Question

For hyperbola $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$ which of the following remains constant with change in $\alpha$

• A. abscissae of vertices
• B. abscissae of foci
• C. eccentricity
• D. directrix

Answer 1

The correct option is B abscissae of foci

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, 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 as vertices are $(\pm \cos \alpha ,0)$
Eccentricity $ae=1$ $\Rightarrow e=\frac{1}{\cos \alpha }$
Hence foci reamins constant with change in ${}^{\prime }{\alpha }^{\prime }$