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

\[m^2 x^2 = 4ax\]
\[ \Rightarrow m^2 x^2 - 4ax = 0\]
\[ \Rightarrow x\left( m^2 x - 4a \right) = 0\]
\[ \Rightarrow x = 0\text{ or }x = \frac{4a}{m^2}\] 
So, the points of intersection of the given parabola and line are O(0, 0) and

\[A\left( \frac{4a}{m^2}, \frac{4a}{m} \right)\]

∴ Area bounded by the given parabola and line
= Area of the shaded region
\[= \int_0^\frac{4a}{m^2} y_{\text{ parabola }} dx - \int_0^\frac{4a}{m^2} y_{\text{ line }} dx\]
\[ = \int_0^\frac{4a}{m^2} \sqrt{4ax}dx - \int_0^\frac{4a}{m^2} mxdx\]
\[ = \left.2\sqrt{a} \times \frac{x^\frac{3}{2}}{\frac{3}{2}}\right|_0^\frac{4a}{m^2} - \left.m \times \frac{x^2}{2}\right|_0^\frac{4a}{m^2} \]
\[ = \frac{4\sqrt{a}}{3}\left[ \left( \frac{4a}{m^2} \right)^\frac{3}{2} - 0 \right] - \frac{m}{2}\left[ \left( \frac{4a}{m^2} \right)^2 - 0 \right]\]
\[ = \frac{4\sqrt{a}}{3} \times \frac{8a\sqrt{a}}{m^3} - \frac{m}{2} \times \frac{16 a^2}{m^4}\]
\[ = \frac{32 a^2}{3 m^3} - \frac{8 a^2}{m^3}\]
\[ = \frac{8 a^2}{3 m^3}\text{ square units }\]
But,
Area bounded by the given parabola and line = \[\frac{a^2}{12}\] sq. units       ............(Given)
\[\therefore \frac{8 a^2}{3 m^3} = \frac{a^2}{12}\]
\[ \Rightarrow m^3 = 32\]
\[ \Rightarrow m = \sqrt[3]{32}\]
Thus, the value of m is \[\sqrt[3]{32}\]