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The locus of the foot of perpendicular drawn from the centre of the ellipse x^2+3y^2=6 on any tangent to it is

JEE Main 2014 — Mathematics Coordinate Geometry

2014mcqhard

The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6{x}^{2}+3{y}^{2}=6x2+3y2=6 on any tangent to it is

Official previous-year question

Held on 6 Apr 2014 · Verified 6 Jul 2026.

Options

  1. A

    (x2+y2)2=6x2+2y2{({x}^{2}+{y}^{2})}^{2}=6{x}^{2}+2{y}^{2}(x2+y2)2=6x2+2y2

  2. B

    (x2+y2)2=6x2−2y2{({x}^{2}+{y}^{2})}^{2}=6{x}^{2}-2{y}^{2}(x2+y2)2=6x2−2y2

  3. C

    (x2−y2)2=6x2+2y2{({x}^{2}-{y}^{2})}^{2}=6{x}^{2}+2{y}^{2}(x2−y2)2=6x2+2y2

  4. D

    (x2−y2)2=6x2−2y2{({x}^{2}-{y}^{2})}^{2}=6{x}^{2}-2{y}^{2}(x2−y2)2=6x2−2y2

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