Find the area enclosed between the parabola y | vedclass

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(A) The area enclosed between the parabola $y^{2}=4ax$ and the line $y=mx$ is found by determining the points of intersection.Substituting $y=mx$ into

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$\frac{8a^{2}}{3m^{3}}$

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(A) The area enclosed between the parabola $y^{2}=4ax$ and the line $y=mx$ is found by determining the points of intersection.Substituting $y=mx$ into $y^{2}=4ax$,we get:$(mx)^{2}=4ax \implies m^{2}x^{2}-4ax=0 \implies x(m^{2}x-4a)=0$.Thus,the points of intersection are $x=0$ and $x=\frac{4a}{m^{2}}$.The area $A$ is given by the integral of the difference between the upper curve (parabola) and the lower curve (line) from $x=0$ to $x=\frac{4a}{m^{2}}$:$A = \int_{0}^{\frac{4a}{m^{2}}} (\sqrt{4ax} - mx) dx$$A = 2\sqrt{a} \int_{0}^{\frac{4a}{m^{2}}} x^{1/2} dx - m \int_{0}^{\frac{4a}{m^{2}}} x dx$$A = 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{\frac{4a}{m^{2}}} - m \left[ \frac{x^{2}}{2} \right]_{0}^{\frac{4a}{m^{2}}}$$A = \frac{4\sqrt{a}}{3} \left( \frac{4a}{m^{2}} \right)^{3/2} - \frac{m}{2} \left( \frac{4a}{m^{2}} \right)^{2}$$A = \frac{4\sqrt{a}}{3} \cdot \frac{8a^{3/2}}{m^{3}} - \frac{m}{2} \cdot \frac{16a^{2}}{m^{4}}$$A = \frac{32a^{2}}{3m^{3}} - \frac{8a^{2}}{m^{3}} = \frac{32a^{2} - 24a^{2}}{3m^{3}} = \frac{8a^{2}}{3m^{3}}$ sq. units.

Find the area enclosed between the parabola $y^{2}=4ax$ and the line $y=mx$.

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