Area bounded (in square units) by the curves   $y=\frac{x^2}{4a}$ and  $y=\frac{8a^3}{x^2+4a^2}$ is

The curve  $y(x^2+4a^2)=8a^3$  is symmetrical about Y-axis and cut it at A (0,2a).

 Tangent at A is parallel to X-axis and X-axis is asymptote
 This curve meets  $x^2=4ay$, where  $\frac{x^2}{4a}=\frac{8a^3}{x^2+4a^2}$
 $\Rightarrow x^4+4a^2{x}^2-32a^4=0\Rightarrow (x^2+4a^2)(x^2+8a^2)=0\Rightarrow x=\pm 2a$

]$$\begin{gathered} \therefore \text{Required\hspace{0.17em}area\hspace{0.17em}}=2[{\int }_0^{2a}\frac{8a^3}{x^2+4a^2}dx-{\int }_0^{2a}\frac{x^2}{4a}dx] \\ =16a^3\cdot \frac{1}{2a}{\left[{\tan }^{-1}\frac{x}{2a}\right]}_0^{2a}-\frac{1}{2a}{\left[\frac{x^3}{3}\right]}_0^{2a} \end{gathered}$$