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If the points of intersection of the ellipse x^2 16+ y^2 b^2=1 and the circle x^2+y^2=4b,b>4 lie on the curve y^2=3x^2, then b is equal to :

JEE Main 2021 — Mathematics Coordinate Geometry

2021mcqeasy

If the points of intersection of the ellipse x216+y2b2=1\frac{{x}^{2}}{16}+\frac{{y}^{2}}{{b}^{2}}=116x2​+b2y2​=1 and the circle x2+y2=4b,b>4{x}^{2}+{y}^{2}=4b,b>4x2+y2=4b,b>4 lie on the curve y2=3x2{y}^{2}=3{x}^{2}y2=3x2, then bbb is equal to :

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

Options

  1. A

    121212

  2. B

    555

  3. C

    666

  4. D

    101010

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