If the eccentricity of the hyperbola x^{2} − y^{2} sec^{2}α = 5 is \[\sqrt{3}\] times the eccentricity of the ellipse x^{2} sec^{2} α + y^{2} = 25, then α =

#### Options

\[\frac{\pi}{6}\]

\[\frac{\pi}{4}\]

\[\frac{\pi}{3}\]

\[\frac{\pi}{2}\]

#### Solution

\[\frac{\pi}{4}\]

The hyperbola

\[x^2 - y^2 se c^2 \alpha = 5\] can be rewritten in the following way:

\[\frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1\]

This is the standard form of a hyperbola, where

\[\Rightarrow b^2 = a^2 \left( {e_1}^2 - 1 \right)\]

\[ \Rightarrow 5 \cos^2 \alpha = 5\left( {e_1}^2 - 1 \right)\]

\[ \Rightarrow {e_1}^2 = \cos^2 \alpha + 1 . . . . . \left( 1 \right)\]

The ellipse

\[x^2 se c^2 \alpha + y^2 = 25\] can be rewritten in the following way:

\[\frac{x^2}{25 \cos^2 \alpha} + \frac{y^2}{25} = 1\]

This is the standard form of an ellipse, where

\[b^2 = a^2 \left( 1 - {e_2}^2 \right)\]

\[ \Rightarrow {e_2}^2 = 1 - \cos^2 \alpha\]

\[ \Rightarrow {e_2}^2 = \sin^2 \alpha . . . . . . (2)\]

According to the question,

\[\cos^2 \alpha + 1 = 3\left( \sin^2 \alpha \right)\]

\[ \Rightarrow 2 = 4 \sin^2 \alpha\]

\[ \Rightarrow \sin\alpha = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \alpha = \frac{\pi}{4}\]