For 0<θ<π / 2 , if the eccentricity of the hyperbola x 2 -y 2 cosec 2 θ=5 is √(7) times eccentricity of the ellipse x 2 cosec 2 θ+y 2 =5 , then the value of θ is :

Mathematics · Hyperbola

JEE Main 2024 — 1 February, Shift 1 — Question 13

For $0<\theta<\pi / 2$ , if the eccentricity of the hyperbola $x^{2}-y^{2} \operatorname{cosec}^{2} \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^{2} \operatorname{cosec}^{2} \theta+y^{2}=5$ , then the value of $\theta$ is :

Option A:
$\frac{\pi}{6}$
Option B:
$\frac{5 \pi}{12}$
Option C:
$\frac{\pi}{3}$
Correct
Option D:
$\frac{\pi}{4}$

Answer: C

Step-by-step solution

$e_{h}=\sqrt{1+\sin ^{2} \theta}$

$e_{c}=\sqrt{1-\sin ^{2} \theta}$

$e_{h}=\sqrt{7} e_{c}$

$1+\sin ^{2} \theta=7\left(1-\sin ^{2} \theta\right)$

$\sin ^{2} \theta=\frac{6}{8}=\frac{3}{4}$

$\sin \theta=\frac{\sqrt{3}}{2}$

$\theta=\frac{\pi}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola