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
$=\underset{0}{\overset{16m^2}{\int }}\left(\sqrt{16x}-mx\right)dx=\frac{2}{3}$


$\Rightarrow {\left[4\times \frac{2}{3}{x}^{\frac{3}{2}}-m\left(\frac{x^2}{2}\right)\right]}_0^{\frac{16}{m^2}}=\frac{2}{3}$

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

$\Rightarrow m=4$