Question

If $e,e_1$ are the eccentricities of hyperbola and its conjugate hyperbola prove that $\frac{1}{e^2}+\frac{1}{e_1^2}=1$.

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, $e$ and $e_1$ are the eccentricities of (1) and (2) respectively.

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

$⟹\frac{1}{1+\frac{b^2}{a^2}}+\frac{1}{1+\frac{a^2}{b^2}}$

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

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

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

$=1$ (Proved)