Question

For the hyperbola $\frac{x^2}{\cos^2\,\alpha}-\frac{y^2}{\sin^2{\alpha}}=1$ which of the following remains constant when $\alpha$ varies

• A. Directrix
• B. Abscissae of vertices
• C. Abscissae of foci
• D. Eccentricity

Answer 1

Answer: (C)

Given equation of hyperbola is
$\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1$
Here, $ a^{2}=\cos ^{2} \alpha$ and $b^{2}=\sin ^{2} \alpha$
[i.e. comparing with standard equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ]
We know that, foci $=(\pm a e, 0)$
where, $a e=\sqrt{a^{2}+b^{2}}=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}=1$
$\Rightarrow$ Foci $=(\pm 1,0)$
where, vertices are $(\pm \cos \alpha, 0)$.
Eccentricity, $ a e=1$ or $e=\frac{1}{\cos \alpha}$
Hence, foci remains constant with change in $\alpha$.