If origin is the centre, X-axis is the major axis and 25 is the eccentricity of an ellipse which passes through (–3,1), then theequation of that ellipse is

e=25 b2=a21-25 b2=3a25 eq of ellipse x2a2+y2b2=1 passes through (-3, 1) 9a2+1b2=1 9a2+53a2=1 27+53a2=1 a2=323 b2=325 ∴eq 3x232+y232 =1

If origin is the centre, X-axis is the major axis and 25 is the eccentricity of an ellipse which passes through (–3,1), then theequation of that ellipse is

e=25 b2=a21-25 b2=3a25 eq of ellipse x2a2+y2b2=1 passes through (-3, 1) 9a2+1b2=1 9a2+53a2=1 27+53a2=1 a2=323 b2=325 ∴eq 3x232+y232 =1

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If origin is the centre, 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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$2 x^{2} + y^{2} = 19$

$x^{2} + 2 y^{2} = 11$

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

$x^{2} + 23 y^{2} = 32$

answer is A.

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Detailed Solution

$e = \frac{\sqrt{2}}{5} b^{2} = a^{2} (1 - \frac{2}{5}) b^{2} = \frac{3 a^{2}}{5} eq of ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 passes through (- 3, 1) \frac{9}{a^{2}} + \frac{1}{b^{2}} = 1 \frac{9}{a^{2}} + \frac{5}{3 a^{2}} = 1 \frac{27 + 5}{3 a^{2}} = 1 a^{2} = \frac{32}{3} b^{2} = \frac{32}{5} ∴ eq \frac{3 x^{2}}{32} + \frac{\sqrt{y^{2}}}{32} = 1$