Q. The area out off by latusrectum form the parabola $y^{2} = 4ax$ is
A) $(8/3)\ a$ sq units
B) $(8/3)\sqrt{a}$ sq units
C) $(3/8)\ a^{2}$ sq units
D) $(8/3)a^{2}$ sq units
Explanation
(d)
$Required\ area = 2\int_{0}^{a}{\sqrt{4ax}\ dx}$
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$= 2.2\sqrt{a}\left\lbrack \frac{x^{3/2}}{3/2} \right\rbrack_{0}^{a} = \frac{8}{3}a^{2}\ sq\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.