Q. The area cut off from a parabola by any double ordinate is $k$ times the corresponding rectangle contained by that double ordinate and its distance from the vertex, then $k$ is
A) $\frac{2}{3}$
B) $\frac{1}{3}$
C) $\frac{3}{2}$
D) 3
Explanation
(a)
Required area $= 2\int_{0}^{\alpha}{\sqrt{4ax}dx}$
$= k(\alpha)(2\sqrt{4a\alpha})$
[image:8_image14.emf]
$\frac{8\sqrt{a}}{3}\alpha^{3/2} = 4\sqrt{a}k\alpha^{3/2}$
$\Rightarrow \ k = \frac{2}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.