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 :
A
5(21/3)
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B
(10)2/3
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C
5
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D
5√5
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Solution
The correct option is C5 Consider the parabola Area(ΔABC)=Area(ΔAOB)+Area(ΔAOC) =12×a2×2a+12×a2×2a =2a3 ∴2a3=250 ⇒a3=125 ⇒a=5