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(D) The equation of the parabola is ${y^2} = -4(x - {a^2})$.The vertex of the parabola is $V = ({a^2}, 0)$.To find the intersection points with the $y

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$5$

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(D) The equation of the parabola is ${y^2} = -4(x - {a^2})$.The vertex of the parabola is $V = ({a^2}, 0)$.To find the intersection points with the $y$-axis,set $x = 0$ in the equation:${y^2} = -4(0 - {a^2}) = 4{a^2}$.Thus,$y = \pm 2a$. The intersection points are $P = (0, 2a)$ and $Q = (0, -2a)$.The triangle has vertices $V({a^2}, 0)$,$P(0, 2a)$,and $Q(0, -2a)$.The base of the triangle along the $y$-axis is the distance between $P$ and $Q$,which is $|2a - (-2a)| = |4a| = 4|a|$.The height of the triangle from the vertex $V$ to the $y$-axis is the distance from $({a^2}, 0)$ to the line $x = 0$,which is $|{a^2}| = {a^2}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4|a| 	imes {a^2} = 2|a|^3$.Given the area is $250$,we have $2|a|^3 = 250$,so $|a|^3 = 125$.Therefore,$|a| = 5$,which implies $a = 5$ or $a = -5$.

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 $250 \text{ sq. units}$,then a value of $a$ is

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