Question

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

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

Answer 1

Answer: (B)

Let eccentricity of the hyperbola $\frac{x^2}{5}-\frac{y^2}{5co{s}^2\alpha }=1$ is
$e_1^2=1+\frac{b^2}{a^2}=1+\frac{5co{s}^2\alpha }{5}=1+co{s}^2\alpha ;$
Similarly, eccentricity of the ellipse $\frac{x^2}{25co{s}^2\alpha }+\frac{y^2}{25}=1$ is
$e_2^2=1-\frac{25co{s}^2\alpha }{25}$ $=si{n}^2\alpha ;$
Putting $e_1=\sqrt{3}{e}_2\Rightarrow e_1^2=3e_2^2$
$\Rightarrow 1+co{s}^2\alpha =3si{n}^2\alpha \Rightarrow 2=4si{n}^2\alpha$
$\Rightarrow sin\alpha =\frac{1}{\sqrt{2}}\Rightarrow ta{n}^2\alpha =1$