The area bounded by the parabola y2 = 4ax, latusrectum and x-axis is ___________ .

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KollegeApply · Accepted Answer

$\frac{4}{3}$ a^2 Clearly, the latusrectum passes x-axis through the point D(a, 0). Therefore, the required area ABCD, A = \int_0^a y d x ...........\left(\text{Where, } y = 2$\sqrt$(ax) \right) = \int_0^1 2$\sqrt$(ax) d x = \left[ 4$\sqrt$(a)/3 \left( x \right)^$\frac{3}{2}$ \right]_0^a = \left[ 4$\sqrt$(a)/3 \left( a \right)^$\frac{3}{2}$ \right] - \left[ 4$\sqrt$(a)/3 \left( 0 \right)^$\frac{3}{2}$ \right] = $\frac{4}{3}$ a^2\text{ square units } [Solution diagram: 44020_solution_1.png]

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