Let , a > b be an ellipse, whose eccentricity is and the length of the latusrectum is . Then the square of the eccentricity of is: [2024]

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Solution

Q.
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 latusrectum is $\sqrt{14}$ . Then the square of the eccentricity of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is: [2024]

1 3

2 3/2

3 7/2

4 5/2

Ans.
(2)

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

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

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 : e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{1}{2} = \frac{3}{2}$

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