Question

If $e$ and $e_1$ are the eccentricities of a hyperbola and its conjugate, then $\frac{1}{e^2}+\frac{1}{e_1^2}$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

Consider Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1..\left(1\right)$
So it's conjugate hyperbola is $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1..\left(2\right)$
Given $e$ and $e_1$ are eccentricties of (1) and (2) respectively
$\Rightarrow e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2}$
& $e_1^2=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}$
Thus $\frac{1}{e^2}+\frac{1}{e_1^2}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$
Hence, option 'C' is correct.