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 $=(\pm ae,0)$
where, $ae=\sqrt{a^2+b^2}=\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }=1$
$\Rightarrow$ Foci $=(\pm 1,0)$
where, vertices are $(\pm \cos \alpha ,0)$.
Eccentricity, $ae=1$ or $e=\frac{1}{\cos \alpha }$
Hence, foci remains constant with change in $\alpha$.