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If the area of the triangle whose one vertex is at the vertex of the parabola, y^2+4 (x-a^2 )=0 and the other two vertices are the points of…

JEE Main 2019 — Mathematics Coordinate Geometry

2019mcqeasy

If the area of the triangle whose one vertex is at the vertex of the parabola, y2+4(x−a2)=0y^{2}+4\left(x-a^{2}\right)=0y2+4(x−a2)=0 and the other two vertices are the points of intersection of the parabola and yyy -axis, is 250 sq. units, then a value of 'a' is :

Official previous-year question

Held on 11 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    555 \sqrt{5}55​

  2. B

    5(21/3)5\left(2^{1 / 3}\right)5(21/3)

  3. C

    (10)33(10)^{33}(10)33

  4. D

    5

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