Let the foci and length of the latus rectum of an ellipse , a > b be (5, 0) and , respectively. Then, the square of the eccentricity of the hyperbola equals [2024]

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Solution

Q.
Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

Ans.
(51)

Here ae = 5 and $\frac{2 b^{2}}{a} = \sqrt{50}$

$\frac{b^{2}}{a} = \frac{5 \sqrt{2}}{2}$

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

$25 = a^{2} - \frac{5 \sqrt{2}}{2} a \Rightarrow 2 a^{2} - 5 \sqrt{2} a - 50 = 0$

$\Rightarrow [a - 5 \sqrt{2}] [2 a + 5 \sqrt{2}] = 0 \Rightarrow a = 5 \sqrt{2} or a = \frac{- 5 \sqrt{2}}{2}$

$\Rightarrow e = \frac{5}{a} = \frac{1}{\sqrt{2}}$

Also, $b^{2} = 5 \sqrt{2} \times \frac{5 \sqrt{2}}{2} = 25 \Rightarrow b = 5$

Now, $e = \sqrt{1 + \frac{a^{2} b^{2}}{b^{2}}} = \sqrt{1 + a^{2}} \Rightarrow e^{2} = 1 + a^{2} = 51$ .

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