Question

If the eccentricity of the hyperbola, $x^2-y^2{\sec }^2\alpha =5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2{\sec }^2\alpha +y^2-16$, then a value of $\alpha$ is

• A. $\pi /6$
• B. $\pi /4$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

The correct option is B $\pi /4$

Equation of hyperbola $\Rightarrow x^2-y^2{\sec }^2\alpha =5$

$\Rightarrow \frac{x^2}{5}-\frac{y^2}{5{\cos }^2\alpha }=1$

$e_1.\sqrt{1+\frac{5{\cos }^2\alpha }{5}}=\sqrt{1+{\cos }^2\alpha }$

Equation of ellipse $\Rightarrow x^2{\sec }^2\alpha +y^2=16$

$\Rightarrow \frac{x^2}{16{\cos }^2\alpha }+\frac{y^2}{16}=1$

$e_2.\sqrt{1-\frac{16{\cos }^2\alpha }{16}}=\sqrt{1-{\cos }^2\alpha }$

A/Q $e_1=\sqrt{3}.e_2$

$=\sqrt{1+{\cos }^2\alpha }=\sqrt{3}.\sqrt{1-{\cos }^2\alpha }$

$\Rightarrow 1+{\cos }^2\alpha =3-3{\cos }^2\alpha$

$\Rightarrow 4{\cos }^2\alpha =2$

$\Rightarrow {\cos }^2\alpha =1/2$

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

$\Rightarrow \alpha =\pi /4$