If the area bounded by the curves y2 = 4ax and y = mx is a23 then m =

A=∫04am2 (2ax−mx)dx2ax3232−mx2204am2=a234a34am232−m24am22−0−0=a2332a23m3−8a2m3=a2383a2m3=a23m3=8m=2

If the area bounded by the curves y2 = 4ax and y = mx is a23 then m =

A=∫04am2 (2ax−mx)dx2ax3232−mx2204am2=a234a34am232−m24am22−0−0=a2332a23m3−8a2m3=a2383a2m3=a23m3=8m=2

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If the area bounded by the curves y² = 4ax and y = mx is $\frac{a^{2}}{3}$ then m =

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$\frac{1}{2}$

answer is B.

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Detailed Solution

$A = \int_{0}^{\frac{4 a}{m^{2}}} (2 \sqrt{a} \sqrt{x} - mx) dx$

$\begin{array}{l} {[2 \sqrt{a} \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - m \frac{x^{2}}{2}]}_{0}^{4 \frac{a}{m^{2}}} = \frac{a^{2}}{3} \\ \frac{4 \sqrt{a}}{3} {(4 \frac{a}{m^{2}})}^{\frac{3}{2}} - \frac{m}{2} {(4 \frac{a}{m^{2}})}^{2} - 0 - 0 = \frac{a^{2}}{3} \\ \frac{32 a^{2}}{3 m^{3}} - \frac{8 a^{2}}{m^{3}} = \frac{a^{2}}{3} \\ \frac{8}{3} \frac{a^{2}}{m^{3}} = \frac{a^{2}}{3} \\ m^{3} = 8 \\ m = 2 \end{array}$