Skip to main content

Let the maximum area of the triangle that can be inscribed in the ellipse x^2 a^2+ y^2 4=1,a>2, having one of its vertices at one end of the major…

JEE Main 2022 — Mathematics Coordinate Geometry

2022mcqhard

Let the maximum area of the triangle that can be inscribed in the ellipse x2a2+y24=1,a>2\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{4}=1,a>2a2x2​+4y2​=1,a>2, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the yyy-axis, be 636\sqrt{3}63​. Then the eccentricity of the ellipse is:

Official previous-year question

Held on 24 Jun 2022 · Verified 6 Jul 2026.

Options

  1. A

    32\frac{\sqrt{3}}{2}23​​

  2. B

    12\frac{1}{2}21​

  3. C

    12\frac{1}{\sqrt{2}}2​1​

  4. D

    34\frac{\sqrt{3}}{4}43​​

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Coordinate Geometry in other years

More JEE Main Coordinate Geometry questions

Other Mathematics topics for JEE Main

Keep practicing Coordinate Geometry

Work through every JEE Main Coordinate Geometry PYQ, year by year.

Browse all Mathematics topics