 1}$ (Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$

(III) $y^{2} = 4 a x$ (iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$ (R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$

(IV) $x^{2} - a^{2} y^{2} = a^{2}$ (iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$ (S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

Q. The tangent to a suitable conic (Column 1) at $(\sqrt{3}, \frac{1}{2})$ is found to be $\sqrt{3} x + 2 y = 4,$ then which of the following options is the only correct combination?

(IV)(iii)(S)
(IV)(iv)(S)
(II)(iii)(R)
(II)(iv)(R)

1

2

3

4

(4)

Point of contact $(\sqrt{3}, \frac{1}{2})$ and tangent $\sqrt{3} x + 2 y = 4$

$∴$ $m = - \frac{\sqrt{3}}{2}$

$∴$ $Both the coordinates are positive and m is negative.$

$The possibilities for points are$

$Q (\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$

or $R (\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$

For point $Q$ $(\frac{\sqrt{3} a}{\sqrt{7}}, \frac{2 a}{\sqrt{7}}) = (\sqrt{3}, \frac{1}{2})$

We get $a = \sqrt{7}$ and $a = \frac{\sqrt{7}}{4},$ which is not possible.

For point $R$ $(\frac{a^{2} \sqrt{3}}{\sqrt{3 a^{2} + 4}}, \frac{2}{\sqrt{3 a^{2} + 4}}) = (\sqrt{3}, \frac{1}{2})$

$\Rightarrow \frac{a^{2}}{\sqrt{3 a^{2} + 4}} = 1$ and $\frac{2}{\sqrt{3 a^{2} + 4}} = \frac{1}{2}$

$\Rightarrow a^{4} - 3 a^{2} - 4 = 0$ and $3 a^{2} = 12$

$∴$ $a^{2} = 4$

Also for $a^{2} = 4$ , equation of ellipse

$x^{2} + a^{2} y^{2} = a^{2}$ is satisfied for the point $(\sqrt{3}, \frac{1}{2})$

$∴$ $II, (iv), R is the correct combination.$

Q 21 :

Match the conics in Column I with the statements/expressions in Column II. [2009]

Column I Column II

(A) Circle (p) The locus of the point $(h, k)$ for which the line $h x + k y = 1$ touches the circle $x^{2} + y^{2} = 4$

(B) Parabola (q) Points $z$ in the complex plane satisfying $| z + 2 |$ $-$ $| z - 2 |$ $= \pm 3$

(C) Ellipse (r) Points of the conic have parametric representation $x = \sqrt{3} (\frac{1 - t^{2}}{1 + t^{2}}), y = \frac{2 t}{1 + t^{2}}$

(D) Hyperbola (s) The eccentricity of the conic lies in the interval $1 \leq x < \infty$

(t) Points $z$ in the complex plane satisfying $Re {(z + 1)}^{2} = | z |^{2} + 1$

$A \to (p); B \to (s, t); C \to (r); D \to (q, s)$
$A \to (q, s); B \to (s, t); C \to (r); D \to (p)$
$A \to (q, s); B \to (r); C \to (s, t); D \to (p)$
$A \to (r); B \to (q, s); C \to (s, t); D \to (p)$

1

2

3

4

(1)

$(p) As the line h x + k y = 1, touches the circle x^{2} + y^{2} = 4$

$∴$ $Length of perpendicular from centre (0, 0) of circle to the line$ $= radius of the circle$

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

$∴$ $Locus of (h, k) is x^{2} + y^{2} = \frac{1}{4}, which is a circle.$

$(q) We know that if | z - z_{1} | - | z - z_{2} | = k,$

where $| k | < | z_{1} - z_{2} |,$ then $z$ traces a hyperbola.

Here $| z + 2 | - | z - 2 | = \pm 3$

$∴$ $Locus of z is a hyperbola.$

$(r) Given : x = \sqrt{3} (\frac{1 - t^{2}}{1 + t^{2}}), y = \frac{2 t}{1 + t^{2}}$

$\Rightarrow \frac{x}{\sqrt{3}} = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$

On squaring and adding, we get

$\frac{x^{2}}{3} + y^{2} = \frac{{(1 - t^{2})}^{2} + 4 t^{2}}{{(1 + t^{2})}^{2}} = 1$ $\Rightarrow \frac{x^{2}}{3} + \frac{y^{2}}{1} = 1$

which is the equation of an ellipse.

$(s) We know, eccentricity of a parabola = 1$

$and eccentricity of an ellipse < 1$

$and eccentricity of a hyperbola > 1 .$

$Hence, the conics whose eccentricity lies in$ $1 \leq x < \infty$ $are parabola and hyperbola.$

$(t) Let z = x + i y$

$∴$ $Re {[(x + 1) + i y]}^{2} = x^{2} + y^{2} + 1$

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

$∴$ $y^{2} = x,$ which is a parabola.

Q 22 :

Match the statements in Column I with the properties in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2007]

Column I Column II

(A) Two intersecting circles (p) have a common tangent

(B) Two mutually external circles (q) have a common normal

(C) Two circles, one strictly inside the other (r) do not have a common tangent

(D) Two branches of a hyperbola (s) do not have a common normal

$(A) \to (q, r); (B) \to (p, q); (C) \to (q, r); (D) \to (p, q)$
$(A) \to (p, q); (B) \to (p, q); (C) \to (q, r); (D) \to (q, r)$
$(A) \to (q, r); (B) \to (q, r); (C) \to (p, q); (D) \to (p, q)$
$(A) \to (p, q); (B) \to (p, q); (C) \to (q, r); (D) \to (q, r)$

1

2

3

4

(2)

$(A) - p, q$

It is clear from the figure that two intersecting circles have a common tangent and a common normal joining the centres.

$(B) - p, q$

$(C) - q, r$

Two circles, when one is strictly inside the other, have a common normal $C_{1} C_{2}$ but no common tangent.

$(D) - q, r$

Two branches of hyperbola have no common tangent but have a common normal joining $S_{1} S_{2} .$

Q 23 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of the circle with $A B$ as its diameter is

$x^{2} + y^{2} - 12 x + 24 = 0$
$x^{2} + y^{2} + 12 x + 24 = 0$
$x^{2} + y^{2} + 24 x - 12 = 0$
$x^{2} + y^{2} - 24 x - 12 = 0$

1

2

3

4

(1)

Given a circle

$x^{2} + y^{2} - 8 x = 0 ...(i)$

and a hyperbola $4 x^{2} - 9 y^{2} - 36 = 0 ...(ii)$

To find their point of intersection, substitute the value of $y^{2}$ from equation (i) in equation (ii), we get

$4 x^{2} - 9 (8 x - x^{2}) = 36$ $\Rightarrow 13 x^{2} - 72 x - 36 = 0$

$\Rightarrow x = 6, - \frac{6}{13}$ $\Rightarrow y^{2} = 12, \frac{- 48}{13} - \frac{36}{169} (not possible)$

$∴$ $(6, 2 \sqrt{3}) and (6, - 2 \sqrt{3}) are points of intersection.$

$∴$ $Equation of required circle is$

$(x - 6) (x - 6) + (y - 2 \sqrt{3}) (y + 2 \sqrt{3}) = 0$

$\Rightarrow x^{2} + y^{2} - 12 x + 24 = 0$

Q 24 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

$2 x - \sqrt{5} y - 20 = 0$
$2 x - \sqrt{5} y + 4 = 0$
$3 x - 4 y + 8 = 0$
$4 x - 3 y + 4 = 0$

1

2

3

4

(2)

Any tangent to $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ is $\frac{x \sec α}{3} - \frac{y \tan α}{2} = 1$

It touches circle with center $(4, 0)$ and radius $= 4$

$∴$ $\frac{\frac{4 \sec α - 3}{3}}{\sqrt{\frac{\sec^{2} α}{9} + \frac{\tan^{2} α}{4}}} = 4$

$\Rightarrow 16 \sec^{2} α - 24 \sec α + 9 = 144 (\frac{\sec^{2} α}{9} + \frac{\tan^{2} α}{4})$

$\Rightarrow 12 \sec^{2} α + 8 \sec α - 15 = 0$ $\Rightarrow \sec α = \frac{5}{6} or - \frac{3}{2}$

since $\sec α = \frac{5}{6} < 1$ is not possible.

$∴$ $\sec α = - \frac{3}{2}$ $\Rightarrow \tan α = \pm \frac{\sqrt{5}}{2}$

$∴$ $Slope of tangent = \frac{2 \sec α}{3 \tan α} = \frac{2 (- 3 / 2)}{3 (- \sqrt{5} / 2)} = \frac{2}{\sqrt{5}}$ $(for +ve value of \tan α)$

$∴$ $Equation of tangent is$ $\frac{- x}{2} + \frac{y \sqrt{5}}{4} = 1$

$\Rightarrow 2 x - \sqrt{5} y + 4 = 0$

Topic Question Set

Q 1 :

The locus of the orthocentre of the triangle formed by the lines

$(1 + p) x - p y + p (1 + p) = 0,$

$(1 + q) x - q y + q (1 + q) = 0,$

and $y = 0,$ where $p \neq q$ , is [2009]

Q 2 :

Consider a branch of the hyperbola $x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$ with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is [2008]

Q 3 :

Let $a$ and $b$ be non-zero real numbers. Then, the equation $(a x^{2} + b y^{2} + c) (x^{2} - 5 x y + 6 y^{2}) = 0$ represents [2008]

Q 4 :

A hyperbola, having the transverse axis of length $2 \sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12 .$ Then its equation is [2007]

Q 5 :

If the line $2 x + \sqrt{6} y = 2$ touches the hyperbola $x^{2} - 2 y^{2} = 4,$ then the point of contact is [2004]

Q 6 :

For hyperbola $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = 1,$ which of the following remains constant with change in $' α'$ [2003]

Q 7 :

The equation of the common tangent to the curves $y^{2} = 8 x$ and $x y = - 1$ is [2002]

Q 9 :

The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ If this line passes through the point of intersection of the nearest directrix and the $x$ -axis, then the eccentricity of the hyperbola is ____. [2010]

Q 11 :

If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{16} = 1,$ then which of the following cannot be sides of a right angled triangle? [2017]

Q 13 :

Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1,$ parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are [2012]

Q 14 :

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4 .$ If the hyperbola passes through a focus of the ellipse, then [2011]

Q 15 :

An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

Q 23 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of the circle with $A B$ as its diameter is

Q 24 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

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	List I		List II
P.	The length of the conjugate axis of $H$ is	1.	$8$
Q.	The eccentricity of $H$ is	2.	$\frac{4}{\sqrt{3}}$
R.	The distance between the foci of $H$ is	3.	$\frac{2}{\sqrt{3}}$
S.	The length of the latus rectum of $H$ is	4.	$4$

	Column 1		Column 2		Column 3
(I)	$x^{2} + y^{2} = a^{2}$	(i)	$m y = m^{2} x + a$	(P)	$(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II)	$x^{2} + a^{2} y^{2} = a^{2}$	(ii)	$y = m x + a \sqrt{m^{2} + 1}$	(Q)	$(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III)	$y^{2} = 4 a x$	(iii)	$y = m x + \sqrt{a^{2} m^{2} - 1}$	(R)	$(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV)	$x^{2} - a^{2} y^{2} = a^{2}$	(iv)	$y = m x + \sqrt{a^{2} m^{2} + 1}$	(S)	$(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

	Column I		Column II
(A)	Circle	(p)	The locus of the point $(h, k)$ for which the line $h x + k y = 1$ touches the circle $x^{2} + y^{2} = 4$
(B)	Parabola	(q)	Points $z$ in the complex plane satisfying $\| z + 2 \|$ $-$ $\| z - 2 \|$ $= \pm 3$
(C)	Ellipse	(r)	Points of the conic have parametric representation $x = \sqrt{3} (\frac{1 - t^{2}}{1 + t^{2}}), y = \frac{2 t}{1 + t^{2}}$
(D)	Hyperbola	(s)	The eccentricity of the conic lies in the interval $1 \leq x < \infty$
		(t)	Points $z$ in the complex plane satisfying $Re {(z + 1)}^{2} = \| z \|^{2} + 1$

	Column I		Column II
(A)	Two intersecting circles	(p)	have a common tangent
(B)	Two mutually external circles	(q)	have a common normal
(C)	Two circles, one strictly inside the other	(r)	do not have a common tangent
(D)	Two branches of a hyperbola	(s)	do not have a common normal

Topic Question Set

Q 1 : The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0, (1+q)x-qy+q(1+q)=0, and y=0, where p≠q, is [2009]

Q 2 : Consider a branch of the hyperbola x2-2y2-22x-42y-6=0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is [2008]

Q 3 : Let a and b be non-zero real numbers. Then, the equation (ax2+by2+c)(x2-5xy+6y2)=0 represents [2008]

Q 4 : A hyperbola, having the transverse axis of length 2sinθ, is confocal with the ellipse 3x2+4y2=12. Then its equation is [2007]

Q 5 : If the line 2x+6y=2 touches the hyperbola x2-2y2=4, then the point of contact is [2004]

Q 6 : For hyperbola x2cos2α-y2sin2α=1, which of the following remains constant with change in 'α' [2003]

Q 7 : The equation of the common tangent to the curves y2=8x and xy=-1 is [2002]

Q 9 : The line 2x+y=1 is tangent to the hyperbola x2a2-y2b2=1. If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is ____. [2010]

Q 11 : If 2x-y+1=0 is a tangent to the hyperbola x2a2-y216=1, then which of the following cannot be sides of a right angled triangle? [2017]

Q 13 : Tangents are drawn to the hyperbola x29-y24=1, parallel to the straight line 2x-y=1. The points of contact of the tangents on the hyperbola are [2012]

Q 14 : Let the eccentricity of the hyperbola x2a2-y2b2=1 be reciprocal to that of the ellipse x2+4y2=4. If the hyperbola passes through a focus of the ellipse, then [2011]

Q 15 : An ellipse intersects the hyperbola 2x2-2y2=1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

Q 16 : Let a hyperbola passes through the focus of the ellipse x225+y216=1. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then [2006]

Q 23 : The circle x2+y2-8x=0 and hyperbola x29-y24=1 intersect at the points A and B. [2010] Q. Equation of the circle with AB as its diameter is

Q 24 : The circle x2+y2-8x=0 and hyperbola x29-y24=1 intersect at the points A and B. [2010] Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

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Q 1 :

The locus of the orthocentre of the triangle formed by the lines

$(1 + p) x - p y + p (1 + p) = 0,$

$(1 + q) x - q y + q (1 + q) = 0,$

and $y = 0,$ where $p \neq q$ , is [2009]

Q 2 :

Consider a branch of the hyperbola $x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$ with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is [2008]

Q 3 :

Let $a$ and $b$ be non-zero real numbers. Then, the equation $(a x^{2} + b y^{2} + c) (x^{2} - 5 x y + 6 y^{2}) = 0$ represents [2008]

Q 4 :

A hyperbola, having the transverse axis of length $2 \sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12 .$ Then its equation is [2007]

Q 5 :

If the line $2 x + \sqrt{6} y = 2$ touches the hyperbola $x^{2} - 2 y^{2} = 4,$ then the point of contact is [2004]

Q 6 :

For hyperbola $\frac{x^{2}}{\cos^{2} α} - \frac{y^{2}}{\sin^{2} α} = 1,$ which of the following remains constant with change in $' α'$ [2003]

Q 7 :

The equation of the common tangent to the curves $y^{2} = 8 x$ and $x y = - 1$ is [2002]

Q 9 :

The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ If this line passes through the point of intersection of the nearest directrix and the $x$ -axis, then the eccentricity of the hyperbola is ____. [2010]

Q 11 :

If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{16} = 1,$ then which of the following cannot be sides of a right angled triangle? [2017]

Q 13 :

Tangents are drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1,$ parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are [2012]

Q 14 :

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4 .$ If the hyperbola passes through a focus of the ellipse, then [2011]

Q 15 :

An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

Q 23 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of the circle with $A B$ as its diameter is

Q 24 :

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is