Question

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

Answer 1

$y^{2} = - 4\left(x - a^{2}\right)$

Area $= \frac{1}{2} \left(4a\right)\left(a^{2}\right) = 2a^{3}$
Since $2a^{3} = 250$
$\Rightarrow a = 5$