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

The parabola $y^2=4ax$ opens towards the positive x-axis and its focus is (a, 0).

The line y = mx passes through the origin (0, 0).

Solving $y^2=4ax$ and y = mx, we get

$$\begin{gathered} m^2{x}^2=4ax \\ \Rightarrow m^2{x}^2-4ax=0 \\ \Rightarrow x\left(m^{2}x-4a\right)=0 \\ \Rightarrow x=0\mathrm{or}x=\frac{4a}{m^2} \end{gathered}$$
So, the points of intersection of the given parabola and line are O(0, 0) and $\mathrm{A}\left(\frac{4a}{m^2},\frac{4a}{m}\right)$.


∴ Area bounded by the given parabola and line

= Area of the shaded region

$$\begin{gathered} ={\int }_0^{\frac{4a}{m^2}}{y}_{\mathrm{parabola}}dx-{\int }_0^{\frac{4a}{m^2}}{y}_{\mathrm{line}}dx \\ ={\int }_0^{\frac{4a}{m^2}}\sqrt{4ax}dx-{\int }_0^{\frac{4a}{m^2}}mxdx \\ =2\sqrt{a}\times {\overline{)\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}}}_0^{\frac{4a}{m^2}}-m\times {\overline{)\frac{x^{{\displaystyle 2}}}{{\displaystyle 2}}}}_0^{\frac{4a}{m^2}} \\ =\frac{4\sqrt{a}}{3}\left[{\left(\frac{4a}{m^2}\right)}^{\frac{3}{2}}-0\right]-\frac{m}{2}\left[{\left(\frac{4a}{m^2}\right)}^2-0\right] \\ =\frac{4\sqrt{a}}{3}\times \frac{8a\sqrt{a}}{m^3}-\frac{m}{2}\times \frac{16a^2}{m^4} \\ =\frac{32a^2}{3m^3}-\frac{8a^2}{m^3} \\ =\frac{8a^2}{3m^3}\mathrm{square}\mathrm{units} \end{gathered}$$

But,
Area bounded by the given parabola and line = $\frac{a^2}{12}$ sq. units (Given)
$$\begin{gathered} \therefore \frac{8a^2}{3m^3}=\frac{a^2}{12} \\ \Rightarrow m^3=32 \\ \Rightarrow m=\sqrt[3]{32} \end{gathered}$$

Thus, the value of m is $\sqrt[3]{32}$.