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Mathematics · Ellipse

JEE Main 2024 — 30 January, Shift 1 — Question 15

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

  1. Option A:

    53\frac{\sqrt{5}}{3}

  2. Option B:

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

  3. Option C:

    13\frac{1}{\sqrt{3}}

  4. Option D:

    25\frac{2}{\sqrt{5}}

    Correct

Answer: D

Step-by-step solution

2 b=ac2 \mathrm{~b}=\mathrm{ac}

ba=e2\frac{b}{a}=\frac{e}{2}

e=1−e24e=\sqrt{1-\frac{\mathrm{e}^{2}}{4}}

e=25e=\frac{2}{\sqrt{5}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
30 January, Shift 1
Subject
Mathematics
Chapter
Ellipse
Topic
Introduction to Ellipse
If the length of the minor axis of ellipse is equal to half of the… | JEE Main 2024 PYQ with Solution · DhiX AI