If the area bounded by the curve \(y^2=16x\) and the line \(y = mx\) is \(\displaystyle\frac{2}{3}\) sq. units, then the value of m is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
Take m > 0 (all the options are positive). The points of intersection of \(y^2 = 16x\) and y = mx:
\(m^2x^2 = 16x\)
\(x = 0\) or \(x = \frac{16}{m^2}\)
Between these, the parabola \(y = 4\sqrt{x}\) lies above the line y = mx, so the area is
\( \displaystyle A = \int_{0}^{16/m^2}\left(4\sqrt{x} - mx\right)dx \)
\( \displaystyle = \left[\frac{8}{3}x^{3/2} - \frac{m}{2}x^2\right]_{0}^{16/m^2} \)
\( \displaystyle = \frac{8}{3}\cdot\frac{64}{m^3} - \frac{m}{2}\cdot\frac{256}{m^4} \)
\( \displaystyle = \frac{512}{3m^3} - \frac{128}{m^3} = \frac{128}{3m^3} \)
Setting the area equal to \(\frac{2}{3}\):
\( \displaystyle \frac{128}{3m^3} = \frac{2}{3} \)
\(m^3 = 64\)
\(m = 4\)