Question

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

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

Answer 1

Answer: (B)

$1$

Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\because b^2=a^2(e^2-1)$
$\Rightarrow e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2}$
Equation of conjugate hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$
$\Rightarrow \frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
$\therefore e^{{}^{\prime }2}=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}$
Now, $\frac{1}{e^2}+\frac{1}{e^{{}^{\prime }2}}=\frac{a^2+b^2}{a^2+b^2}=1$