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If the area of the triangle whose one ve...

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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The correct Answer is:
B
Vertex of parabola `y^(2) = - 4(x - a^(2))` is `(a^(2) , 0)`
For point of intersection with Y-axis, put x = 0 in the given equation of parabola.
This gives, `y^(2) = 4 a^(2) implies` `y = +- 2a`
Thus, the point of intersection are (0, 2a) and (0, -2a).

From the given condition, we have
Area of `Delta ABC = 250`
`:. (1)/(2) (4a) a^(2) = 250` `implies a^(3) = 125 = 5^(3)`
`:. a = 5`
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