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Mathematics Question on Hyperbola

If the eccentricity of the hyperbola x2−y2cosec2α=25is5x^{2}-y^{2} cos ec^{2} \alpha=25 is \sqrt{5} times the eccentricity of the ellipse x2cosec2α+y2=5,thenαx^{2} cos ec^{2} \alpha+y^{2}=5, then \alpha is equal to :

A

tan−12tan^{-1} \sqrt{2}

B

sin−134 sin^{-1} \sqrt{\frac{3}{4}}

C

tan−125 tan^{-1} \sqrt{\frac{2}{5}}

D

sin−125 sin^{-1}\sqrt{\frac{2}{5}}

Answer

tan−12tan^{-1} \sqrt{2}

Explanation

Solution

Eccentrcity ofx225−y225sin2α=1is 1+sin2α. Eccentrcity \, of \frac{x^{2}}{25}-\frac{y^{2}}{25 sin^{2}\alpha}=1 is\, \sqrt{1+sin^{2}\alpha.}
Eccentricity ofx25sin2α+y25=1is 1−sin2αEccentricity \, of \frac{x^{2}}{5 sin^{2}\alpha}+\frac{y^{2}}{5}=1 is\, \sqrt{1-sin^{2}\alpha}
Given,1+sin2α=51−sin2αGiven, \sqrt{1+sin^{2}\alpha}=\sqrt{5}\sqrt{1-sin^{2}\alpha}
⇒sin2α=23\Rightarrow sin^{2} \alpha =\frac{2}{3}
⇒α=sin−123=tan−12\Rightarrow\alpha=sin^{-1} \sqrt{\frac{2}{3}}=tan^{-1}\sqrt{2}