If $e_{1}$ & $e_{2}$ are eccentricities of hyperbola & its conjugate hyperbola then find $\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}}$

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Solution

Let the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow$ conjugate hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$

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

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

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

$\Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = 1$

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