1. ## Prove this!!!!

If $e$ and $e'$are the eccentricity of a hyperbola and its conjugate, prove that
$\left(\frac{1}{e^2}\right)+\left(\frac{1}{e'^2}\ri ght)=1$

2. ## Ugly Algebra Proof

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

$\rightarrow e'=\frac{\sqrt{b^2+a^2}}{b}$

$\rightarrow \left( \frac{1}{e^2} \right) + \left( \frac{1}{{e'}^2} \right) = {\left( \frac{1}{\frac{\sqrt{a^2+b^2}}{a}} \right)}^2 + {\left( \frac{1}{\frac{\sqrt{b^2+a^2}}{b}} \right)}^2$

$={\left( \frac{a}{\sqrt{a^2+b^2}} \right)}^2 + {\left( \frac{b}{\sqrt{b^2+a^2}} \right)}^2$

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