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(A) Given,the major axis of the ellipse lies on the $X$-axis,so its equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 > b^2$.Since it pas

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

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(A) Given,the major axis of the ellipse lies on the $X$-axis,so its equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 > b^2$.Since it passes through $(-3, 1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$  $(i)$.The eccentricity is $e = \sqrt{\frac{2}{5}}$,so $e^2 = \frac{2}{5}$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = a^2(1 - \frac{2}{5}) = a^2(\frac{3}{5})$,which implies $b^2 = \frac{3a^2}{5}$ (ii).Substituting (ii) into $(i)$: $\frac{9}{a^2} + \frac{5}{3a^2} = 1$.Multiplying by $3a^2$: $27 + 5 = 3a^2$,so $3a^2 = 32$,which gives $a^2 = \frac{32}{3}$.Then $b^2 = \frac{3}{5} 	imes \frac{32}{3} = \frac{32}{5}$.The equation of the ellipse is $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$,which simplifies to $3x^2 + 5y^2 = 32$.

If the origin is the centre,the $X$-axis is the major axis,and $\sqrt{\frac{2}{5}}$ is the eccentricity of an ellipse which passes through $(-3, 1)$,then the equation of that ellipse is:

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