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 =

(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{2}$

Answer 1

(b) $\frac{\mathrm{\pi }}{4}$
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 $a^2=5\mathrm{and}{b}^2=5{\cos }^2\alpha$.

$$\begin{gathered} \Rightarrow b^2=a^2\left({e_1}^2-1\right) \\ \Rightarrow 5{\cos }^2\alpha =5\left({e_1}^2-1\right) \\ \Rightarrow {e_1}^2={\cos }^2\alpha +1.....\left(1\right) \end{gathered}$$

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 $a^2=25\mathrm{and}{b}^2=25{\cos }^2\alpha$.

$$\begin{gathered} b^2=a^2\left(1-{e_2}^2\right) \\ \Rightarrow {e_2}^2=1-{\cos }^2\alpha \\ \Rightarrow {e_2}^2={\sin }^2\alpha ......\left(2\right) \end{gathered}$$

According to the question,

$$\begin{gathered} {\cos }^2\alpha +1=3\left({\sin }^2\alpha \right) \\ \Rightarrow 2=4{\sin }^2\alpha \\ \Rightarrow \sin \alpha =\frac{1}{\sqrt{2}} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \end{gathered}$$