Mathematics · Ellipse

JEE Main 2024 — 27 January, Shift 2 — Question 18

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 $\mathrm{e}_{1} \mathrm{e}_{2}=1$ , then the length of the chord of the ellipse parallel to the x -axis and passing through $(0,2)$ is :

Option A:
$4 \sqrt{5}$
Option B:
$\frac{8 \sqrt{5}}{3}$
Option C:
$\frac{10 \sqrt{5}}{3}$
Correct
Option D:
$3 \sqrt{5}$

Answer: C

Step-by-step solution

$H: \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \quad$

$e_{1}=\frac{5}{4}$

$\therefore e_{1} e_{2}=1$

$\Rightarrow e_{2}=\frac{4}{5}$

Also, ellipse is passing through $( \pm 5,0)$

$\therefore \mathrm{a}=5$ and $\mathrm{b}=3$

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

End point of chord are $\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$

$\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Mathematics
Chapter: Ellipse
Topic: Chords connected with an Ellipse