Q79 / 155 · WBJEE 2019 · Mathematics

For the hyperbola $\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1,$ which of the following remains fixed when $\alpha$ varies?

  • A
    directrix
  • B
    vertices
  • Correct answer
    C
    foci
  • D
    eccentricity

Solution

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 $$ \text { equation } \left.\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\right] $$ We know that, foci $=(\pm a e, 0)$ where, $a e=\sqrt{a^{2}+b^{2}}=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}=1$ $\Rightarrow$ foci $(\pm 1,0),$ where vertices are $(\pm \cos \alpha, 0)$ Eccentricity, $a e=1$ or $e=\frac{1}{\cos \alpha}$ Hence, foci remains constant with change in $\alpha$.

Hence, the correct answer is option: 'C'.

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