If $e_{1}, e_{2}$ be respectively the eccentricities of ellipse $9 x^{2} + 4 y^{2} = 36$ and hyperbola $9 x^{2} - 4 y^{2} = 36,$ then:

e_{1}^{2} + e_{2}^{2} > 3

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

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

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

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Solution

The correct options are
C $e_{1}^{2} + e_{2}^{2} > 3$
D $e_{2}^{1} + e_{2}^{2} < 4$
$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1,$ $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1,$ Major axis along y-axis, $a^{2} = 9, b^{2} = 4$
Trans. axes along x-axis a $a^{2} = 4, b^{2} = 9$
$e_{1}^{2} = 1 - \frac{4}{9} = \frac{5}{9}$ ,
$e_{2}^{2} = 1 + \frac{9}{4} = \frac{13}{4}$
$e_{1}^{2} + e_{2}^{2} = \frac{5}{9} + \frac{13}{4} = \frac{137}{36} > 3$ but $< 4.$

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Similar questions

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

Q. If e₁ is the eccentricity of the conic 9x² + 4y² = 36 and e₂ is the eccentricity of the conic 9x² − 4y² = 36, then
(a) e₁² − e₂² = 2
(b) 2 < e₂² − e₁² < 3
(c) e₂² − e₁² = 2
(d) e₂² − e₁² > 3

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