If the area bounded by the parabola `y^2 = 4ax` and the line y=mx is `a^2/12` sq. units, by using integration find the value of m.

Explanation

`y^{2}=4ax`


Intersection between points


`x=frac{4a}{ m^{2}} `

` y=frac{4a}{m}`

`2 sqrt{a} int_{0}^{frac{4a}{m^{2}}} sqrt{x}dx-frac{1}{2} times frac{4a}{m} times frac{4a}{m^{2}} `

`frac{2 sqrt{a}}{3/2}[4^{3 / 2}]_{0}^{4 am^{2}}-frac{8a^{2}}{m^{3}}=frac{a^{2}}{12}`
`[frac{8 times 4}{3}] times frac{a^{1 / 2} times(a)^{3 / 2}}{m^{3}}-frac{8a^{2}}{m^{3} `

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