JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The eccentricity of an ellipse

\frac{x^{2}}{a^2}+\frac{y^2}{b^2}=1

whose latusrectum is half of its minor axis is

View answer and explanation

Correct option: C

\frac{\sqrt{3}}{2}

Option C (

\frac{{\sqrt{{3}}}}{{2}}

) is correct because the given condition (latus rectum is half of its minor axis) leads to the relationship

a = 2b

, which allows us to calculate the eccentricity as

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

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