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