The equation of the ellipse whose centre is at the origin and the X...

Let equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Since it passes through the points $(-3,1)$ and $(2-2)$$\begin{aligned}&\therefore \quad \frac{(-3)^{2}}{a^{2}}+\frac{(1)^{2}}{b^{2}}=1 \text { and } \frac{(2)^{2}}{a^{2}}+\frac{(-2)^{2}}{b^{2}}=1 \\ &\Rightarrow \quad \frac{9}{a^{2}}+\frac{1}{b^{2}}=1 \quad \text{...(i)}\\ &\Rightarrow \quad \frac{4}{a^{2}}+\frac{4}{b^{2}}=1 \quad \text{...(ii)} \end{aligned}$Multiply by 4 in Eq. (i) and subtracting Eq. (i) from Eq. (ii), 3 a^{2}=32 \Rightarrow a^{2}=$\frac{32}{3}$Substituting in Eq. (i), gives\begin{aligned}& \frac{9 \times 3}{32}+\frac{1}{b^{2}}=1 \\ \Rightarrow \quad & \frac{1}{b^{2}}=1-\frac{27}{32}=\frac{5}{32} \Rightarrowb^{2}=\frac{32}{5} \end{aligned}$$\therefore$ Required equation is $\frac{3 x^{2}}{32}+\frac{5 y^{2}}{32}=$1 $\text { or }$3 x^{2}+5 y^{2}=32