If the area bounded by the curves {y^2} = 4ax | vedclass

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Question

(A) The curves are ${y^2} = 4ax$ and $y = mx$.To find the intersection points,substitute $y = mx$ into ${y^2} = 4ax$:$(mx)^2 = 4ax \Rightarrow m^2x^2

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$2$

Vedclass · Answer

(A) The curves are ${y^2} = 4ax$ and $y = mx$.To find the intersection points,substitute $y = mx$ into ${y^2} = 4ax$:$(mx)^2 = 4ax \Rightarrow m^2x^2 - 4ax = 0 \Rightarrow x(m^2x - 4a) = 0$.Thus,$x = 0$ or $x = \frac{4a}{m^2}$.The area $A$ is given by $\int_0^{4a/m^2} (\sqrt{4ax} - mx) \,dx = \frac{a^2}{3}$.Integrating,we get $\left[ 2\sqrt{a} \cdot \frac{2}{3}x^{3/2} - \frac{mx^2}{2} \right]_0^{4a/m^2} = \frac{a^2}{3}$.Substituting the limits: $\frac{4\sqrt{a}}{3} \left( \frac{4a}{m^2} \right)^{3/2} - \frac{m}{2} \left( \frac{4a}{m^2} \right)^2 = \frac{a^2}{3}$.$\frac{4\sqrt{a}}{3} \cdot \frac{8a^{3/2}}{m^3} - \frac{m}{2} \cdot \frac{16a^2}{m^4} = \frac{a^2}{3}$.$\frac{32a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{a^2}{3}$.$\frac{a^2}{m^3} \left( \frac{32}{3} - 8 \right) = \frac{a^2}{3} \Rightarrow \frac{a^2}{m^3} \left( \frac{8}{3} \right) = \frac{a^2}{3}$.$\frac{8}{m^3} = 1 \Rightarrow m^3 = 8 \Rightarrow m = 2$.

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

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