Question

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

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

Answer 1

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

$y^2=4a(x+a)$

$y=\pm \sqrt{4ax+4a^2}$

As the area bounded by this curve is in -ve $x$ quadrant . Therfore

$y=-\sqrt{-4ax+4a^2}$

Area bounded by this curve will be ${\int }_{-a}^0\sqrt{4ax+4a^{2}dx}$

Area bounded by this curve will be $\frac{8}{3}{a}^2$

Similarly , The area bounded by $y^2=-4a(x-a)$ curve will be

${\int }_0^a\sqrt{-4ax+4a^{2}dx}$

Area = $\frac{8}{3}{a}^2$

thus , total bounded area will be $\frac{16}{3}{a}^2$