Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$…

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
  1. $5 x^2+3 y^2-48=0$
  2. $3 x^2+5 y^2-15=0$
  3. $5 x^2+3 y^2-32=0$
  4. $3 x^2+5 y^2-32=0$

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 $3 x^2+5 y^2-32=0$

Asked in: JEE Main 2011

Practice more Circle questions on Aicharya