If $e_{1}$ and $e_{2}$ are the eccentricities of a hyperbola $3 x^{2} - 3 y^{2} = 25$ and its conjugate respectively, then

e_{1}^{2} + e_{2}^{2} = 2

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e_{1}^{2} + e_{2}^{2} = 4

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e_{1} + e_{2} = 4

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e_{1} + e_{2} = \sqrt{2}

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Solution

The correct option is B $e_{1}^{2} + e_{2}^{2} = 4$
Given hyperbola is $3 x^{2} - 3 y^{2} = 25$ i.e. $\frac{x^{2}}{\frac{25}{3}} - \frac{y^{2}}{\frac{25}{3}} = 1$ , we can see that $a^{2}, b^{2}$ are equal,
i.e. $a^{2} = \frac{25}{3} = b^{2}$
Eccentricity of a hyperbola is $= e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = \frac{a^{2} + b^{2}}{a^{2}} = 2$

And eccentricity of it's conjugate $= e_{2}^{2} = 1 + \frac{a^{2}}{b^{2}} = \frac{a^{2} + b^{2}}{b^{2}} = 2$
So, $e_{1}^{2} + e_{2}^{2} = 4$
Hence, option B is correct.

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Q. If

e_{1}

e_{2}

are eccentricities of hyperbola & its conjugate hyperbola then find

\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}}

$I f e_{1} a n d e_{2} a r e r e s p e c t i v e l y t h e e c c e n t r i c i t i e s o f t h e e l l i p s e \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1 a n d t h e h y p e r b o l a \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1, t h e n t h e r e l a t i o n b e t w e e n e_{1} a n d e_{2} i s$

Q. If e₁ and e₂ are respectively the eccentricities of the ellipse

\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1

and the hyperbola

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

, then the relation between e₁ and e₂ is
(a) 3 e₁² + e₂² = 2
(b) e₁² + 2 e₂² = 3
(c) 2 e₁² + e₂² = 3
(d) e₁² + 3 e₂² = 2

$I f e_{1} i s t h e e c c e n t r i c i t y o f t h e c o n i c 9 x^{2} + 4 y^{2} = 36 a n d e^{2} i s t h e e c c e n t r i c i t y o f t h e c o n i c 9 x^{2} - 4 y^{2} = 36, t h e n$