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10-2. Parabola, Ellipse, Hyperbola
hard
If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of the ellipse, $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ and the hyperbola, $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ respectively and $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ is a point on the ellipse, $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ then $\mathrm{k}$ is equal to
A
$15$
B
$14$
C
$17$
D
$16$
(JEE MAIN-2020)
Solution
$e_{1}=\sqrt{1-\frac{4}{18}}=\frac{\sqrt{7}}{3}$
$\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$
$\because \quad\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ lies on $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k}$
$\Rightarrow \quad 15 \mathrm{e}_{1}^{2}+3 \mathrm{e}_{2}^{2}=\mathrm{k}$
$\Rightarrow \quad k=16$
Standard 11
Mathematics
