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(D) Since the major axis is along the $Y$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$. Given eccentricit

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$5x^2 + 3y^2 - 48 = 0$

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(D) Since the major axis is along the $Y$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$. Given eccentricity $e = \sqrt{\frac{2}{5}}$,we have $e^2 = 1 - \frac{a^2}{b^2} = \frac{2}{5}$. $\frac{a^2}{b^2} = 1 - \frac{2}{5} = \frac{3}{5} \implies b^2 = \frac{5a^2}{3}$. The ellipse passes through $(-3, 1)$,so $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1$. $\frac{9}{a^2} + \frac{1}{b^2} = 1$. Substituting $b^2 = \frac{5a^2}{3}$,we get $\frac{9}{a^2} + \frac{3}{5a^2} = 1$. $\frac{45 + 3}{5a^2} = 1 \implies 5a^2 = 48 \implies a^2 = \frac{48}{5}$. Then $b^2 = \frac{5}{3} 	imes \frac{48}{5} = 16$. The equation is $\frac{x^2}{48/5} + \frac{y^2}{16} = 1$. $\frac{5x^2}{48} + \frac{y^2}{16} = 1 \implies 5x^2 + 3y^2 = 48$.

An ellipse having the coordinate axes as its axes and its major axis along the $Y$-axis,passes through the point $(-3, 1)$ and has eccentricity $e = \sqrt{\frac{2}{5}}$. Then its equation is:

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