Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

For hyperbola $\frac{x^2}{5}-\frac{y^2}{5co{s}^2\alpha }=1$

we have

$e_1^2=1+\frac{b^2}{a^2}=1+\frac{5co{s}^2\alpha }{5}$

$=1+co{s}^2\alpha$

For ellipse

$\frac{x^2}{25co{s}^2\alpha }+\frac{y^2}{25}=1$

We have, $e_2^2=1-\frac{25co{s}^2\alpha }{25}=si{n}^2\alpha$

$\Rightarrow$ $e_1=\sqrt{3}{e}_2$

$\Rightarrow$ $e_1^2=3e_2^2$

$\Rightarrow$ $1+co{s}^2\alpha =3si{n}^2\alpha$

$\Rightarrow$ $2a=4si{n}^2\alpha$

$\Rightarrow$ $sin\alpha =\frac{1}{\sqrt{2}}$

$\Rightarrow$ $\alpha =\frac{\pi }{4}$