Q74 / 120 · VITEEE 2017 · Mathematics
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 :
- Correct answerA$\tan ^{-1} \sqrt{2}$
- B$\sin ^{-1} \sqrt{\frac{3}{4}}$
- C$\tan ^{-1} \sqrt{\frac{2}{5}}$
- D$\sin ^{-1} \sqrt{\frac{2}{5}}$
Solution
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, $\sqrt{1+\sin ^{2} \alpha}=\sqrt{5} \sqrt{1-\sin ^{2} \alpha}$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \alpha=\frac{2}{3} \\
\Rightarrow \alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}
\end{array}
$$
Hence, the correct answer is option: 'A'.