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Q.
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$ is
Conic Sections
Solution:
Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
It passes through $(-3,1)$
so $\frac{9}{a^{2}}+\frac{1}{b^{2}}=1 \,\,\,\,\,\,\,\, (i)$
Also, $b^{2}=a^{2}(1-2 / 5) $
$\Rightarrow 5 b^{2}=3 a^{2} \,\,\,\,\,\,\,\, ....(ii)$
Solving (i) and (ii) we get $a ^{2}-\frac{32}{3}, b^{2}=\frac{32}{5}$
So the equation of the ellipse is $3 x^{2}+5 v^{2}=32$