Question

The area bounded by the parabola $y^2=4a(x+a)$ and $y^2=-4a(x-a)$ is

• A. $\frac{16}{3}{a}^2$
• B. $\frac{8}{3}{a}^2$
• C. $\frac{4}{3}{a}^2$
• D. None of these

Answer 1

The correct option is A $\frac{16}{3}{a}^2$

Required area $=4{\int }_0^4\sqrt{4a(a-x)}dx$
$=4\times 2\sqrt{a}{\left[\frac{-2{(a-x)}^{3/2}}{3}\right]}_0^4=\frac{16}{3}{a}^2$