If e_{1} and e_{2} are the eccentricities of | vedclass

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(D) For the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$,we have $a^{2}=18$ and $b^{2}=4$. The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \s

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$16$

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(D) For the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$,we have $a^{2}=18$ and $b^{2}=4$. The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{4}{18}} = \sqrt{\frac{14}{18}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$.For the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$,we have $a^{2}=9$ and $b^{2}=4$. The eccentricity $e_{2} = \sqrt{1+\frac{b^{2}}{a^{2}}} = \sqrt{1+\frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.Since the point $(e_{1}, e_{2})$ lies on the ellipse $15x^{2}+3y^{2}=k$,we substitute the values of $e_{1}$ and $e_{2}$ into the equation:$15(\frac{\sqrt{7}}{3})^{2} + 3(\frac{\sqrt{13}}{3})^{2} = k$$15(\frac{7}{9}) + 3(\frac{13}{9}) = k$$\frac{105}{9} + \frac{39}{9} = k$$\frac{144}{9} = k$$k = 16$.

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 $15x^{2}+3y^{2}=k$,then $k$ is equal to:

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