Consider an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$ .Let a hyperbola is having its vertices at the extremities of minor axis of an ellipse and length of major axis of an ellipse is equal to the distance between the foci of hyperbola. Let e₁ and e₂ be the eccentricities of an ellipse and hyperbola respectively. Then the relation between e₁ and e₂ is

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$e_{1}^{2} (e_{2}^{2} - 1) = 1$

$e_{1} e_{2} (1 - e_{1}^{2}) = 1$

$e_{1} e_{2} = 1$

$e_{2}^{2} (1 - e_{1}^{2}) = 1$

answer is B.

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Detailed Solution

$G i v e n e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b) {e^{2}}_{1} = 1 - \frac{b^{2}}{a^{2}} \Rightarrow 1 - {e_{1}}^{2} = \frac{b^{2}}{a^{2}} G i v e n t h a t v e r t i c e s o f h y p e r o l a a r e (0, b) a n d (0, - b) G i v e n t h a t 2 a = 2 b e_{2} \Rightarrow a = b e_{2} \Rightarrow e_{2} = \frac{a}{b} N o w {e_{2}}^{2} (1 - {e_{1}}^{2}) = \frac{a^{2}}{b^{2}} \times \frac{b^{2}}{a^{2}} = 1$

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