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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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The correct answer is:
5
$y^{t}=-4\left(x-a^{2}\right)$
$\mathrm{Area}=\frac{1}{2}(4 a)\left(a^{2}\right)=2 a^{3}$
Since $2 a^{3}=250 \Rightarrow a=5$
$\mathrm{Area}=\frac{1}{2}(4 a)\left(a^{2}\right)=2 a^{3}$
Since $2 a^{3}=250 \Rightarrow a=5$
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