13. 6.3. $\star \star$ Line $l$ passes through point $A(2,0)$, and intersects the parabola $y=x^{2}+2$ at points $B$ and $C$. $B B_{1} \perp$ $x$-axis, $C C_{1} \perp x$-axis, $B_{1}$ and $C_{1}$ are the feet of the perpendiculars, point $P$ lies on segment $B C$, and $\overrightarrow{B P}=\frac{\left|B B_{1}\right|}{\left|C C_{1}\right|} \cdot \overrightarrow{P C}$, find the equation of the trajectory of the centroid $G$ of $\triangle P O A$ (where $O$ is the origin).
According to the problem, we can set $l: y=k(x-2), k$ exists and $k \neq 0$. From $\left\{\begin{array}{l}y=k(x-2), \\ y=x^{2}+2,\end{array}\right.$ we get $y^{2}+\left(4 k-k^{2}\right) y+6 k^{2}=0$, hence $\Delta=\left(4 k-k^{2}\right)^{2}-$ $24 k^{2}>0$, which means $k^{2}\left(k^{2}-8 k-8\right)>0$, and since $k \neq 0$, solving this yields $k>4+2 \sqrt{6}$, or $k<4-2 \sqrt{6}$. Also, $\Delta>0$ and $\left\{\begin{array}{l}y_{1}+y_{2}=k^{2}-4 k, \\ y_{1} y_{2}=6 k^{2},\end{array}\right.$, thus $\frac{\left|B B_{1}\right|}{\left|C C_{1}\right|}=$ $\frac{y_{1}}{y_{2}}$. By the section formula, we have
$$
y_{0}=\frac{y_{1}+y_{2} \cdot \frac{y_{1}}{y_{2}}}{1+\frac{y_{1}}{y_{2}}}=\frac{2 y_{1} y_{2}}{y_{1}+y_{2}}=\frac{12 k^{2}}{k^{2}-4 k}=\frac{12 k}{k-4},
$$

then $x_{0}=\frac{y_{0}}{k}+2=\frac{12}{k-4}+2$. Let the centroid of $\triangle P O A$ be $G(x, y)$, so
$$
\left\{\begin{array}{l}
x=\frac{1}{3}\left(x_{0}+2\right)=\frac{4}{k-4}+\frac{4}{3}, \\
y=\frac{1}{3} y_{0}=\frac{4 k}{k-4}=\frac{16}{k-4}+4,
\end{array}\right.
$$

eliminating the parameter $k$, we get $y-4=4\left(x-\frac{4}{3}\right)$. That is, $12 x-3 y-4=0$, and from $k>4+2 \sqrt{6}$ or $k<4-2 \sqrt{6}$, we solve to get $4-\frac{4}{3} \sqrt{6}<y<4+\frac{4}{3} \sqrt{6}$ and $y \neq 4$.