The ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (a > b) and the hyperbola $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ are given to be confocal and length of minor axis of the ellipse is same as the conjugate axis of the hyperbola. If $e_{1}$ and $e_{2}$ represents the eccentricities of ellipse and hyperbola respectively, then the value of $e_{1}^{- 2} + e_{2}^{- 2}$ is

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answer is 2.

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Detailed Solution

(2) $a e_{1} = \sqrt{a^{2} - b^{2}} \Rightarrow a^{2} e_{1}^{2} = a^{2} - b^{2}$

$\begin{array}{l} \Rightarrow A e_{2} = \sqrt{A^{2} + B^{2}} = \sqrt{A^{2} + b^{2}} \Rightarrow A^{2} . e_{2}^{2} = A^{2} + b^{2} \\ \Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = \frac{a^{2}}{a^{2} - b^{2}} + \frac{A^{2}}{a^{2} + b^{2}} \\ B u t a e_{1} = A e_{2} \Rightarrow a^{2} e_{1}^{2} = A^{2} e_{2}^{2} \Rightarrow a^{2} - b^{2} = A^{2} + b^{2} \\ \Rightarrow A^{2} = a^{2} - 2 b^{2} \\ \Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = \frac{a^{2}}{a^{2} - b^{2}} + \frac{a^{2} - 2 b^{2}}{a^{2} - b^{2}} = 2 \end{array}$