The area between the parabolas y^2 = 4ax and | vedclass

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(C) To find the area between the parabolas $y^2 = 4ax$ and $x^2 = 8ay$,we first find their points of intersection.From the first equation,$y = \sqrt{4

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$\frac{32}{3}a^2$

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(C) To find the area between the parabolas $y^2 = 4ax$ and $x^2 = 8ay$,we first find their points of intersection.From the first equation,$y = \sqrt{4ax} = 2\sqrt{a}\sqrt{x}$.Substituting this into the second equation: $x^2 = 8a(2\sqrt{a}\sqrt{x}) = 16a^{3/2}x^{1/2}$.$x^2 / x^{1/2} = 16a^{3/2} \implies x^{3/2} = 16a^{3/2}$.Squaring both sides to the power of $2/3$,we get $x = (16a^{3/2})^{2/3} = 16^{2/3} a = (2^4)^{2/3} a = 2^{8/3}a$.Now,the area $A$ is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = 2^{8/3}a$:$A = \int_0^{2^{8/3}a} (\sqrt{4ax} - \frac{x^2}{8a}) dx$$A = [2\sqrt{a} \cdot \frac{2}{3}x^{3/2} - \frac{x^3}{24a}]_0^{2^{8/3}a}$$A = \frac{4\sqrt{a}}{3} (2^{8/3}a)^{3/2} - \frac{(2^{8/3}a)^3}{24a}$$A = \frac{4\sqrt{a}}{3} (2^4 a^{3/2}) - \frac{2^8 a^3}{24a}$$A = \frac{4 \cdot 16}{3} a^2 - \frac{256}{24} a^2 = \frac{64}{3} a^2 - \frac{32}{3} a^2 = \frac{32}{3} a^2$.

The area between the parabolas $y^2 = 4ax$ and $x^2 = 8ay$ is

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