Mathematics | VITEEE 2017 | Discussion Page

If the eccentricity of the hyperbola

x²-y² cosec²α= 25 is $\sqrt{5}$ times the

eccentricity of the elipse x² cosec²α+y²=5 then α is equal to;

$\tan^{-1}\sqrt{2}$

$\sin^{-1}\sqrt{\frac{3}{4}}$

$\tan^{-1}\sqrt{\frac{2}{5}}$

$\sin^{-1}\sqrt{\frac{2}{5}}$

Option A

Eccentrcity of $\frac{x^{2}}{25}+\frac{y^{2}}{25\sin^{2}\alpha}=1$ is

$\sqrt{1+\sin^{2}\alpha}$

Eccentrcity of $\frac{x^{2}}{5\sin^{2}\alpha}+\frac{y^{2}}{5}=1$ is

$\sqrt{1-\sin^{2}\alpha}$

Given, $\sqrt{1+\sin^{2}\alpha}=\sqrt{5}\sqrt{1-\sin^{2}\alpha}$

⇒ $\sin^{2}\alpha=\frac{2}{3}$

⇒ $\alpha=\sin^{-1}\sqrt{\frac{2}{3}}=\tan^{-1}\sqrt{2}$

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