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

Answer 1

Answer: (B)

Given Hyperbola $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$
given that the angle $\alpha$ varies
We have $a=\cos \alpha ,b=\sin \alpha$
now eccentricity $e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}{\cos \alpha }=\frac{1}{\cos \alpha }$
Thus eccentricity varies with $\alpha$.
Now foci $(\pm ae,0)=(1,0)$ Independent of $\alpha$.
Vertex $(\pm a,0)=(\cos \alpha ,0)$ dependent on $\alpha$.
Directrix is given by $x=\pm \frac{a}{e}={\cos }^2\alpha$
so dependent on $\alpha$.
So abscissae of foci are independent of $\alpha$.