Let frac{x^2}{a^2}+frac{y^2}{b^2}=1, a>b be an ellipse, whose eccentricity is frac{1}{√{2}} and

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10-2. Parabola, Ellipse, Hyperbola

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Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ 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 :

A

$3$

B

$7 / 2$

C

$3 / 2$

D

$5 / 2$

(JEE MAIN-2024)

Solution

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

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

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

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

Standard 11

Mathematics

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