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

Options

A. `5 sqrt5`
B. 5
C. `(10) ^(2//3)`
D. `5 (2 ^(1//3))`

Answer

B

Explanation

Parabola is `y ^(2) =-4 (x -a ^(2))`
`V (a ^(2) ,0),` let point of intersection of parabola with y-axis be A and B.Then `A (0,2a), B(0,-2a)`
`Delta =1/2|{:(a ^(2), 0,1),(0, 2a, 1),(0, -2a,1):}|=250," " 4a ^(3) =500implies a =5.`