The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is:
Answer & explanation
Correct answer: option 2
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