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)

Abscissae of foci

Answer 2

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 $(\pm a e, 0)=(1,0)$ 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 $\alpha$.

The correct answer is (B): Abscissae of foci