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

e_{2}^{2} - e_{1}^{2} = 3

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

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e_{2}^{2} - e_{1}^{2} = - 2

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e_{2}^{2} - e_{1}^{2} = - 3

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Solution

The correct option is B $e_{2}^{2} + e_{1}^{2} = 2$
Ellipse $: 9 x^{2} + 4 y^{2} = 36 \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
$\Rightarrow a = 3, b = 2 ∴ e_{1}^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{4}{9} = \frac{5}{9}$
Hyperbola $: 9 x^{2} - 4 y^{2} = 36 \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
$\Rightarrow a = 3, b = 2 ∴ e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{4}{9} = \frac{13}{9}$
Hence $e_{2}^{2} + e_{1}^{2} = \frac{13}{9} + \frac{5}{9} = 2$

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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

Q. Let

e_{1}

and

e_{2}

are the eccentricities of

16 x^{2} + 9 y^{2} = 144

and

16 x^{2} - 9 y^{2} = 144

respectively, then which of the following is/are correct

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

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

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If e1 and e2 are the eccentricities of the ellipse 9x2+4y2=36 and the hyperbola 9x2−4y2=36 respectively, then

The correct option is B e22+e21=2Ellipse:9x2+4y2=36⇒x24+y29=1⇒a=3,b=2∴e21=1−b2a2=1−49=59Hyperbola :9x2−4y2=36⇒x24−y29=1⇒a=3,b=2∴e22=1+b2a2=1+49=139Hence e22+e21=139+59=2

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