Explanation
Firstly we have to find the point of intersection.
$$\dfrac { { x }^{ 2 } }{ 4a } =\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } $$
On solving the equation We get $$x^{2}=4a^{2}$$ and $$x^{2}= -8a^{2}$$ (which is not possible)
We have $$x=2a$$ and $$x=-2a$$
Now area between these 2 curves is
$$\displaystyle \int _{ -2a }^{ 2a }{ \left(\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ 4a } \right)}dx$$
$$\displaystyle =\int _{ -2a }^{ 2a }{ \dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } dx } -\int _{ -2a }^{ 2a }{ \dfrac { { x }^{ 2 } }{ 4a } } dx$$
$$=\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { x }{ 2a } } -\dfrac { { x }^{ 3 } }{ 12a } $$
After putting the limits we get
$$\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { 2a }{ 2a } } -\dfrac { { 8a }^{ 3 } }{ 12a } -(\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { -2a }{ 2a } } -\dfrac { { -8a }^{ 3 } }{ 12a } )$$
$$=\pi { a }^{ 2 }-\dfrac { 4 }{ 3 } { a }^{ 2 }$$
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