Question

Area bounded by the curve $y^2=16x$ and line $y=mx$ is $\frac{2}{3}$ is equal to

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is B

$4$

Explanation for the correct option:

Step 1. Draw the curves $y^2=16x$ and $y=mx$:

Given that $y^2=16x$

$\Rightarrow$ $y=\sqrt{16x}$

From the figure, we can see

Intersection points of both the curves are $\left(0,0\right)$ and $\left(\frac{16}{m^2},\frac{16}{m}\right)$

Step 2. Area bounded by the curves $={\int }_0^{\frac{16}{m^2}}\left(\sqrt{16x}-mx\right)dx$

$\Rightarrow$ ${\int }_0^{\frac{16}{m^2}}\left(\sqrt{16x}-mx\right)dx=\frac{2}{3}$

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

$\Rightarrow$ $\frac{8}{3}\times {\left(\frac{16}{m^2}\right)}^{\frac{3}{2}}-\frac{m}{2}{\left(\frac{16}{m^2}\right)}^2=\frac{2}{3}$

$\Rightarrow$ $\frac{8}{3}\times \frac{64}{m^3}-\frac{16\times 16}{2m^3}=\frac{2}{3}$

$\Rightarrow$ $\frac{1}{m^3}\left[\frac{512}{3}-\frac{256}{2}\right]=\frac{2}{3}$

$\Rightarrow$ $m^3=\frac{128}{3}\times \frac{3}{2}$

$\Rightarrow$ $m^3=64$

$\mathbf{\therefore }\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}$

Hence, Option ‘B’ is Correct.