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. eccentricity
• B. directrix
• C. abscissa of vertices
• D. abscissa of foci

Answer 1

The correct option is D abscissa of foci

The given equation of hyperbola is $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$

Here $a^2={\cos }^2\alpha ,b^2={\sin }^2\alpha$

Coordinates of foci are $(\pm ae,0)$

$\because e=\sqrt{1+\frac{b^2}{a^2}}\Rightarrow \sqrt{1+\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}=\sqrt{1+{\tan }^2\alpha }\Rightarrow e=\sec \alpha$

Hence, abscissae of foci remain constant when $\alpha$ varies