Q46 / 90 · JEE Main 2011 · Mathematics
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$ is
- A$5 x^2+3 y^2-48=0$
- B$3 x^2+5 y^2-15=0$
- C$5 x^2+3 y^2-32=0$
- Correct answerD$3 x^2+5 y^2-32=0$
Solution
$$
\begin{aligned}
& \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2\left(1-\frac{2}{5}\right)=\mathrm{a}^2 \frac{3}{5}=\frac{3 \mathrm{a}^2}{5} \\
& \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \Rightarrow \frac{9}{\mathrm{a}^2}+\frac{5}{3 \mathrm{a}^2}=1 \\
& \mathrm{a}^2=\frac{32}{3} \\
& \mathrm{~b}^2=\frac{32}{5}
\end{aligned}
$$
$\therefore$ Required equation of ellipse $3 x^2+5 y^2-32=0$
Hence, the correct answer is option: 'D'.