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The area between `x=y^2`and `x = 4`is divided into two equal parts by the line`x = a`, find the value of a.

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The graph of `y^(2)=4ax` is symmetric about X-axis which is shown below. `x=a` lies between the lines `x=0` and `x=4.` According to the given condition.
image
`2 int_(0)^(a)ydx=2int_(a)^(4) ydx`
`implies int_(0)^(a)2sqrt(a)sqrt(x)dx=int_(a)^(4)2sqrt(a)sqrt(x)dx`
`implies (2)/(3)[x^(3//2)]_(0)^(a)=(2)/(3)[x^(3//2)]_(a)^(4)`
`implies a^(3//2)=4^(3//2)-a^(3//2)`
`implies 2a^(3//2)=8impliesa^(3//2)=4`
`implies a=(4)^(2//3)`

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