Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The area bounded by the parabola $y^2 = 4ax$ and the line $x = a$ and $x = 4a$ is

BITSATBITSAT 2006

Solution:

Required area = area of curve PSMNRQP
$=2$ area of curve PSRQP
$=2 \int\limits_{a}^{4 a} \sqrt{4 a x} d x$
$=4 \sqrt{a}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{a}^{4 a}=\frac{8}{3} \sqrt{a}\left[(4 a)^ \frac{3}{2}-a^{\frac{3}{2}}\right]$
$=\frac{8}{3} \sqrt{a}\left[8 a^{\frac{3}{2}}-a \frac{3}{2}\right)=\frac{56}{3} a^{2}$

Solution Image