Tangent
$\begin{aligned}
& y=m(x+2) \
& y^2=x-2 \
& (m(n+2))^2=n-2 \
& m^2 x^2+\left(4 m^2-1\right) x+\left(4 m^2+2\right)=0 \
& D=0 \
& \left(4 m^2-1\right)^2-4 m^2\left(4 m^2+2\right)=0 \
& m=\frac{1}{4} \
& y=\frac{1}{4}(n+2)
\end{aligned}$
and point of tangency (6,2)

$\begin{aligned}
& \text { Area } A=\int_0^2\left(\left(y^2+2\right)-(4 y-2)\right) d y \
& A=\frac{8}{3}
\end{aligned}$
option (3)