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Mathematics · Ellipse

JEE Main 2026 — 2 April, Morning Shift — Question 28

Let an ellipse x2a2+y2b2=1\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}} +\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}} = 1 , a < b, pass through the point (4, 3) and have eccentricity 53\frac{\sqrt{5}}{3} . Then the length of its latus rectum is :

  1. Option A:

    453\frac{4\sqrt{5}}{3}

  2. Option B:

    252\sqrt{5}

  3. Option C:

    753\frac{7\sqrt{5}}{3}

  4. Option D:

    853\frac{8\sqrt{5}}{3}

    Correct

Answer: D

Step-by-step solution

e2=1−a2 b2⇒a2 b2=1−e2=1−59=49\mathrm{e}^{2}=1-\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}} \Rightarrow \frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=1-\mathrm{e}^{2}=1-\frac{5}{9}=\frac{4}{9}

\Rightarrow \frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}=\frac{4}{9} \end{gathered}$$ Passes through $(4,3) \Rightarrow \frac{16}{\mathrm{a}^{2}}+\frac{9}{\mathrm{~b}^{2}}=1$ from (1) and (2) $\mathrm{a}^{2}=20$ and $\mathrm{b}^{2}=45$ $\mathrm{LR}=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2(20)}{3 \sqrt{5}}=\frac{8 \sqrt{5}}{3}$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Mathematics
Chapter
Ellipse
Topic
Introduction to Ellipse