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

Q 31 :

Let $m_{1}$ and $m_{2}$ be the slopes of the tangents drawn from the point P(4, 1) to the hyperbola $H : \frac{y^{2}}{25} - \frac{x^{2}}{16} = 1 .$ If Q is the point from which the tangents drawn to H have slopes $| m_{1} |$ and $| m_{2} |$ and they make positive intercepts $α$ and $β$ on the $x$ -axis, then $\frac{{(P Q)}^{2}}{α β}$ is equal to _________ . [2023]

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(8)

$Equation of tangent to the hyperbola \frac{y^{2}}{25} - \frac{x^{2}}{16} = 1$

$y = m x \pm \sqrt{25 - 16 m^{2}}$ ...(i)

$Equation (i) passes through (4, 1),$

$1 = 4 m \pm \sqrt{25 - 16 m^{2}}$ $\Rightarrow 1 - 4 m = \pm \sqrt{25 - 16 m^{2}}$

$\Rightarrow 1 + 16 m^{2} - 8 m = 25 - 16 m^{2} \Rightarrow 32 m^{2} - 8 m - 24 = 0$

$\Rightarrow 4 m^{2} - m - 3 = 0 \Rightarrow 4 m^{2} - 4 m + 3 m - 3 = 0$

$\Rightarrow 4 m (m - 1) + 3 (m - 1) = 0$

$\Rightarrow (4 m + 3) (m - 1) = 0 \Rightarrow m = 1, - \frac{3}{4}$

It is given that $| m_{1} |$ and $| m_{2} |$ are the slopes of tangents, so

$| m_{1} |$ $= 1$ and $| m_{2} |$ $= \frac{3}{4}$

Equations of tangents with positive slope are:

$y = x \pm \sqrt{25 - 16} \Rightarrow y = x \pm 3$

and $y = \frac{3}{4} x \pm \sqrt{25 - 16 \times \frac{9}{16}} \Rightarrow 4 y = 3 x \pm 16$

$For positive intercept on the x -axis, y = x - 3$ and $4 y = 3 x - 16$

$∴$ $α = \frac{16}{3}$ and $β = 3$

$Solve y = x - 3 and 4 y = 3 x - 16 to find their intersection point.$

$\Rightarrow x = - 4$ and $y = - 7$

$So, Q \equiv (- 4, - 7)$

${(P Q)}^{2} = {(4 + 4)}^{2} + {(1 + 7)}^{2} = 64 + 64 = 128$

$∴$ $\frac{{(P Q)}^{2}}{α β} = \frac{128}{16} = 8$

Q 32 :

The foci of a hyperbola are $(\pm 2, 0)$ and its eccentricity is $\frac{3}{2}$ . A tangent perpendicular to the line $2 x + 3 y = 6$ is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$ - and $y$ -axes are $a$ and $b$ respectively, then $| 6 a |$ $+$ $| 5 b |$ is equal to _________ . [2023]

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(12)

$e = \frac{3}{2}, a e = 2$

$\Rightarrow a = \frac{4}{3} \Rightarrow e^{2} = 1 + \frac{b^{2}}{a^{2}}$

$\Rightarrow \frac{5}{4} = \frac{9 b^{2}}{16} \Rightarrow b^{2} = \frac{5}{4} \times \frac{16}{9} = \frac{20}{9}$

$Equation of tangent:$ $y = m x + \sqrt{a^{2} m^{2} - b^{2}}$

$y = \frac{3}{2} x + \sqrt{\frac{16}{9} \cdot \frac{9}{4} - \frac{20}{9}}$

$y = \frac{3}{2} x + \frac{4}{3}$

$y-intercept = \frac{4}{3},$ $x-intercept = \frac{- 8}{9}$

$| 6 a |$ $+$ $| 5 b |$ $=$ $| 6 \times \frac{- 8}{9} |$ $+$ $| 5 \times \frac{4}{3} |$ $= \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$

Q 33 :

The vertices of a hyperbola H are $(\pm 6, 0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$ . Let N be the normal to H at a point in the first quadrant and parallel to the line $\sqrt{2} x + y = 2 \sqrt{2}$ . If $d$ is the length of the line segment of N between H and the y-axis, then $d^{2}$ is equal to _________ . [2023]

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(216)

$Equation of hyperbola = \frac{x^{2}}{36} - \frac{y^{2}}{b^{2}} = 1$

$Let H be (x_{1}, y_{1})$

$b^{2} = 36 (\frac{5}{4} - 1) = 9$

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

$Equation of normal: \frac{36 x}{x_{1}} + \frac{9 y}{y_{1}} = 45$ $\Rightarrow - \frac{4 y_{1}}{x_{1}} = - \sqrt{2}$

$x_{1} = 2 \sqrt{2} y_{1}$ and $\frac{{(2 \sqrt{2} y_{1})}^{2}}{36} - \frac{y_{1}^{2}}{9} = 1$

$\frac{8 y_{1}^{2}}{36} - \frac{y_{1}^{2}}{9} = 1$ $\Rightarrow y_{1} = 3 (y_{1} > 0)$ $\Rightarrow x_{1} = 6 \sqrt{2}$

$Equation of normal: y = - \sqrt{2} x + 15$

$This intersects the y–axis at (0, 15)$

$d^{2} = {(6 \sqrt{2} - 0)}^{2} + {(3 - 15)}^{2}$

$= 72 + 144 = 216$

Q 34 :

Let the domain of the function $f (x)$ $= \log_{3} \log_{5} \log_{7} (9 x - x^{2} - 13)$ be the interval $(m, n)$ . Let the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ have eccentricity $\frac{n}{3}$ and the length of the latus rectum $\frac{8 m}{3} .$ Then $b^{2} - a^{2}$ is equal to: [2026]

9
11
5
7

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2

3

4

(4)

$\log_{5} (\log_{7} (9 x - x^{2} - 13)) > 0$

$\Rightarrow 9 x - x^{2} - 13 > 7$

$x^{2} - 9 x + 20 < 0 \Rightarrow 4 < x < 5$

$m = 4, n = 5$

$\Rightarrow e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \frac{5}{3} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{25}{9} - 1 = \frac{16}{9}$

$\frac{b}{a} = \frac{4}{3}$

$\Rightarrow \frac{2 b^{2}}{a} = \frac{8 m}{3} \Rightarrow \frac{2 b^{2}}{a} = \frac{32}{3}$

$\Rightarrow 2 b^{2} = \frac{32}{3} \times \frac{3 b}{4} \Rightarrow b = 4, a = 3$

$b^{2} - a^{2} = 16 - 9 = 7$

Q 35 :

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If the eccentricity of the hyperbola is 5 then the length of its latus rectum is: [2026]

$24 \sqrt{5}$
$\frac{96}{\sqrt{5}}$
$12$
16

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2

3

4

(2)

$Let e_{1} be eccentricity of ellipse$

$\Rightarrow e_{1} = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$

$So a e_{1} = 6 \cdot \frac{\sqrt{5}}{3} = 2 \sqrt{5}$

$Now H : \frac{x^{2}}{p^{2}} - \frac{y^{2}}{q^{2}} = 1$

$p . e = a e_{1}$

$p \cdot 5 = 2 \sqrt{5}$

$p = \frac{2}{\sqrt{5}} \Rightarrow e^{2} = 1 + \frac{q^{2}}{p^{2}} \Rightarrow 25 = 1 + \frac{5 q^{2}}{4} \Rightarrow q^{2} = \frac{96}{5}$

$So length of LR = \frac{2 q^{2}}{p} = \frac{96}{\sqrt{5}}$

Q 36 :

If the line $α x + 2 y = 1$ , where $α \in ℝ$ , does not meet the hyperbola $x^{2} - 9 y^{2} = 9,$ then a possible value of $α$ is: [2026]

0.5
0.6
0.7
0.8

1

2

3

4

(4)

$y = \frac{1 - α x}{2}$

Put this in equation of hyperbola

$∴$ $x^{2} - 9 {(\frac{1 - α x}{2})}^{2} = 9$

$(4 - 9 α^{2}) x^{2} + 18 α x - 45 = 0$

$∵$ $line does not intersect hyperbola$

$∴$ $D < 0$

$\Rightarrow α^{2} - \frac{5}{9} > 0$

$\Rightarrow α \in (- \infty, - \frac{\sqrt{5}}{3}) \cup (\frac{\sqrt{5}}{3}, \infty)$

$Here \frac{\sqrt{5}}{3} \approx 0.74$

Q 37 :

Let the ellipse $E : \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H : \frac{x^{2}}{16} - \frac{y^{2}}{λ^{2}} = - 1$ have the same foci. If e and L respectively denote the eccentricity and the length of latus rectum of H, then the value of 24(e+L) is : [2026]

148
126
296
67

1

2

3

4

(3)

Q 38 :

Let PQ be a chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{b^{2}} = 1$ perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$ then the area of the triangle OPQ is. [2026]

$\frac{9}{5}$
$\frac{8 \sqrt{3}}{5}$
$\frac{11}{5}$
$2 \sqrt{3}$

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(2)

$e = \sqrt{1 + \frac{b}{4}} = \sqrt{3} \Rightarrow b = 8$

$∴$ $Hyperbola: \frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$

[IMAGE 191]

$\frac{P M}{O M} = \tan 30 °$

$\Rightarrow \frac{2 \sqrt{2} \tan θ}{2 \sec θ} = \frac{1}{\sqrt{3}} \Rightarrow \sin θ = \frac{1}{\sqrt{6}}$

$Area = 2 \times \frac{1}{2} \times O M \times M P$

$= 2 \sec θ \times 2 \sqrt{2} \tan θ$

$= 4 \sqrt{2} \frac{\sin θ}{\cos^{2} θ}$

$= 4 \sqrt{2} \times \frac{1}{\sqrt{6} (1 - \frac{1}{6})}$

$= \frac{8 \sqrt{3}}{5}$

Q 39 :

Let $P (10, 2 \sqrt{15})$ be a point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ whose foci are S and S'. If the length of its latus rectum is 8 then the square of the area of $∆ P S S'$ is equal to: [2026]

4200
900
1462
2700

1

2

3

4

(4)

$P (10, 2 \sqrt{15}) lies on \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$∴$ $\frac{100}{a^{2}} - \frac{60}{b^{2}} = 1 . . . (1)$

$∵$ $length of latus rectum = 8$

$\frac{2 b^{2}}{a} = 8 \Rightarrow \frac{b^{2}}{a} = 4 . . . (2)$

$From (1) & (2)$

$\frac{100}{a^{2}} - \frac{60}{4 a} = 1$

$400 - 60 a = 4 a^{2}$

$4 a^{2} + 60 a - 400 = 0$

$a^{2} + 15 a - 100 = 0$

$a = 5 & - 20 (rejected)$

$\Rightarrow b = \sqrt{20}$

$∴$ $Hyperbola is \frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$

$∴$ $Focal length S_{1} S_{2} = 2 a e$ $= 2 \cdot 5 (\sqrt{1 + \frac{4}{5}}) = 6 \sqrt{5}$

$∴$ $Area of ∆ P S_{1} S_{2} = \frac{1}{2} \cdot 6 \sqrt{5} \cdot 2 \sqrt{15} = 30 \sqrt{3} = A$

$∴$ $A^{2} = 2700$

Q 40 :

For some $θ \in ((0, \frac{π}{2}))$ , let the eccentricity and the length of the latus rectum of the hyperbola $x^{2} - y^{2} \sec^{2} θ = 8$ be $e_{1} a n d l_{1},$ respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^{2} \sec^{2} θ + y^{2} = 6$ be $e_{2} a n d l_{2}$ , respectively. If $e_{1}^{2} = e_{2}^{2} ((s {ec}^{2} θ + 1)),$ then $((\frac{l_{1} l_{2}}{e_{1} e_{2}})) \tan^{2} θ$ is equal to______. [2026]

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Q 35 :

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If the eccentricity of the hyperbola is 5 then the length of its latus rectum is: [2026]

Q 36 :

If the line $α x + 2 y = 1$ , where $α \in ℝ$ , does not meet the hyperbola $x^{2} - 9 y^{2} = 9,$ then a possible value of $α$ is: [2026]

Q 37 :

Let the ellipse $E : \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H : \frac{x^{2}}{16} - \frac{y^{2}}{λ^{2}} = - 1$ have the same foci. If e and L respectively denote the eccentricity and the length of latus rectum of H, then the value of 24(e+L) is : [2026]

Q 38 :

Let PQ be a chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{b^{2}} = 1$ perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$ then the area of the triangle OPQ is. [2026]

Q 39 :

Let $P (10, 2 \sqrt{15})$ be a point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ whose foci are S and S'. If the length of its latus rectum is 8 then the square of the area of $∆ P S S'$ is equal to: [2026]

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