Question:
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 $\left(e_{1}, e_{2}\right)$ is
a point on the ellipse, $15 x^{2}+3 y^{2}=k$, then $k$ is equal to
Correct Option: 1
Solution:
Eccentricity of ellipse
$e_{1}=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}$
Eccentricity of hyperbola
$e_{2}=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}$
Since, the point $\left(e_{1}, e_{2}\right)$ is on the ellipse
$15 x^{2}+3 y^{2}=k$
Then, $15 e_{1}^{2}+3 e_{2}^{2}=k$
$\Rightarrow \quad k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right)$
$\Rightarrow \quad k=16$