Question

If e and e' be the eccentricities of a hyperbola and its conjugate, then $\frac{1}{e^2}+\frac{1}{e^{\prime 2}}$ is equal to

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

The correct option is B 1

Suppose $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be a hyperbola and let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ be its conjugate.
Then their eccentricities are given by $e^2=\frac{a^2+b^2}{a^2}$ and $e^{\prime 2}=\frac{a^2+b^2}{b^2}$ respectively.
$\therefore \frac{1}{e^2}+\frac{1}{e^{\prime 2}}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$