Question

If the area bounded by the parabola $y^2=4ax$ and the line $y=mx$ is $\frac{a^2}{12}$sq. units, then using integration, find the value of m.

Answer 1

$y^2=4ax$

Intersection between points

$x=\frac{4a}{m^2}$

$y=\frac{4a}{m}$

$2\sqrt{a}{\int }_0^{\frac{4a}{m^2}}\sqrt{x}dx-\frac{1}{2}\times \frac{4a}{m}\times \frac{4a}{m^2}$

$\Rightarrow$ $\frac{2\sqrt{a}}{\frac{3}{2}}{\left[4^{3/2}\right]}_0^{4a/m^2}-\frac{8a^2}{m^3}=\frac{a^2}{12}$

$\Rightarrow$ $\left[\frac{8\times 4}{3}\right]\times \frac{(a^{1/2}\times (a{)}^{3/2}}{m^3}-\frac{8a^2}{m^3}$

$\Rightarrow$ $[\frac{32}{3}{a}^2-8a^2]\frac{1}{m^3}=\frac{a^2}{m^3}$

$\Rightarrow$ $\frac{8a^2}{m^3}=\frac{a^2}{m^3}$

$\Rightarrow$ $m^3=32$

$\Rightarrow$ $m=(32{)}^{1/3}$