Let H : frac x 2 a 2 -frac y 2 b 2 =1 , where a b 0 , be a hyperbola in the xy-plane whose conjugate axis LM subtends an angle of 60 ° at one of its vertices N . Let the area of the triangle LMN be 4 √(3) . 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 √(3) R. The distance between the foci of H is 3. frac 2 √(3) S. The length of the latus rectum of H is 4. 4

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

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

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

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

Mathematics · Hyperbola

JEE Advanced 2018 — Paper 2 — Question 44

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

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

Option A:
$\mathbf{P} \rightarrow \mathbf{4 ; Q} \rightarrow \mathbf{2 ; R}\rightarrow \mathbf{1 ; S \rightarrow 3}$
Option B:
$\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 2$
Correct
Option C:
$\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow \mathbf{1} ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 2$
Option D:
$\mathbf{P} \rightarrow \mathbf{3} ; \mathbf{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow \mathbf{1}$

Answer: B

Step-by-step solution

Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a > b > 0$ . Endpoints of conjugate axis: $L(0,b), M(0,-b)$ ; one vertex $N(a,0)$ . Angle subtended by LM at N is $60^\circ$ . Since $\angle LNM = 2\theta$ with $\tan\theta = \frac{b}{a}$ , we get $\theta = 30^\circ$ , so $\tan30^\circ = \frac{1}{\sqrt{3}} = \frac{b}{a}$ → $a = b\sqrt{3}$ . Area of $\triangle LMN = \frac{1}{2} \times (2b) \times a = ab = 4\sqrt{3}$ . Substitute $a = b\sqrt{3}$ : $b^2\sqrt{3} = 4\sqrt{3}$ → $b^2 = 4$ → $b = 2$ , $a = 2\sqrt{3}$ . Length of conjugate axis: $2b = 4$ . Eccentricity: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ . Distance between foci: $2ae = 2 \times 2\sqrt{3} \times \frac{2}{\sqrt{3}} = 8$ . $(\mathrm{P}) \rightarrow(4),(\mathrm{Q}) \rightarrow(3),(\mathrm{R}) \rightarrow(1),(\mathrm{S}) \rightarrow(2)$

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola