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 :-
Correct Option: , 4
Solution:
Vertex is $\left(a^{2}, 0\right)$
$\mathrm{y}^{2}=-\left(\mathrm{x}-\mathrm{a}^{2}\right)$ and $\mathrm{x}=0 \Rightarrow(0, \pm 2 \mathrm{a})$
Area of triangle is $=\frac{1}{2}, 4 a \cdot\left(a^{2}\right)=250$
$\Rightarrow a^{3}=125$ or $a=5$
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