Q. The area lying between parabola $y^{2} = 4ax$ and it’s latusrectum is
A) $\frac{4}{3}a^{2}$ sq unit
B) $\frac{16}{3}a^{2}$ sq unit
C) $\frac{8}{3}a^{2}$ sq unit
D) None of these
Explanation
(c)
Required area $= 2\int_{0}^{a}{\sqrt{4ax}dx}$
$= 4\sqrt{a} \times \frac{2}{3}\left\lbrack x^{3/2} \right\rbrack_{0}^{a} = \frac{8}{3}a^{2}\ sq\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.