Question

If $e_1$ and $e_2$ are respectively the eccentricities of a hyperbola and its conjugate. Prove that $\frac{1}{e_1^2}+\frac{1}{e_2^2}=1$.

Answer 1

Let,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .....(1) and

$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ .....(2) are two hyperbola conjugate to each other.

Also let, $e_1$ and $e_2$ are the eccentricities of hyperbola (1) and (2) respectively.

Then, ${e_1}^2=1+\frac{b^2}{a^2}and{e_2}^2=1+\frac{a^2}{b^2}$

$\therefore ,\frac{1}{e_1^2}+\frac{1}{e_2^2}$

$=\frac{1}{(1+\frac{b^2}{a^2})}+\frac{1}{(1+\frac{a^2}{b^2})}$

$=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}$

$=\frac{a^2+b^2}{a^2+b^2}$

$=1.$

Hence, proved.