If the eccentricity of the hyperbola x2 − y2 sec2α = 5 is \[\sqrt{3}\] times the eccentricity of the ellipse x2 sec2 α + y2 = 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,
\[\cos^2 \alpha + 1 = 3\left( \sin^2 \alpha \right)\]
\[ \Rightarrow 2 = 4 \sin^2 \alpha\]
\[ \Rightarrow \sin\alpha = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \alpha = \frac{\pi}{4}\]
