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

3 x^{2} + 5 y^{2} - 32 = 0

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5 x^{2} + 3 y^{2} - 48 = 0

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3 x^{2} + 5 y^{2} - 15 = 0

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5 x^{2} + 3 y^{2} - 32 = 0

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Solution

The correct option is B $3 x^{2} + 5 y^{2} - 32 = 0$
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, $b^{2} = a^{2} (1 - e^{2}) = a^{2} (1 - \frac{2}{5})$

$∴ 5 b^{2} = 3 a^{2} \dots$ (ii)

Solving we get $a^{2} = \frac{32}{3}, b^{2} = \frac{32}{5}$

So, the ellipse is $3 x^{2} + 5 y^{2} = 32$ .

Hence, option 'A' is correct.

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