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 the 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 $16$

Here, $e_1^2=1-\frac{4}{18}=\frac{14}{18}$
$e_2^2=1+\frac{4}{9}=\frac{13}{9}$
As $(e_1,e_2)$ lies on the ellipse $15x^2+3y^2=k$
$\therefore 15e_1^2+3e_2^2=k$
$\Rightarrow 15\cdot \frac{14}{18}+3\cdot \frac{13}{9}=k$
$\Rightarrow k=16$