Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 1 :

Consider a hyperbola H having centre at the origin and foci on the x-axis. Let $C_{1}$ be the circle touching the hyperbola H and having the centre at the origin. Let $C_{2}$ be the circle touching the hyperbola H at its vertex and having the centre at one of its foci. If areas (in sq. units) of $C_{1}$ and $C_{2}$ are 36 $π$ and 4 $π$ , respectively, then the length (in units) of latus rectum of H is [2024]

$\frac{28}{3}$
$\frac{14}{3}$
$\frac{10}{3}$
$\frac{11}{3}$

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(1)

$C_{1} : x^{2} + y^{2} = a^{2}$

$\Rightarrow Area = {πa}^{2} = 36 π \Rightarrow a = 6$

$C_{2} : {(x - a e)}^{2} + y^{2} = {(a e - a)}^{2}$

$\Rightarrow Area = π {(a e - a)}^{2} = 4 π \Rightarrow 36 {(e - 1)}^{2} = 4$

$\Rightarrow e - 1 = \frac{1}{3} \Rightarrow e = \frac{4}{3} \Rightarrow b^{2} = 28$

$∴$ Length of Latus Rectum = $\frac{2 b^{2}}{a} = \frac{2 \times 28}{6} = \frac{28}{3}$ units.

Q 2 :

Let $H : \frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4 \sqrt{3}$ . Suppose the point ( $α$ , 6), $α$ > 0 lies on H. If $β$ is the product of the focal distances of the point ( $α$ , 6), then $α^{2} + β$ is equal to [2024]

169
171
172
170

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(2)

$H : \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ and $e = \sqrt{3}$

Now, $e = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{3}$

$\Rightarrow \frac{a^{2}}{b^{2}} = 2 \Rightarrow a^{2} = 2 b^{2}$

Length of latus rectum $= \frac{2 a^{2}}{b} = \frac{4 b^{2}}{b} = 4 b = 4 \sqrt{3}$ [Given]

$\Rightarrow b = \sqrt{3} \Rightarrow a = \sqrt{6}$ [ $∵$ a > 0]

Now, P( $α$ , 6) lies on $H : \frac{y^{2}}{3} - \frac{x^{2}}{6} = 1$

$\Rightarrow 12 - \frac{α^{2}}{6} = 1 \Rightarrow α^{2} = 66$

Now, foci of hyperbola = (0, $\pm$ be) = (0, $\pm$ 3)

Let $d_{1}$ and $d_{2}$ be focal distancee of ( $α$ , 6)

$\Rightarrow d_{1} = \sqrt{α^{2} + {(6 - 3)}^{2}}, d_{2} = \sqrt{α^{2} + {(6 + 3)}^{2}}$

$\Rightarrow d_{1} = \sqrt{66 + 9}, d_{2} = \sqrt{66 + 81}$

$d_{1} = \sqrt{75}, d_{2} = \sqrt{147}$

Now, $β = d_{1} d_{2} = \sqrt{11025} = 105$

Now, $α^{2} + β = 66 + 105 = 171$ .

Q 3 :

Let the foci of a hypebola H coincide with the foci of the ellipse $E : \frac{{(x - 1)}^{2}}{100} + \frac{{(y - 1)}^{2}}{75} = 1$ and the eccentriticy of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is $α$ and the length of its conjugate axis is $β$ , then $3 α^{2} + 2 β^{2}$ is equal to [2024]

237
225
205
242

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(2)

Eccentricity of ellipse E i.e., $e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$

$= \sqrt{1 - \frac{75}{100}} = \sqrt{\frac{25}{100}} = \frac{1}{2}$

So, eccentricity of hyperbola = 2

Now, foci of ellipse = (1 $\pm$ ae, 1) = (1 $\pm$ 5, 1)

= (6, 1) and (–4, 1) = Foci of hyperbola

Now, distance between foci = $2 a e_{H}$

$\Rightarrow \sqrt{10^{2} + 0} = 4 a \Rightarrow a = \frac{5}{2}$

Also, $e_{H}^{2} = 1 + \frac{b^{2}}{a^{2}}$

$\Rightarrow {(2)}^{2} = 1 + \frac{4 b^{2}}{25} \Rightarrow \frac{75}{4} = b^{2} \Rightarrow b = \frac{\sqrt{75}}{2}$

Length of transverse axis of H = 2a = $2 \times \frac{5}{2}$

$\Rightarrow α = 5$

Length of conjugate axis of H = 2b = $\sqrt{75}$

$\Rightarrow β = \sqrt{75}$

Now, $3 α^{2} + 2 β^{2} = 75 + 150 = 225$ .

Q 4 :

For 0 < $θ$ < $π$ /2, if the eccentricity of the hyperbola $x^{2} - y^{2} {cosec}^{2} θ = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} {cosec}^{2} θ + y^{2} = 5$ , then the value of $θ$ is [2024]

$π / 6$
$π / 3$
$5 π / 12$
$π / 4$

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(2)

Given, $\frac{x^{2}}{5} - \frac{y^{2}}{5 \sin^{2} θ} = 1$

$a = \sqrt{5} > b = \sqrt{5} \sin θ$

$e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \sin^{2} θ}$ ... (i)

Also, $\frac{x^{2}}{5 \sin^{2} θ} + \frac{y^{2}}{5} = 1$ (Given)

$a = \sqrt{5} \sin θ < b = \sqrt{5}$

$e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \sin^{2} θ}$ ... (ii)

Given, eqn. (i) = $\sqrt{7}$ eqn. (ii)

$\sqrt{1 + \sin^{2} θ} = \sqrt{7} \sqrt{1 - \sin^{2} θ}$

Squaring on both sides.

$1 + \sin^{2} θ = 7 (1 - \sin^{2} θ) \Rightarrow 1 + \sin^{2} θ = 7 - 7 \sin^{2} θ$

$\Rightarrow 8 \sin^{2} θ = 6$

$\sin^{2} θ = \frac{6}{8} = \frac{3}{4} \Rightarrow \sin θ = \frac{\sqrt{3}}{2}$

So, $θ = \frac{π}{3}$ .

Q 5 :

Let $e_{1}$ be the eccentricity of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ and $e_{2}$ be the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b, which passes through the foci of the hyperbola. If $e_{1} e_{2} = 1$ , then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is [2024]

$4 \sqrt{5}$
$\frac{10 \sqrt{5}}{3}$
$\frac{8 \sqrt{5}}{3}$
$3 \sqrt{5}$

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(2)

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

$e_{1}$ = eccentricity of hyperbola = $\frac{\sqrt{16 + 9}}{4} = \frac{5}{4}$

Foci of hyperbola = ( $\pm$ 5, 0)

Given $e_{1} e_{2} = 1$

$\Rightarrow e_{2} = \frac{4}{5}$ = eccentricity of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow$ $\frac{\sqrt{a^{2} - b^{2}}}{a} = \frac{4}{5}$ ... (i)

$∴$ The ellipse also passes through the foci of the hyperbola.

$\Rightarrow a^{2} = 25$

From (i), $b^{2} = 9$

Thus, the equation of ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ ... (ii)

Now, equation of chord of ellipse which passes through (0, 2) and parallel to x-axis is y = 2.

From (ii),

$\frac{x^{2}}{25} = \frac{5}{9} \Rightarrow x^{2} = \frac{25 \times 5}{9} \Rightarrow x = \pm \frac{5 \sqrt{5}}{3}$

Thus, the end point of chord are $(\frac{- 5 \sqrt{5}}{3}, 2) and (\frac{5 \sqrt{5}}{3}, 2)$ .

$∴$ Length of chord = $\sqrt{{(\frac{5 \sqrt{5}}{3} + \frac{5 \sqrt{5}}{3})}^{2} + {(2 - 2)}^{2}} = \frac{10 \sqrt{5}}{3}$ .

Q 6 :

Let P be a point on the hyperbola $H : \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ in the first quadrant such that the area of triangle formed by P and the two foci of H is $2 \sqrt{13}$ . Then, the square of the distance of P from the origin is [2024]

20
18
26
22

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(4)

We have, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

Here, $a^{2} = 9, b^{2} = 4$

$b^{2} = a^{2} (e^{2} - 1) \Rightarrow e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow e^{2} = 1 + \frac{4}{9} = \frac{13}{9}$

$\Rightarrow e = \frac{\sqrt{13}}{3} \Rightarrow s_{1} s_{2} = 2 a e = 2 \times 3 \times \frac{\sqrt{13}}{3} = 2 \sqrt{13}$

Area of $△ P S_{1} S_{2} = \frac{1}{2} \times β \times s_{1} s_{2} = 2 \sqrt{13}$

$\Rightarrow \frac{1}{2} \times β \times (2 \sqrt{13}) = 2 \sqrt{13} \Rightarrow β = 2$

$\frac{α^{2}}{9} - \frac{β^{2}}{4} = 1 \Rightarrow \frac{α^{2}}{9} = 2 \Rightarrow α^{2} = 18 \Rightarrow α = 3 \sqrt{2}$

Distance of P from origin = $\sqrt{α^{2} + β^{2}} = \sqrt{18 + 4} = \sqrt{22}$ .

Q 7 :

If the foci of a hyperbola are same as that of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}})$ on the hyperbola, is equal to [2024]

$7 \sqrt{\frac{2}{5}} - \frac{8}{3}$
$14 \sqrt{\frac{2}{5}} - \frac{16}{5}$
$14 \sqrt{\frac{2}{5}} - \frac{4}{3}$
$7 \sqrt{\frac{2}{5}} + \frac{8}{3}$

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(1)

We have, $E : \frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$a = 5, b = 3$

foci : (0, $\pm$ 5e)

Since, for an ellipes, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow 9 = 25 (1 - e^{2})$

$e^{2} = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow e = \frac{4}{5}$

Eccentricity of hyperbola = $e' = \frac{15}{8} \times \frac{4}{5} = \frac{3}{2}$

Focus of hyperbola = (0, $\pm$ 5e) = (0, $\pm$ 4)

$\Rightarrow a \cdot e' = 4 \Rightarrow a = 4 \times \frac{2}{3} = \frac{8}{3}$

Since, $e'^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow \frac{9}{4} = 1 + \frac{b^{2}}{1} \times \frac{9}{64} \Rightarrow b^{2} = \frac{80}{9}$

$∴$ The equation of hyperbola $\frac{9 y^{2}}{64} - \frac{9 x^{2}}{80} = 1$

PS = $e' p M$

$= \frac{3}{2} \times (\frac{14}{3} \sqrt{\frac{2}{5}} - \frac{16}{9})$

$= 7 \sqrt{\frac{2}{5}} - \frac{8}{3}$ .

Q 8 :

Let A be a square matrix of order 2 such that |A| = 2 and the sum of its diagonal elements is –3. If the points (x, y) satisfying $A^{2}$ + xA + y $I$ = O lie on a hyperbola, whose transverse axis is parallel to the x-axis, eccentricity is e and the length of the latus rectum is $l$ , then $e^{4} + l^{4}$ is equal to ___________. [2024]

(*)

The given data is inadequate.

Q 9 :

The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $x = \pm \frac{4}{\sqrt{3}}$ , respectively. Let the line $y - \sqrt{3} x + \sqrt{3} = 0$ touch this hyperbola at $(x_{0}, y_{0})$ . If m is the product of the focal distances of the point $(x_{0}, y_{0})$ , then $4 e^{2} + m$ is equal to __________. [2024]

(*)

Equation of tangent to hyperbola is given by $y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$

$\Rightarrow m = \sqrt{3}$

and $a^{2} m^{2} - b^{2} = 3$

$\Rightarrow 3 a^{2} - b^{2} = 3$ ... (i)

Now, length of latus rectum of hyperbola = 9

$\Rightarrow \frac{2 b^{2}}{a} = 9$

$\Rightarrow b^{2} = \frac{9 a}{2}$ ... (ii)

From equation (i) and (ii), we get

$3 a^{2} - \frac{9 a}{2} = 3$

$\Rightarrow 6 a^{2} - 9 a - 6 = 0 \Rightarrow 2 a^{2} - 3 a - 2 = 0$

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

$\Rightarrow a = 2, \frac{- 1}{2} (ignore) \Rightarrow a = 2 and b = 3$

Equation of hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

Solving for tangent $y = \sqrt{3} x - \sqrt{3}$ , we get

$\frac{x^{2}}{4} - \frac{3 x^{2} + 3 - 6 x}{9} = 1$

$\Rightarrow 9 x^{2} - 12 x^{2} - 12 + 24 x = 36$

$\Rightarrow 3 x^{2} - 24 x + 48 = 0$

$\Rightarrow x^{2} - 8 x + 16 = 0 \Rightarrow {(x - 4)}^{2} = 0$

$\Rightarrow x = 4 and y = 3 \sqrt{3}, e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}$

So, $(x_{0}, y_{0}) \equiv (4, 3 \sqrt{3})$ is the point of contact.

Focus of hyperbola is $(\pm \sqrt{13}, 0)$

$∴ P F_{1} \cdot P F_{2}$

$= (\sqrt{{(4 - \sqrt{13})}^{2} + {(3 \sqrt{3})}^{2}}) (\sqrt{{(4 + \sqrt{13})}^{2} + {(3 \sqrt{3})}^{2}})$

$= \sqrt{2304} = 48 = m$

$∴ 4 e^{2} + m = 4 \times 13 / 4 + 48 = 61$

Note: Here equation of directrix should be $x = \pm \frac{4}{\sqrt{13}}$ , but in given question it is given $x = \pm \frac{4}{\sqrt{3}}$ which is wrong because eccentricity should be greater than 1, So, this question is bonus.

Q 10 :

Let S be the focus of the hyperbola $\frac{x^{2}}{3} - \frac{y^{2}}{5} = 1$ , on the positive x-axis. Let C be the circle with its centre at $A (\sqrt{6}, \sqrt{5})$ and passing through the point S. If O is the origin and SAB is a diameter of C, then the square of the area of the triangle OSB is equal to __________. [2024]

(40)

$\frac{x^{3}}{3} - \frac{y^{2}}{5} = 1 \Rightarrow a = \sqrt{3}, b = \sqrt{5}$

$∴ e = \sqrt{1 + \frac{5}{3}} \Rightarrow e = \sqrt{\frac{8}{3}}$

$∴ a e = \sqrt{8} = 2 \sqrt{2} ∴ S = (2 \sqrt{2}, 0)$

A is mid-point of BS $\Rightarrow B (2 \sqrt{6} - 2 \sqrt{2}, 2 \sqrt{5)}$

Area of $△ (O S B)$ , $A =$ $|$ $\frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ 2 \sqrt{2} & 0 & 1 \\ 2 \sqrt{6} - 2 \sqrt{2} & 2 \sqrt{5} & 1 \end{matrix} |$ $|$ $= 2 \sqrt{10}$

$\Rightarrow A^{2} = 40$ .

Q 11 :

Let the latus rectum of the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{b^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + \sqrt{n})$ , where $l$ and m are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to __________. [2024]

(182)

In right triangle OFP, we have $\tan 30 ° = \frac{P F}{O F}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{b^{2}}{a^{2} e} \Rightarrow e = \frac{\sqrt{3} b^{2}}{a^{2}}$

$e = \frac{\sqrt{3} b^{2}}{9}$ (Given $a^{2} = 9$ )

Squaring both sides, we get

$e^{2} = \frac{3 b^{4}}{81}$ ... (i)

and $e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow e^{2} = 1 + \frac{b^{2}}{9}$ ... (ii)

Comparing equation (i) and (ii), we get

$\frac{3 b^{4}}{81} = 1 + \frac{b^{2}}{9} \Rightarrow b^{4} = 27 + 3 b^{2}$

$\Rightarrow b^{4} - 3 b^{2} - 27 = 0 \Rightarrow b^{2} = \frac{3}{2} (1 + \sqrt{13})$

According to given condition, $b^{2} = \frac{l}{m} (1 + \sqrt{n})$

On comparing, we get $l$ = 3, m = 2 and n = 13

$\Rightarrow l^{2} + m^{2} + n^{2} = {(3)}^{2} + {(2)}^{2} + {(13)}^{2} = 9 + 4 + 160 = 182$ .

Q 12 :

Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

(51)

Here ae = 5 and $\frac{2 b^{2}}{a} = \sqrt{50}$

$\frac{b^{2}}{a} = \frac{5 \sqrt{2}}{2}$

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

$25 = a^{2} - \frac{5 \sqrt{2}}{2} a \Rightarrow 2 a^{2} - 5 \sqrt{2} a - 50 = 0$

$\Rightarrow [a - 5 \sqrt{2}] [2 a + 5 \sqrt{2}] = 0 \Rightarrow a = 5 \sqrt{2} or a = \frac{- 5 \sqrt{2}}{2}$

$\Rightarrow e = \frac{5}{a} = \frac{1}{\sqrt{2}}$

Also, $b^{2} = 5 \sqrt{2} \times \frac{5 \sqrt{2}}{2} = 25 \Rightarrow b = 5$

Now, $e = \sqrt{1 + \frac{a^{2} b^{2}}{b^{2}}} = \sqrt{1 + a^{2}} \Rightarrow e^{2} = 1 + a^{2} = 51$ .

Q 13 :

Let one focus of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$ . If e and $l$ respectively are the eccentricity and the length of the latus rectum of H, then $9 (e^{2} + l)$ is equal to : [2025]

16
15
12
14

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(1)

We have, $a e = \sqrt{10} and \frac{a}{e} = \frac{9}{\sqrt{10}}$

$\Rightarrow a^{2} = 9 and e = \frac{\sqrt{10}}{3}$

Now, ${(a e)}^{2} = a^{2} + b^{2} \Rightarrow 10 = 9 + b^{2} \Rightarrow b^{2} = 1$

and $l = \frac{2 b^{2}}{a} = \frac{2 (1)}{3} = \frac{2}{3}$

$∴ 9 (e^{2} + l) = 9 (\frac{10}{9} + \frac{2}{3}) = 10 + 6 = 16$

Q 14 :

Let the sum of the focal distances of the point P(4, 3) on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be $8 \sqrt{\frac{5}{3}}$ . If for H, the length of the latus rectum is $l$ and the product of the focal distances of the point P is m, then $9 l^{2} + 6 m$ is equal to : [2025]

187
184
185
186

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(3)

Given, hyperbola $(H) : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Sum of focal distance of P(4, 3) = $8 \sqrt{\frac{5}{3}}$

$\Rightarrow 2 e x = 8 \sqrt{\frac{5}{3}} \Rightarrow 2 e (4) = 8 \sqrt{\frac{5}{3}} \Rightarrow e^{2} = \frac{5}{3}$

$∵ b^{2} = a^{2} (\frac{5}{3} - 1) \Rightarrow b^{2} = \frac{2}{3} a^{2}$ ... (i)

$\Rightarrow \frac{16}{a^{2}} - \frac{9}{b^{2}} = 1$ ... (ii)

Using (i) and (ii), we get

$\Rightarrow a^{2} = \frac{5}{2}, b^{2} = \frac{5}{3}$

Now, length of latus rectum, $l = \frac{2 b^{2}}{a} \Rightarrow l^{2} = \frac{4 b^{4}}{a^{2}}$

$\Rightarrow l^{2} = \frac{4 \times {(5 / 3)}^{2}}{5 / 2} \Rightarrow 9 l^{2} = 40$

Also, $m = (e x + a) (e x - a) = {(e x)}^{2} - {(a)}^{2}$

$= \frac{5}{3} (16) - \frac{5}{2} = \frac{145}{6}$

$\Rightarrow 6 m = 145$

$∴ 9 l^{2} + 6 m = 40 + 145 = 185$ .

Q 15 :

Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{25} = 1$ and the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1$ , respectively. If b < 5 and $e_{1} e_{2} = 1$ , then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is : [2025]

$\frac{\sqrt{3}}{2}$
$\frac{4}{5}$
$\frac{3}{5}$
$\frac{\sqrt{7}}{4}$

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(3)

We have, $e_{1}^{2} = 1 - \frac{b^{2}}{25} and e_{2}^{2} = 1 + \frac{b^{2}}{16}$

$∵ e_{1} e_{2} = 1$ (Given)

$∴ e_{1}^{2} e_{2}^{2} = 1$

$\Rightarrow (1 - \frac{b^{2}}{25}) (1 + \frac{b^{2}}{16}) = 1$

$\Rightarrow 1 + \frac{b^{2}}{16} - \frac{b^{2}}{25} - \frac{b^{4}}{400} = 1$

$\Rightarrow \frac{9 b^{2}}{400} = \frac{b^{4}}{400}$

$\Rightarrow b^{2} = 9$

We get, $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1 and \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

Here, $e_{1} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} and e_{2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$

Foci : $(0, \pm 4) and (\pm 5, 0)$

The ellipse passing through all four foci is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$

Now, the eccentricity is given by $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ .

Q 16 :

Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

$\frac{25}{6}$
$\frac{24}{5}$
$\frac{144}{5}$
$\frac{288}{5}$

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(4)

Distance between foci $\sqrt{{(1 - 1)}^{2} + {(14 + 12)}^{2}}$

$\Rightarrow 2 b e = 26 \Rightarrow b e = 13$

Mid-point of foci $= (\frac{1 + 1}{2}, \frac{14 - 12}{2}) = (1, 1)$

Equation of hyperbola is $\frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y - 1)}^{2}}{b^{2}} = 1$

Hyperbola passes through (1, 6)

$∴ \frac{{(6 - 1)}^{2}}{b^{2}} = 1 \Rightarrow b^{2} = 25$

Now, $a^{2} = b^{2} (e^{2} - 1) = b^{2} e^{2} - b^{2} = 169 - 25 = 144$

$∴ a^{2} = 144 and b^{2} = 25$

$∴$ Length of latus rectum $= \frac{2 a^{2}}{b} = \frac{2 \times 144}{5} = \frac{288}{5}$

Q 17 :

Let $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ . Let the distane between the foci of E and the foci of H be $2 \sqrt{3}$ . If a – A = 2, and the ratio of the eccentricities of E and H is $\frac{1}{3}$ , then the sum of the lengths of their latus rectums is equal to: [2025]

9
7
10
8

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(4)

We have, $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , whose foci are ( $\pm$ ae, 0) and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ , whose foci are ( $\pm$ Ae', 0).

Given, $2 a e = 2 \sqrt{3} and 2 A e' = 2 \sqrt{3}$

$\Rightarrow a e = \sqrt{3} and A e' = \sqrt{3}$

$\Rightarrow \frac{a e}{A e'} = 1 \Rightarrow \frac{e}{e'} = \frac{A}{a}$

$\Rightarrow \frac{1}{3} = \frac{A}{a} \Rightarrow a = 3 A$ $[Given \frac{e}{e'} = \frac{1}{3}]$

Now, a – A = 2 $\Rightarrow$ 2A = 1 [a = 3A]

$\Rightarrow A = 1, a = 3$

$∴ e = \frac{1}{\sqrt{3}} and e' = \sqrt{3}$

Also, $b^{2} = a^{2} (1 - e^{2}) and B^{2} = A^{2} ({(e')}^{2} - 1)$

$\Rightarrow b^{2} = 9 (1 - {(\frac{1}{\sqrt{3}})}^{2}) and B^{2} = {(1)}^{2} ({(\sqrt{3})}^{2} - 1)$

$\Rightarrow b^{2} = 6 and B^{2} = 2$

$∴$ Sum of latus rectum $= \frac{2 b^{2}}{a} + \frac{2 B^{2}}{A} = 8$ .

Q 18 :

If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

x + 9y = 36
9x – 9y = 32
4x – 9y = 12
6x – 9y = 20

1

2

3

4

(4)

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

and $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 \Rightarrow 4 x^{2} - 9 y^{2} = 36$ ... (ii)

Solving (i) and (ii), we get

$4 x^{2} - 9 (8 x - x^{2}) = 36$

$\Rightarrow 13 x^{2} - 72 x - 36 = 0$

$\Rightarrow (13 x + 6) (x - 6) = 0$

$\Rightarrow x = \frac{- 6}{13}, 6$

$∴ x = 6$ $[∵ x = \frac{- 6}{13} not possible]$

From (i), $y^{2} = 8 (6) - {(6)}^{2} = 48 - 36 = 12 \Rightarrow y = \pm \sqrt{12}$

$∴$ $(6, \sqrt{12})$ and $(6, - \sqrt{12})$ are the points of intersection of circle and hyperbola.

Let $P (α, β)$ be the point moves on the line 2x – 3y + 4 = 0 such that

$2 α - 3 β + 4 = 0$ ... (iii)

Centroid of $△$ PAB is given by $(\frac{α + 6 + 6}{3}, \frac{β}{3}) = (h, k)$

$∴ α = 3 h - 12 and β = 3 k$

From (iii), 2(3h –12) – 3(3k) + 4 = 0

$\Rightarrow 6 h - 24 - 9 k + 4 = 0 \Rightarrow 6 h - 9 k = 20$

i.e., 6x – 9y = 20.

Q 19 :

Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

(120)

We have, $P (4, 2 \sqrt{3})$

Now, $P S_{1} \cdot P S_{2} = 32$ ... (i)

where $S_{1} = (a e, 0) and S_{2} = (- a e, 0)$

ALso, $| P S_{1} - P S_{2} | = 2 a$

Now, $P (4, 2 \sqrt{3})$ lies on H

$∴ \frac{16}{a^{2}} - \frac{12}{b^{2}} = 1$

$\Rightarrow 16 b^{2} - 12 a^{2} = a^{2} b^{2}$ ... (ii)

Now, $| P S_{1} - P S_{2} |^{2} = 4 a^{2}$

$\Rightarrow P S_{1}^{2} + P S_{2}^{2} - 2 P S_{1} \cdot P S_{2} = 4 a^{2}$

$\Rightarrow {(a e - 4)}^{2} + 12 + {(a e + 4)}^{2} + 12 - 64 = 4 a^{2}$

$\Rightarrow 2 a^{2} e^{2} - 8 = 4 a^{2} \Rightarrow a^{2} + b^{2} - 4 = 2 a^{2}$

$\Rightarrow b^{2} - a^{2} = 4$ ... (iii)

From equation (ii) and (iii)

$\Rightarrow 16 (a^{2} + 4) - 12 a^{2} = a^{2} (a^{2} + 4)$

$\Rightarrow 16 a^{2} + 64 - 12 a^{2} = a^{4} + 4 a^{2}$

$\Rightarrow a^{4} = 64 \Rightarrow a^{2} = 8 ∴ b^{2} = 12$

Now, $p^{2} + q^{2} = 4 b^{2} + \frac{4 b^{4}}{a^{2}} = 120$ .

Q 20 :

If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]

(141)

Equation of hyperbola is $\frac{{(x - 6)}^{2}}{a^{2}} - \frac{{(y - 2)}^{2}}{4 - a^{2}} = 1$

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

$\Rightarrow (4 - a^{2}) x^{2} - a^{2} y^{2} + (12 a^{2} - 48) x + 4 a^{2} y + 144 - 44 a^{2} + a^{4} = 0$

On comparing with $3 x^{2} - y^{2} - α x + β y + γ = 0$ , we get $a^{2} = 1$ and $α = 36$ , $β = 4$ and $γ = 101$

$∴ α + β + γ = 141$

Q 21 :

Consider the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and $a^{2} b^{2} = α \sqrt{2} - β, α, β \in N$ , then $α + β$ is ___________. [2025]

(1944)

We have, $F_{1} \equiv (- a e, 0) \equiv P (- 3, 0) \Rightarrow a e = 3$

$L_{1} L_{2} = \frac{2 b^{2}}{a}$

In $△ F_{1} F_{2} L_{2}$ ,

$\tan 45 ° = \frac{L_{2} F_{2}}{F_{1} F_{2}} = \frac{b^{2} / a}{2 a e}$

$\Rightarrow 2 a e = \frac{b^{2}}{a}$

$\Rightarrow b^{2} = 6 a$ [ $∵$ ae = 3]

Also, $a^{2} e^{2} = a^{2} + b^{2}$

$\Rightarrow 9 = a^{2} + 6 a$ [ $∵ a e = 3 and b^{2} = 6 a$ ]

$\Rightarrow a^{2} + 6 a - 9 = 0$

$a = - 3 \pm 3 \sqrt{2} = - 3 (1 \pm \sqrt{2})$

Now, $a^{2} b^{2} = a^{2} \cdot 6 a = 6 a^{3}$ [ $∵ b^{2} = 6 a$ ]

$= 6 (135 \sqrt{2} - 189)$

On comparing, we get $α$ = 810 and $β$ = 1134

$∴ α + β = 1944$ .

Q 22 :

Let the lengths of the transverse and conugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (–5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point $(α, 2 \sqrt{5})$ on the hyperbola is p, then 4p is equal to __________. [2025]

(189)

Given, focus = (–5, 0)

$\Rightarrow a e = 5 and \frac{a}{e} = \frac{9}{5}$

Also, $a = 3, e = \frac{5}{3} and b = 4$

$\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Since, hyperbola passes through $(α, 2 \sqrt{5})$ , we get

$\frac{α^{2}}{9} - \frac{4 (5)}{16} = 1$

$\Rightarrow α^{2} = 9 (\frac{36}{16}) = \frac{81}{4}$

Now, $P = e^{2} α^{2} - a^{2} = \frac{25}{9} \times \frac{81}{4} - 9 = \frac{189}{4}$

$∴ 4 p = 189$ .

Q 23 :

Let the circle C touch the line x – y + 1 = 0, have the centre on the positive x-axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line –3x + 2y = 1. Let H be the hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ , whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then $2 α^{2} + 3 β^{2}$ is equal to __________. [2025]

(19)

Since, centre of circle lies on positive x-axis and one of the foci of hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ are same.

$∴$ Centre of circle = $(α, 0)$

Since, x – y + 1 = 0 is tangent to the circle.

$∴$ $| \frac{α + 1}{\sqrt{2}} |$ $= r$ [where 'r' is the radius of circle]

$\Rightarrow {(α + 1)}^{2} = 2 r^{2}$ ... (i)

Also, –3x + 2y = 1 is the chord of the circle

$∴$ $| \frac{- 3 α - 1}{\sqrt{9 + 4}} |^{2}$ ${+ (\frac{2}{\sqrt{13}})}^{2} = r^{2}$ $[∵ Length of chord = \frac{4}{\sqrt{13}}]$

$∴ \frac{{(3 α + 1)}^{2}}{13} + \frac{4}{13} = r^{2}$

$\Rightarrow \frac{9 α^{2} + 1 + 6 α + 4}{13} = \frac{(α^{2} + 1 + 2 α)}{2}$

$\Rightarrow 5 α^{2} - 14 α - 3 = 0$

$\Rightarrow α = 3$ [ $∵ α > 0$ ]

$∴ r = 2 \sqrt{2}$

Now, $α e = 3 and 2 α = 4 \sqrt{2}$

$∴ α^{2} e^{2} = 9 and α = 2 \sqrt{2}$

$\Rightarrow α^{2} (1 + \frac{β^{2}}{α^{2}}) = 9 \Rightarrow 8 + β^{2} = 9 \Rightarrow β^{2} = 1$

$∴ 2 α^{2} + 3 β^{2} = 2 (8) + 3 \times 1 = 16 + 3 = 19$ .

Q 24 :

Let $H_{1} : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $H_{2} : - \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their eccentricities be $e_{1} = \sqrt{\frac{5}{2}}$ and $e_{2}$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$ , then $25 e_{2}^{2}$ is equal to __________. [2025]

(55)

Given, Hyperbola : $H_{1} : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $e_{1} = \sqrt{\frac{5}{2}}$ and $H_{2} : \frac{- x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$

Using $H_{1}$ , length of latus rectum = $15 \sqrt{2}$

$\Rightarrow \frac{2 b^{2}}{a} = 15 \sqrt{2}$ ... (i)

Since, $e_{1} = \sqrt{\frac{5}{2}} \Rightarrow 1 + \frac{b^{2}}{a^{2}} = \frac{5}{2} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{3}{2}$ ... (ii)

Using (i) and (ii), we get $a = 5 \sqrt{2} and b = 5 \sqrt{3}$

Now, for $H_{2}, \frac{2 A^{2}}{B} = 12 \sqrt{5}$ ... (iii)

Since, product of transverse axes is $100 \sqrt{10}$ , then

$2 a \cdot 2 B = 100 \sqrt{10}$

$\Rightarrow 2 \times 5 \sqrt{2} \times 2 B = 100 \sqrt{10} \Rightarrow B = 5 \sqrt{5}$

$∴ A = 5 \sqrt{6}$ [Using (iii)]

Now, eccentricity of $H_{2}$ is given by

$e_{2}^{2} = 1 + \frac{A^{2}}{B^{2}} = 1 + \frac{150}{125} = 1 + \frac{30}{25} = \frac{55}{25}$

$\Rightarrow 25 e_{2}^{2} = 55$ .

Q 25 :

Let $P (x_{0}, y_{0})$ be the point on the hyperbola $3 x^{2} - 4 y^{2} = 36$ , which is nearest to the line $3 x + 2 y = 1 .$ Then $\sqrt{2} (y_{0} - x_{0})$ is equal to [2023]

- 9
3
9
- 3

1

2

3

4

(1)

We have, $3 x^{2} - 4 y^{2} = 36$

$The slope of 3 x + 2 y = 1 is$ $m = - \frac{3}{2}$

$The point of hyperbola is (\sqrt{12} \sec θ, 3 \tan θ) .$

$The equation of normal is$ $\sqrt{12} x \cos θ + 3 y \cot θ = 21$

$\Rightarrow Slope$ $= - \frac{\sqrt{12} \cos θ}{3 \cot θ}$

$According to questions,$ $(- \frac{3}{2}) (\frac{- \sqrt{12} \cos θ}{3 \cot θ}) = - 1$

$\Rightarrow \sqrt{3} \sin θ = - 1$ $\Rightarrow \sin θ = - \frac{1}{\sqrt{3}}$ $\Rightarrow \cos θ = \sqrt{\frac{2}{3}}$

$∴$ $(x_{0}, y_{0}) \equiv (\frac{6}{\sqrt{2}}, - \frac{3}{\sqrt{2}})$ $∴$ $\sqrt{2} (y_{0} - x_{0}) = - 9$

Q 26 :

Let T and C respectively be the transverse and conjugate axes of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ . Then the area of the region above the parabola $x^{2} = y + 4,$ below the transverse axis T and on the right of the conjugate axis C is: [2023]

$4 \sqrt{6} + \frac{44}{3}$
$4 \sqrt{6} - \frac{28}{3}$
$4 \sqrt{6} + \frac{28}{3}$
$4 \sqrt{6} - \frac{44}{3}$

1

2

3

4

(3)

$We have given 16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$

$16 (x^{2} + 4 x) - (y^{2} - 4 y) + 44 = 0$

$16 (x^{2} + 4 x + 4) - 64 - (y^{2} - 4 y + 4) + 4 + 44 = 0$

$16 {(x + 2)}^{2} - 64 - {(y - 2)}^{2} + 48 = 0$

$16 {(x + 2)}^{2} - {(y - 2)}^{2} = 16$

$\frac{{(x + 2)}^{2}}{1} - \frac{{(y - 2)}^{2}}{16} = 1$

$\Rightarrow Centre: (- 2, 2)$

$Vertices: (- 1, 2), (- 3, 2)$

$Equation of parabola: x^{2} = y + 4$

$Required area = \int_{- 2}^{\sqrt{6}} {2 - (x^{2} - 4)} d x$

$= \int_{- 2}^{\sqrt{6}} (6 - x^{2}) d x = {[6 x - \frac{x^{3}}{3}]}_{- 2}^{\sqrt{6}}$

$= 4 \sqrt{6} + \frac{28}{3}$

Q 27 :

Let H be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$ . Then the length of its latus rectum is ___________ . [2023]

$3$
$\frac{5}{2}$
$2$
$\frac{3}{2}$

1

2

3

4

(3)

$Distance between two foci = 2 a e = 2 \sqrt{2}$

$\Rightarrow a e = \sqrt{2}$ $\Rightarrow a (\sqrt{2}) = \sqrt{2} (∵ e = \sqrt{2})$

$\Rightarrow a = 1$

$We know that e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow 2 = 1 + \frac{b^{2}}{1}$

$\Rightarrow b^{2} = 1 \Rightarrow b = 1$

$∴$ $Length of latus rectum = \frac{2 b^{2}}{a} = \frac{2 {(1)}^{2}}{1} = 2$

Q 28 :

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1 .$ If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

(2)

$We have, equation of hyperbola as \frac{x^{2}}{1 / 2} - \frac{y^{2}}{1 / 2} = 1$

$∴$ $e_{H} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{2}$

$So, e_{E} = \frac{1}{\sqrt{2}} (Given that e_{E} = \frac{1}{e_{H}})$

$∵$ $Ellipse and hyperbola intersect at 90 ° .$

$∴$ $They are confocal.$

$Now, foci of hyperbola = (1, 0)$

$∴$ $a e_{E} = 1 \Rightarrow a = \sqrt{2}, b = 1$

So, $Latus rectum of ellipse = \frac{2 b^{2}}{a} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$∴$ $Square of latus rectum is 2 .$

Q 29 :

Let $H_{n} : \frac{x^{2}}{1 + n} - \frac{y^{2}}{3 + n} = 1, n \in N .$ Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_{k}$ is a rational number. If $l$ is the length of the latus rectum of $H_{k},$ then $21 l$ is equal to _________. [2023]

(306)

$H_{n} = \frac{x^{2}}{1 + n} - \frac{y^{2}}{3 + n} = 1 be a hyperbola$

$Eccentricity of hyperbola, e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{3 + n}{1 + n}} = \sqrt{\frac{2 n + 4}{n + 1}}$

$For rational eccentricity, \sqrt{\frac{2 n + 4}{n + 1}} should be a perfect square.$

$∴$ $n = 48 (smallest even value for which e \in Q)$

$∴$ $e = \frac{10}{7}$

$Now, a^{2} = n + 1 = 49, b^{2} = n + 3 = 51$

$l = length of latus rectum = \frac{2 b^{2}}{a}$

$\Rightarrow l = 2 \cdot \frac{51}{7} = \frac{102}{7}$ $∴$ $21 l = 306$

Q 30 :

Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3 .$ Then the square of distance of the point (6, - 4) from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to ____ . [2023]

(116)

$Given, line is 2 x + 2 y = 3$

$\Rightarrow y = - x + \frac{3}{2}$ ...(i)

$Slope of (i) is m = - 1$

$∴$ $Tangent at (3, α) has slope 1 .$

$Now, α^{2} = 12 (3) = 36 \Rightarrow α = 6 [∵ α = - 6 reject]$

$Equation of tangent to y^{2} = 12 x at point (3, 6) is given by$

$y - 6 = x - 3 \Rightarrow x - y + 3 = 0$

$Equation of hyperbola is α^{2} x^{2} - 9 y^{2} = 9 α^{2}$

$\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{36} = 1$

$Equation of normal to the hyperbola at point (α - 1, α + 2),$

$i.e., (5, 8) is \frac{9 x}{5} + \frac{36 y}{8} = 45 \Rightarrow 2 x + 5 y - 50 = 0$

$Now, distance of (6, - 4) from 2 x + 5 y - 50 = 0 is given by$

$d =$ $| \frac{2 (6) - 5 (4) - 50}{\sqrt{4 + 25}} |$ $= \frac{58}{\sqrt{29}}$

$∴$ $Square of distance = {(\frac{58}{\sqrt{29}})}^{2} = 116 .$

Topic Question Set

Q 4 :

For 0 < $θ$ < $π$ /2, if the eccentricity of the hyperbola $x^{2} - y^{2} {cosec}^{2} θ = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} {cosec}^{2} θ + y^{2} = 5$ , then the value of $θ$ is [2024]

Q 6 :

Let P be a point on the hyperbola $H : \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ in the first quadrant such that the area of triangle formed by P and the two foci of H is $2 \sqrt{13}$ . Then, the square of the distance of P from the origin is [2024]

Q 12 :

Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

Q 16 :

Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 18 :

If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

Q 19 :

Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

Q 20 :

If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]

Q 21 :

Consider the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and $a^{2} b^{2} = α \sqrt{2} - β, α, β \in N$ , then $α + β$ is ___________. [2025]

Q 25 :

Let $P (x_{0}, y_{0})$ be the point on the hyperbola $3 x^{2} - 4 y^{2} = 36$ , which is nearest to the line $3 x + 2 y = 1 .$ Then $\sqrt{2} (y_{0} - x_{0})$ is equal to [2023]

Q 26 :

Let T and C respectively be the transverse and conjugate axes of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ . Then the area of the region above the parabola $x^{2} = y + 4,$ below the transverse axis T and on the right of the conjugate axis C is: [2023]

Q 27 :

Let H be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$ . Then the length of its latus rectum is ___________ . [2023]

Q 28 :

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1 .$ If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

Q 29 :

Let $H_{n} : \frac{x^{2}}{1 + n} - \frac{y^{2}}{3 + n} = 1, n \in N .$ Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_{k}$ is a rational number. If $l$ is the length of the latus rectum of $H_{k},$ then $21 l$ is equal to _________. [2023]

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Topic Question Set

Q 2 : Let H:–x2a2+y2b2=1 be the hyperbola, whose eccentricity is 3 and the length of the latus rectum is 43. Suppose the point (α, 6), α > 0 lies on H. If β is the product of the focal distances of the point (α, 6), then α2+β is equal to [2024]

Q 4 : For 0 < θ < π/2, if the eccentricity of the hyperbola x2–y2 cosec2 θ=5 is 7 times eccentricity of the ellipse x2cosec2θ+y2=5, then the value of θ is [2024]

Q 5 : Let e1 be the eccentricity of the hyperbola x216–y29=1 and e2 be the eccentricity of the ellipse x2a2+y2b2=1, a > b, which passes through the foci of the hyperbola. If e1e2=1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is [2024]

Q 6 : Let P be a point on the hyperbola H:x29–y24=1 in the first quadrant such that the area of triangle formed by P and the two foci of H is 213. Then, the square of the distance of P from the origin is [2024]

Q 7 : If the foci of a hyperbola are same as that of the ellipse x29+y225=1 and the eccentricity of the hyperbola is 158 times the eccentricity of the ellipse, then the smaller focal distance of the point (2,14325) on the hyperbola, is equal to [2024]

Q 9 : The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and x=±43, respectively. Let the line y–3x+3=0 touch this hyperbola at (x0, y0). If m is the product of the focal distances of the point (x0, y0), then 4e2+m is equal to __________. [2024]

Q 10 : Let S be the focus of the hyperbola x23–y25=1, on the positive x-axis. Let C be the circle with its centre at A(6,5) and passing through the point S. If O is the origin and SAB is a diameter of C, then the square of the area of the triangle OSB is equal to __________. [2024]

Q 11 : Let the latus rectum of the hyperbola x29–y2b2=1 subtend an angle of π3 at the centre of the hyperbola. If b2 is equal to lm(1+n), where l and m are co-prime numbers, then l2+m2+n2 is equal to __________. [2024]

Q 12 : Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1, a > b be (±5, 0) and 50, respectively. Then, the square of the eccentricity of the hyperbola x2b2–y2a2b2=1 equals [2024]

Q 13 : Let one focus of the hyperbola H : x2a2–y2b2=1 be at (10,0) and the corresponding directrix be x=910. If e and l respectively are the eccentricity and the length of the latus rectum of H, then 9(e2+l) is equal to : [2025]

Q 14 : Let the sum of the focal distances of the point P(4, 3) on the hyperbola H:x2a2–y2b2=1 be 853. If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then 9l2+6m is equal to : [2025]

Q 16 : Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 17 : Let E : x2a2+y2b2=1, a > b and H : x2A2–y2B2=1. Let the distane between the foci of E and the foci of H be 23. If a – A = 2, and the ratio of the eccentricities of E and H is 13, then the sum of the lengths of their latus rectums is equal to: [2025]

Q 18 : If A and B are the points of intersection of the circle x2+y2–8x=0 and the hyperbola x29–y24=1 and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of △PAB lies on the line : [2025]

Q 19 : Let the product of focal distances of the point P(4,23) on the hyperbola H : x2a2–y2b2=1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2+q2 is equal to __________. [2025]

Q 20 : If the equation of the hyperbola with foci (4, 2) and (8, 2) is 3x2–y2–αx+βy+γ=0, then α+β+γ is equal to __________. [2025]

Q 21 : Consider the hyperbola x2a2–y2b2=1 having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and a2b2=α2–β, α, β∈N, then α+β is ___________. [2025]

Q 24 : Let H1:x2a2–y2b2=1 and H2:–x2A2+y2B2=1 be two hyperbolas having length of latus rectums 152 and 125 respectively. Let their eccentricities be e1=52 and e2 respectively. If the product of the lengths of their transverse axes is 10010, then 25e22 is equal to __________. [2025]

Q 25 : Let P(x0,y0) be the point on the hyperbola 3x2-4y2=36, which is nearest to the line 3x+2y=1. Then 2 (y0-x0) is equal to [2023]

Q 26 : Let T and C respectively be the transverse and conjugate axes of the hyperbola 16x2-y2+64x+4y+44=0. Then the area of the region above the parabola x2=y+4, below the transverse axis T and on the right of the conjugate axis C is: [2023]

Q 27 : Let H be the hyperbola, whose foci are (1±2,0) and eccentricity is 2. Then the length of its latus rectum is ___________ . [2023]

Q 28 : Let the eccentricity of an ellipse x2a2+y2b2=1is reciprocal to that of the hyperbola 2x2-2y2=1. If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

Q 29 : Let Hn:x21+n-y23+n=1, n∈N. Let k be the smallest even value of n such that the eccentricity of Hk is a rational number. If l is the length of the latus rectum of Hk, then 21l is equal to _________. [2023]

Q 30 : Let the tangent to the parabola y2=12x at the point (3,α) be perpendicular to the line 2x+2y=3. Then the square of distance of the point (6, - 4) from the normal to the hyperbola α2x2-9y2=9α2 at its point (α-1,α+2) is equal to ____ . [2023]

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Q 4 :

For 0 < $θ$ < $π$ /2, if the eccentricity of the hyperbola $x^{2} - y^{2} {cosec}^{2} θ = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} {cosec}^{2} θ + y^{2} = 5$ , then the value of $θ$ is [2024]

Q 6 :

Let P be a point on the hyperbola $H : \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ in the first quadrant such that the area of triangle formed by P and the two foci of H is $2 \sqrt{13}$ . Then, the square of the distance of P from the origin is [2024]

Q 12 :

Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

Q 16 :

Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 18 :

If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

Q 19 :

Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

Q 20 :

If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]

Q 21 :

Consider the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and $a^{2} b^{2} = α \sqrt{2} - β, α, β \in N$ , then $α + β$ is ___________. [2025]

Q 25 :

Let $P (x_{0}, y_{0})$ be the point on the hyperbola $3 x^{2} - 4 y^{2} = 36$ , which is nearest to the line $3 x + 2 y = 1 .$ Then $\sqrt{2} (y_{0} - x_{0})$ is equal to [2023]

Q 26 :

Let T and C respectively be the transverse and conjugate axes of the hyperbola $16 x^{2} - y^{2} + 64 x + 4 y + 44 = 0$ . Then the area of the region above the parabola $x^{2} = y + 4,$ below the transverse axis T and on the right of the conjugate axis C is: [2023]

Q 27 :

Let H be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$ . Then the length of its latus rectum is ___________ . [2023]

Q 28 :

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1 .$ If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is ________. [2023]

Q 29 :

Let $H_{n} : \frac{x^{2}}{1 + n} - \frac{y^{2}}{3 + n} = 1, n \in N .$ Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_{k}$ is a rational number. If $l$ is the length of the latus rectum of $H_{k},$ then $21 l$ is equal to _________. [2023]