Question

If the eccentricity of the hyperbola $x^{2}-y^{2} cos ec^{2} \alpha=25 is \sqrt{5}$ times the eccentricity of the ellipse $x^{2} cos ec^{2} \alpha+y^{2}=5, then \alpha$ 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

$ Eccentrcity \, of \frac{x^{2}}{25}-\frac{y^{2}}{25 sin^{2}\alpha}=1 is\, \sqrt{1+sin^{2}\alpha.}$
$Eccentricity \, of \frac{x^{2}}{5 sin^{2}\alpha}+\frac{y^{2}}{5}=1 is\, \sqrt{1-sin^{2}\alpha}$
$Given, \sqrt{1+sin^{2}\alpha}=\sqrt{5}\sqrt{1-sin^{2}\alpha}$
$\Rightarrow sin^{2} \alpha =\frac{2}{3}$
$\Rightarrow\alpha=sin^{-1} \sqrt{\frac{2}{3}}=tan^{-1}\sqrt{2}$