Using integration, find the value of `m`. If the area bounded by parabola `y^2=16ax` and the line `y=mx` is `a^2/12` square units.

Explanation

`y^2=16an`

`y=mx`

`(mx)^2=16an`

`x=0,x=(16a)/m^2`

`int_0^(16/m^2)(sqrt(16ax)-mx)dx`

`sqrt(16a)int_0^(16/m^2)x^(1/2)dx-m int_0^((16a)/m^2)*xdx`

`sqrt(16a)*2/3*(8*9sqrta)/m^2-m/2*(256a^2)/m^4`

`64/3*a^2/m^3-(128a^2)/m^3`

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