Let frac x 2 a 2 +frac y 2 b 2 =1, a b be an ellipse, whose eccentricity is frac 1 √(2) and the length of the latus rectum is √(14) . Then the square of the eccentricity of frac x 2 a 2 -frac y 2 b 2 =1 is :

Mathematics · Ellipse

JEE Main 2024 — 1 February, Shift 1 — Question 16

Let $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$ . Then the square of the eccentricity of $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is :

Option A:
3
Option B:
$7 / 2$
Option C:
$3 / 2$
Correct
Option D:
$5 / 2$

Answer: C

Step-by-step solution

$e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^{2}}{a^{2}}}$

$\Rightarrow \frac{1}{2}=1-\frac{b^{2}}{a^{2}}$

$\frac{2 b^{2}}{a}=14$

$e_{H}=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$\left(e_{H}\right)^{2}=\frac{3}{2}$

Answer key and solution verified before publishing.

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