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If the eccentricity of the hyperbola $x^2 - y^2 \sec^2 \alpha = 5$ is $\sqrt 3 $ times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25, $ then a value of $\alpha$ is :
$\pi /6$
$\pi /4$
$\pi /3$
$\pi /2$
Solution
$\frac{{{x^2}}}{5} – \frac{{{y^2}}}{{5{{\cos }^2}\alpha }} = 1$
$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $= $ $1 +$ $\frac{{5{{\cos }^2}\alpha }}{5}$ $ = 1 + cos^2\alpha $ ;
parallelly eccentricity of the ellipse
$\frac{{{x^2}}}{{25{{\cos }^2}\alpha }} + \frac{{{y^2}}}{{25}} = 1$ is $e_2^2 = 1 – \frac{{25{{\cos }^2}\alpha }}{{25}}$ $ = sin^2\alpha$ ;
put $e_1 =\sqrt 3 e_2 \Rightarrow$ $e_1^2 = 3e_2^2$
$\Rightarrow$ $1 + cos^2\alpha = 3\sin^2\alpha$
$ \Rightarrow$ $2 = 4 \sin^2\alpha$
$ \Rightarrow $ $\sin \alpha = \frac{1}{{\sqrt 2 }}$
