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  4. If the area of the triangle whose one vertex is at the vertex of the parabola, y2+4(x-a2)=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

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Q. 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

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Solution:

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