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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Given eccentricity $e = \sqrt{\frac{2}{5}}$.Since $e^2 = 1 - \frac{b^2}

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

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Given eccentricity $e = \sqrt{\frac{2}{5}}$.Since $e^2 = 1 - \frac{b^2}{a^2}$ (assuming $a > b$),we have $\frac{2}{5} = 1 - \frac{b^2}{a^2}$,which implies $\frac{b^2}{a^2} = \frac{3}{5}$,or $a^2 = \frac{5}{3}b^2$.The ellipse passes through $(-3, 1)$,so $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1$,which gives $\frac{9}{a^2} + \frac{1}{b^2} = 1$.Substituting $a^2 = \frac{5}{3}b^2$,we get $\frac{9}{(5/3)b^2} + \frac{1}{b^2} = 1$.$\frac{27}{5b^2} + \frac{1}{b^2} = 1$ $\Rightarrow \frac{27+5}{5b^2} = 1$ $\Rightarrow 5b^2 = 32$ $\Rightarrow b^2 = \frac{32}{5}$.Then $a^2 = \frac{5}{3} 	imes \frac{32}{5} = \frac{32}{3}$.The equation is $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$,which simplifies to $\frac{3x^2}{32} + \frac{5y^2}{32} = 1$.Thus,$3x^2 + 5y^2 = 32$ or $3x^2 + 5y^2 - 32 = 0$.

Find the equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$ and has an eccentricity of $\sqrt{2/5}$.

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