Question

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 =

पर्याय

• \[\frac{\pi}{6}\]

• \[\frac{\pi}{4}\]

• \[\frac{\pi}{3}\]

• \[\frac{\pi}{2}\]

Answer 1

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

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 

\[a^2 = 5 \text { and } b^2 = 5 \cos^2 \alpha\].

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

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

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

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 

\[a^2 = 25 \text { and } b^2 = 25 \cos^2 \alpha\].

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

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

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

According to the question,

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

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

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

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