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Q. Let $e_{1}$ 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. If $\left(e_{1} , e_{2}\right)$ is a point on the ellipse $15x^{2}+3y^{2}=k$ , then the value of $k$ is equal to

NTA AbhyasNTA Abhyas 2020

Solution:

$e_{1}=\sqrt{1 - \frac{4}{18}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}$
$e_{1}=\sqrt{1 + \frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13} }{3}$
Also,
$15e_{1}^{2}+3e_{2}^{2}=k\Rightarrow k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right)$
$\therefore k=16$