Question

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

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

Answer 1

The correct option is B 1

$e^2=\frac{a^2+b^2}{a^2}$ and $e^{\prime 2}=\frac{a^2+b^2}{b^2}$
$\Rightarrow \frac{1}{e^2}+\frac{1}{e^{\prime 2}}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$