The area enclosed by the curves $y^2+4x=4$ and $y-2x=2$ is

Given curves are y² + 4x = 4 and y - 2x = 2
Curves are Intersecting
$\mathrm{x}=\frac{4-{\mathrm{y}}^2}{4}=\frac{\mathrm{y}-2}{2}\Rightarrow \frac{{\mathrm{y}}^2}{4}+\frac{\mathrm{y}}{2}-2$
Area $=\frac{{\mathrm{\Delta }}^{3/2}}{6{\mathrm{a}}^2}=9$