Nice shape

Let Point $C$ be $(p,q)$ .

Using Pythagoras Theorem,

$AC$ can be expressed as $AC=\sqrt { { (p+1) }^{ 2 }+{ q }^{ 2 } }$

$BC$ can be expressed as $BC=\sqrt { { (p-1) }^{ 2 }+{ q }^{ 2 } }$

Given that $BC+AC=4$ , $\sqrt { { (p+1) }^{ 2 }+{ q }^{ 2 } } +\sqrt { { (p-1) }^{ 2 }+{ q }^{ 2 } } =4$

Solving gives $1=\frac { { q }^{ 2 } }{ 3 } +\frac { { p }^{ 2 } }{ 4 }$

Which means the shape is an ellipse.

Therefore, the area of the ellipse is $2\sqrt { 3 } \pi$ , which makes $100(ab)=$ $\boxed{600}$

For those who don't know about ellipses, the dimensions can be easily derived by Pythagoras Theorem again as it is already given that $AC+BC=4$ .