Question

Area bounded by $x^2=4ay$ and $y=\frac{8a^3}{x^2+4a^2}$ is:

• A. $\frac{a^2}{3}(6\pi -4)$
• B. $\frac{\pi a^2}{3}$
• C. $\frac{a^2}{3}(6\pi +4)$
• D. $0$

Answer 1

The correct option is D $0$

points of intersection
$x^2=4ay$ , $y=\frac{8a^3}{x^2+4a^2}$
$y(4ay+4a^2)=8a^3$
$\Rightarrow 4a{y}^2+4a^{2}y-8a^3=0$
$y^2+ay-2a^2=0$
$y=-2a;y=a$
there is only 1 point of intersection for $y=a$
So, they don't enclose area
So, zero