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

$16$

Explanation for the correct option:

Finding the value of $k$:

Eccentricity can be calculated using the formula

$\begin{array}{rcl}e& =& \sqrt{1-\frac{b^2}{a^2}}\end{array}$for ellipse and $\begin{array}{rcl}e& =& \sqrt{1+\frac{b^2}{a^2}}\end{array}$for hyperbola

Hence,

$\begin{array}{rcl}{e}_1& =& \sqrt{1-\frac{4}{18}}\\ & =& \frac{\sqrt{7}}{3}\\ e_2& =& \sqrt{1+\frac{4}{9}}\\ & =& \frac{\sqrt{13}}{3}\end{array}$

Since $(e_1,e_2)$ lies on the ellipse $15x^2+3y^2=k$

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \left(15\times \frac{7}{9}\right)+\left(3\times \frac{13}{9}\right)=k\\ \Rightarrow \mathit{k}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{16}\end{array} \\ \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array} \end{gathered}$$

Hence, The correct answer is option (D).