JEE Math Practice Question
If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :
- A.$$5\sqrt 5 $$
- B.$${\left( {10} \right)^{2/3}}$$
- C.$$5\left( {{2^{1/3}}} \right)$$
- D.5
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