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Let an ellipse x^2 a^2 + y^2 b^2 = 1, a < b, pass through the point (4, 3) and have eccentricity √5 3. Then the length of its latus rectum is :

JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqmedium

Let an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1a2x2​+b2y2​=1, a<ba < ba<b, pass through the point (4,3)(4, 3)(4,3) and have eccentricity 53\dfrac{\sqrt{5}}{3}35​​. Then the length of its latus rectum is :

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

  1. A

    453\dfrac{4\sqrt{5}}{3}345​​

  2. B

    252\sqrt{5}25​

  3. C

    753\dfrac{7\sqrt{5}}{3}375​​

  4. D

    853\dfrac{8\sqrt{5}}{3}385​​

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