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 $tan^{2} \alpha $ is equal to

• A. $2$
• B. $1$
• C. $3$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

Let eccentricity of the hyperbola $\frac{x^{2}}{5}-\frac{y^{2}}{5 cos^{2} \alpha }=1$ is
$e_{1}^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{5 cos^{2} \alpha }{5}=1+cos^{2} ⁡ \alpha ;$
Similarly, eccentricity of the ellipse $\frac{x^{2}}{25 cos^{2} \alpha }+\frac{y^{2}}{25}=1$ is
$e_{2}^{2}=1-\frac{25 cos^{2} \alpha }{25}$ $=sin^{2} \alpha ;$
Putting $e_{1}=\sqrt{3}e_{2}\Rightarrow e_{1}^{2}=3e_{2}^{2}$
$\Rightarrow 1+cos^{2} \alpha =3sin^{2} ⁡ \alpha \Rightarrow 2=4sin^{2} ⁡ \alpha $
$\Rightarrow sin \alpha =\frac{1}{\sqrt{2}}\Rightarrow tan^{2} ⁡ \alpha =1$