Question :
If e and e' are eccentricities of hyperbola and its conjugate, prove that

$\displaystyle
\left(\frac{1}{e^2} \right) + \left( \frac{1}{e'^2} \right) = 1
$

Hello, zorro!

If $\displaystyle e$ and $\displaystyle e'$ are eccentricities of a hyperbola and its conjugate,

prove that: .$\displaystyle \frac{1}{e^2}+ \frac{1}{(e')^2} \:=\: 1$

A hyperbola is: .$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1$
. . Its eccentricity is: .$\displaystyle e \:=\:\frac{c}{a}\:=\:\frac{\sqrt{a^2+b^2}}{a}$

Its conjugate is: .$\displaystyle \frac{y^2}{b^2} - \frac{x^2}{a^2} \:=\:1$
. . Its eccentricity is: .$\displaystyle e' \:=\:\frac{c}{b}\:=\:\frac{\sqrt{a^2+b^2}}{b}$


$\displaystyle \text{Hence: }\;\frac{1}{e^2} + \frac{1}{(e')^2} \;\;=\;\;\frac{1}{\left(\frac{\sqrt{a^2+b^2}}{a}\r ight)^2} + \frac{1}{\left(\frac{\sqrt{a^2+b^2}}{b}\right)^2}$

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

Thanks u very much every one for helping me