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 $\frac{\pi }{4}$

Equation of the hyperbola $:\frac{x^2}{5}-\frac{y^2}{5{\text{cos}}^2\alpha }=1$

$a^2=5,b^2=5{\text{cos}}^2\alpha (a>b)$

Eccentricity of this hyperbola $\left(e_1\right)=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+{\text{cos}}^2\alpha }$

Equation of the ellipse $:\frac{x^2}{25{\text{cos}}^2\alpha }+\frac{y^2}{25}=1$

$a^2=25{\text{cos}}^2\alpha ,b^2=25(a<b)$

Eccentricity of this ellipse $\left(e_2\right)=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-{\text{cos}}^2\alpha }$

Now, it is given that $e_1=\sqrt{3}{e}_2$

$\Rightarrow \sqrt{1+{\text{cos}}^2\alpha }=\sqrt{3(1-{\text{cos}}^2\alpha )}$

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

$\Rightarrow {\text{cos}}^2\alpha =\frac{1}{2}$

A possible value of $\alpha$ is $\frac{\pi }{4}$