Question

The area bounded the curve $y^{2} = 16x$ and line $y =mx$ is $\frac{2}{3}$, then $m$ is equal to

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

We have, $y^{2} = 16x$, a parabola with vertex $(0, 0)$ and line $y = mx$.
Required area = area of shaded region
$=\int\limits_{0}^{16 m^2}\left(\sqrt{16x}-mx\right)dx=\frac{2}{3}$


$\Rightarrow \quad\left[4\times\frac{2}{3}x^\frac{3} {2}-m\left(\frac{x^{2}}{2}\right)\right]_{0}^\frac{16}{m^2}=\frac{2}{3}$

$\Rightarrow \quad\frac{8}{3}\times\frac{64}{m^{3}}-\frac{m}{2}\frac{256}{m^{4}}=\frac{2}{3}\,\Rightarrow \quad\frac{1}{m^{3}}\left[\frac{512}{3}-128\right]=\frac{2}{3}$

$\Rightarrow \quad m=4$