An Ellipse having the cordinate axes as its axes and its major axis along y-axis, passes through the point (–3,1) and has eccentricity $\sqrt{\frac{2}{5}}$ then its equation is

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{ellipse} \mathrm{be} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1(\mathrm{a}<\mathrm{b}) \\ \mathrm{Given} \\ \mathrm{e}=\sqrt{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.} \\ and {\mathrm{a}}^2={\mathrm{b}}^2(1-{\mathrm{e}}^2) \\ {\mathrm{a}}^2={\mathrm{b}}^2\left(1-\frac{2}{5}\right) \\ \Rightarrow {\mathrm{a}}^2=\frac{3{\mathrm{b}}^2}{5} \\ \mathrm{eq}uation \mathrm{of} \mathrm{ellipse} is \\ \frac{5{\mathrm{x}}^2}{3{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ It \mathrm{passes} \mathrm{through} (-3, 1) then \\ \frac{9\times 5}{3{\mathrm{b}}^2}+\frac{1}{{\mathrm{b}}^2}=1 \\ \Rightarrow 16={\mathrm{b}}^2 \\ and {\mathrm{a}}^2=\frac{48}{5} \\ \therefore \mathrm{eq}uation of ellipse is \frac{5{\mathrm{x}}^2}{48}+\frac{{\mathrm{y}}^2}{16}=1 \\ 5{\mathrm{x}}^2+3{\mathrm{y}}^2=48 \end{gathered}$$