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If the eccentricity of the hyperbola $x^{2}-y^{2} \operatorname{cosec}^{2} \alpha=25$ is $\sqrt{5}$ times the eccentricity of the ellipse $x^{2} \operatorname{cosec}^{2} \alpha+y^{2}=5$ then $\alpha$ is equal to :
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Verified Answer
The correct answer is:
$\tan ^{-1} \sqrt{2}$
Eccentrcity of $\frac{x^{2}}{25}-\frac{y^{2}}{25 \sin ^{2} \alpha}=1$ is $\sqrt{1+\sin ^{2} \alpha}$
Eccentricity of $\frac{x^{2}}{5 \sin ^{2} \alpha}+\frac{y^{2}}{5}=1 \quad$ is $\sqrt{1-\sin ^{2} \alpha}$
Given, $\sqrt{1+\sin ^{2} \alpha}=\sqrt{5} \sqrt{1-\sin ^{2} \alpha}$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \alpha=\frac{2}{3} \\
\Rightarrow \alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}
\end{array}
$$
Eccentricity of $\frac{x^{2}}{5 \sin ^{2} \alpha}+\frac{y^{2}}{5}=1 \quad$ is $\sqrt{1-\sin ^{2} \alpha}$
Given, $\sqrt{1+\sin ^{2} \alpha}=\sqrt{5} \sqrt{1-\sin ^{2} \alpha}$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \alpha=\frac{2}{3} \\
\Rightarrow \alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}
\end{array}
$$
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