Question

If the eccentricity of the hyperbola $x^2-y^{2}se{c}^2\alpha =5is\sqrt{3}$ times the eccentricity of the ellipse $x^{2}se{c}^2\alpha +y^2=25$, then the 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

$\frac{\pi }{4}$

$\frac{x^2}{5}-\frac{y^2}{5co{s}^2\alpha }=1e_1^2=1+co{s}^2\alpha \dots \left(1\right)$
and $\frac{x^2}{25co{s}^2\alpha }+\frac{y^2}{25}=1e_2^2=si{n}^2\alpha \dots \left(2\right)$
Given that $e_1=\sqrt{3}{e}_2$
$\Rightarrow e_1^2=3e_2^2$
Putting values from equation (1) and (2)
$\Rightarrow 1+co{s}^2\alpha =3si{n}^2\alpha \Rightarrow sin\alpha =\frac{1}{\sqrt{2}}\alpha =\frac{\pi }{4}$