Hyperbola 1 Question 12
13. For hyperbola $\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1$, which of the following remains constant with change in ’ $\alpha$ ‘? (2003, 1M)
(a) Abscissae of vertices
(b) Abscissae of foci
(c) Eccentricity
(d) Directrix
$(2006,3 M)$
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Answer:
Correct Answer: 13. (b)
Solution:
- Given equation of hyperbola is $\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1$
Here, $\quad a^{2}=\cos ^{2} \alpha$ 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 $=( \pm a e, 0)$
where, $a e=\sqrt{a^{2}+b^{2}}=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}=1$ $\Rightarrow \quad$ Foci $=( \pm 1,0)$
where, vertices are $( \pm \cos \alpha, 0)$.
Eccentricity, $\quad a e=1$ or $e=\frac{1}{\cos \alpha}$
Hence, foci remains constant with change in $\alpha$.
