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

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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(116)

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

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

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

$∴$ $Tangent at (3, α) has slope 1 .$

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

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

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

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

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

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

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

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

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

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

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