Question

If the eccentricity of the hyperbola $x^2-y^2{\text{coesc}}^2\alpha =25$ is $\sqrt{5}$ times the eccentricity of the ellipse $x^{2}co{sec}^2\alpha +y^2=5$, then $\alpha$ is equal to

• A. ${\tan }^{-1}\sqrt{2}$
• B. ${\sin }^{-1}\sqrt{\frac{3}{4}}$
• C. ${\tan }^{-1}\sqrt{\frac{2}{5}}$
• D. ${\sin }^{-1}\sqrt{\frac{2}{5}}$

Answer 1

The correct option is A ${\tan }^{-1}\sqrt{2}$

Given hyperbola is $x^2-y^2{\csc }^2\alpha =25$

This can be written as $\frac{x^2}{25}-\frac{y^2}{25{\sin }^2\alpha }=1$

here we see $a=5,b=5\sin \alpha$

so eccentricity $e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{25+25{\sin }^2\alpha }}{5}=\sqrt{1+{\sin }^2\alpha }=e_1$

Now for the given ellipse, $x^2{\csc }^2\alpha +y^2=5$

or $\frac{x^2}{5{\sin }^2\alpha }+\frac{y^2}{5}=1$

here we get $a=\sqrt{5}\sin \alpha ,b=\sqrt{5}$

so eccentricity $e_2=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{5{\sin }^2\alpha }{5}}=\cos \alpha$

Now we are given that $e_1=\sqrt{5}{e}_2$

thus on solving this we get $\sin \alpha =\sqrt{\frac{2}{3}}$

or $\tan \alpha =\sqrt{2}\Rightarrow \alpha ={\tan }^{-1}\sqrt{2}$