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(A) The standard equation of an ellipse with center $(0, 0)$ and major axis along the $X$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the el

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

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(A) The standard equation of an ellipse with center $(0, 0)$ and major axis along the $X$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(-3, 1)$ and $(2, -2)$,we have:For $(-3, 1)$: $\frac{9}{a^2} + \frac{1}{b^2} = 1$ ... $(i)$For $(2, -2)$: $\frac{4}{a^2} + \frac{4}{b^2} = 1 \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ ... $(ii)$Subtracting $(ii)$ from $(i)$:$\left(\frac{9}{a^2} + \frac{1}{b^2}\right) - \left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 1 - \frac{1}{4}$$\frac{8}{a^2} = \frac{3}{4} \Rightarrow a^2 = \frac{32}{3}$Substituting $a^2$ in $(ii)$:$\frac{1}{\frac{32}{3}} + \frac{1}{b^2} = \frac{1}{4} \Rightarrow \frac{3}{32} + \frac{1}{b^2} = \frac{1}{4}$$\frac{1}{b^2} = \frac{1}{4} - \frac{3}{32} = \frac{8-3}{32} = \frac{5}{32} \Rightarrow b^2 = \frac{32}{5}$Substituting $a^2$ and $b^2$ into the standard equation:$\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1 \Rightarrow \frac{3x^2}{32} + \frac{5y^2}{32} = 1$Therefore,the equation is $3x^2 + 5y^2 = 32$.

Find the equation of the ellipse which passes through the points $(-3, 1)$ and $(2, -2)$,whose center lies at $(0, 0)$ and major axis lies along the $X$-axis.

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