The area bounded by the parabola y2 = 8x, the x-axis and the latusrectum is ___________ .

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KollegeApply · Accepted Answer

$\frac{16}{3}$ y2 = 8x represents a parabola opening side ways , with vertex at O(0, 0) and Focus at B(2, 0) Thus AA' represents the latus rectum of the parabola. The points of intersection of the parabola and latus rectum are A(2, 4) and A'(2, −4) Area bound by curve , x-axis and latus rectum is the area OABO, \text{ The approximating rectangle of width = dx and length }= \left| y \right| \text{ has area }= \left| y \right| dx,\text{ and moves from }x = 0\text{ to }x = 2 \text{ area }\left( OABO \right) = \int_0^2 \left| y \right| dx = \int_0^2 y dx ............\left\{ y > 0 , \Rightarrow \left| y \right| = y \right\} = \int_0^2 $\sqrt$(8x)dx = 2$\sqrt$(2) \int_0^2 $\sqrt$(x)dx = 2$\sqrt$(2) \left[ x^3/2/3/2 \right]_0^2 = 2$\sqrt$(2) $\times$ $\frac{2}{3}$\left( 2^$\frac{3}{2}$ - 0 \right) = 4$\sqrt$(2)/3 $\times$ 2$\sqrt$(2) = $\frac{16}{3}$ \text{ sq units } [Solution diagram: 44129_solution_1.png]

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