The area bounded by the parabola y2 = 4ax and x2 = 4ay is ___________ .

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KollegeApply · Accepted Answer

16 a^$\frac{2}{3}$ To find the point of intersection of the parabolas substitute y = x^$\frac{2}{4}$a in y^2 = 4ax we get x^$\frac{4}{16}$ a^2 = 4ax \Rightarrow x^4 - 64 a^3 x = 0 \Rightarrow x\left( x^3 - 64 a^3 \right) = 0 \Rightarrow x = 0\text{ or }x = 4a \Rightarrow y = 0 \text{ or }y = 4a Therefore, the required area ABCD, A = \int_0^{4a} \left( y_1 - y_2 \right) d x ...........\left(\text{Where, }y_1 = 2$\sqrt$(ax)\text{ and }y_2 = x^$\frac{2}{4}$a \right) = \int_0^{4a} \left( 2$\sqrt$(ax) - x^$\frac{2}{4}$a \right) d x = \left[ 4$\sqrt$(a)/3 x^$\frac{3}{2}$ - x^$\frac{3}{12}$a \right]_0^{4a} = \left[ 4$\sqrt$(a)/3 \left( 4a \right)^$\frac{3}{2}$ - \frac{\left( 4a \right)^3}{12a} \right] - \left[ 4$\sqrt$(a)/3 \left( 0 \right)^$\frac{3}{2}$ - \frac{\left( 0 \right)^3}{12a} \right] = \left[ 4$\sqrt$(a)/38 a^$\frac{3}{2}$ - 64 a^$\frac{3}{12}$a \right] - 0 = 32 a^$\frac{2}{3}$ - 16 a^$\frac{2}{3}$ = 16 a^$\frac{2}{3}$\text{ square units } [Solution diagram: 44004_solution_1.png]

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