Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (-3, 1) and has eccentricity √(2/5) is: - Mathematics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (- 3, 1) and has eccentricity 2/5 is:

Solution & Step-by-step Explanation

e^{2} = 1 - b^{2} / a^{2} = 2/5 ⟹ b^{2} / a^{2} = 3/5 ⟹ b^{2} = \frac{3}{5} a^{2} .Ellipse: \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 .Passes through (- 3, 1) : \frac{9}{a ^{2}} + \frac{1}{( 3/5 ) a ^{2}} = 1 ⟹ \frac{9}{a ^{2}} + \frac{5}{3 a ^{2}} = 1 \frac{27 + 5}{3 a ^{2}} = 1 ⟹ 3 a^{2} = 32 ⟹ a^{2} = 32/3 b^{2} = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5} .Equation: \frac{x ^{2}}{32/3} + \frac{y ^{2}}{32/5} = 1 ⟹ \frac{3 x ^{2}}{32} + \frac{5 y ^{2}}{32} = 1 3 x^{2} + 5 y^{2} = 32 .

Try it yourself before checking the explanation above.

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (- 3, 1) and has eccentricity 2/5 is:

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

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

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

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

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