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The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is: |
$\frac{4a^2}{3}$ sq. unit $\frac{8a^2}{3}$ sq. unit $\frac{2a^2}{3}$ sq. unit $\frac{9a^2}{5}$ sq. unit |
$\frac{8a^2}{3}$ sq. unit |
area of region I = area of region II by symmetry area = 2 × area of region I so, area = $2 ×\int_0^a2\sqrt{ax}dx=4\sqrt{a}\left[\frac{2x^{3/2}}{3}\right]_0^a$ $=\frac{8a^2}{3}$ sq. unit |
