If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29−y24=1 respectively and (e1, e2) is a point on the ellipse 15x2+3y2=k then ‘k’ is

e1=1−418=73; e2=1+49=133∵ (e1, e2) lies on 15x2+3y2=k⇒15e12+3e22=k⇒k=16

If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29−y24=1 respectively and (e1, e2) is a point on the ellipse 15x2+3y2=k then ‘k’ is

e1=1−418=73; e2=1+49=133∵ (e1, e2) lies on 15x2+3y2=k⇒15e12+3e22=k⇒k=16

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If $e_{1} and e_{2}$ are the eccentricities of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse $15 x^{2} + 3 y^{2} = k$ then ‘k’ is

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$e_{1} = \sqrt{1 - \frac{4}{18}} = \frac{\sqrt{7}}{3}; e_{2} = \sqrt{1 + \frac{4}{9}} = \frac{\sqrt{13}}{3}$

$∵ (e_{1}, e_{2}) l i e s o n 15 x^{2} + 3 y^{2} = k$

$\Rightarrow 15 e_{1}^{2} + 3 e_{2}^{2} = k$

$\Rightarrow k = 16$

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If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29−y24=1 respectively and (e1, e2) is a point on the ellipse 15x2+3y2=k then ‘k’ is

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If $e_{1} and e_{2}$ are the eccentricities of the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and $(e_{1}, e_{2})$ is a point on the ellipse $15 x^{2} + 3 y^{2} = k$ then ‘k’ is