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Coordinate Geometry

JEE Main level, single correct, medium

2 of 5

Question 2 of 5

If the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is half of its minor axis, the eccentricity $e$ is:

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Answer: (B) $\sqrt{3}/2$

\frac{2b^2}{a} = b \implies \frac{b}{a} = \frac{1}{2}

e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 1/4} = \frac{\sqrt{3}}{2}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Mathematics practice question on Coordinate Geometry, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

If the latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1a2x2​+b2y2​=1 is half of its minor axis, the eccentricity eee is:

JEE Main question paper PDF

More Mathematics practice

If the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is half of its minor axis, the eccentricity $e$ is: