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An ellipse passes through the point $(-3,1)$ and its eccentricity is $\sqrt{\frac{2}{5}}$. The equation of the ellipse is
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The correct answer is:
$3 x^2+5 y^2=32$
Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\because$ It passes through $(-3,1)$
So, $\frac{9}{a^2}+\frac{1}{b^2}=1 \Rightarrow 9+\frac{a^2}{b^2}=a^2$ ...(i)
Given eccentricity is $\sqrt{2 / 5}$
So, $\frac{2}{5}=1-\frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2}=\frac{3}{5}$ ...(ii)
From equation (i) and (ii), $a^2=\frac{32}{3}, b^2=\frac{32}{5}$
Hence required equation of ellipse is $3 x^2+5 y^2=32$.
$\because$ It passes through $(-3,1)$
So, $\frac{9}{a^2}+\frac{1}{b^2}=1 \Rightarrow 9+\frac{a^2}{b^2}=a^2$ ...(i)
Given eccentricity is $\sqrt{2 / 5}$
So, $\frac{2}{5}=1-\frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2}=\frac{3}{5}$ ...(ii)
From equation (i) and (ii), $a^2=\frac{32}{3}, b^2=\frac{32}{5}$
Hence required equation of ellipse is $3 x^2+5 y^2=32$.
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