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 $(e_1,e_2)$ is a point on th ellipse , $15x^2+3y^2=k$. Then $k$ is equal to :

• A. $14$
• B. $15$
• C. $17$
• D. $16$

Answer 1

The correct option is D $16$

$e_1=\sqrt{1-\frac{4}{18}}=\frac{\sqrt{7}}{3}$ & $e_2=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$

$\because (e_1,e_2)$ lies on the ellipse $15x^2+3y^2=k$
$\therefore 15e_1^2+3e_2^2=k\Rightarrow 15\times \frac{7}{9}+3\times \frac{13}{9}=k\Rightarrow k=16$