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

Question

Dr. Abubacker · Accepted Answer

(b)
Required area $= \int_{0}^{16/m^{2}}{\left( \sqrt{16x} - mx ight)dx = \frac{2}{3}}$ (given)
$\Rightarrow \ \left\lbrack 4 	imes \frac{2}{3}x^{\frac{3}{2}} - m\frac{x^{2}}{2} ightbrack_{0}^{16/m^{2}} = \frac{2}{3}$
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$\Rightarrow \ \frac{8}{3} 	imes \frac{64}{m^{3}} - \frac{m}{2}\frac{256}{m^{4}} = \frac{2}{3}$
$\Rightarrow \ \frac{1}{m^{3}}\left\lbrack \frac{512}{3} - 128 ightbrack = \frac{2}{3}$
$\Rightarrow \ m = 4$

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

Explanation

Reviewed by Dr. Abubacker

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