If the area of the triangle whose one vertex is at the vertex of the parabola,…
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 :
$5 \sqrt{5}$
$5\left(2^{1 / 3}\right)$
$(10)^{33}$
5
Solution
$y^{t}=-4\left(x-a^{2}\right)$
$\mathrm{Area}=\frac{1}{2}(4 a)\left(a^{2}\right)=2 a^{3}$
Since $2 a^{3}=250 \Rightarrow a=5$
