If the eccentricity of the hyperbola $x^{2} - y^{2}\sec^{2}\alpha = 5$ is $\sqrt{3}$ times the eccen

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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 :

Accepted Answer

Sol. (b) $\frac{x^{2}}{1} - \frac{y^{2}}{\cos^{2}\alpha} = 5$;$$\begin{matrix} & \cos^{2}\alpha + 1 = e_{1}^{2} \ & \frac{x^{2}}{\cos^{2}\alpha} + \frac{y^{2}}{1} = 25 \ & 1\left( 1 - e_{2}^{2} ight) = \cos^{2}\alpha \Rightarrow e_{2}^{2} = \sin^{2}\alpha \ & e_{1}^{2} = 3e_{2}^{2} \Rightarrow \cos^{2}\alpha + 1 \ & \ = 3 - 3\cos^{2}\alpha \Rightarrow \cos^{2}\alpha = \frac{1}{2} \end{matrix}$$$$\Rightarrow cos\alpha = \frac{1}{\sqrt{2}}$$

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