II. (25 points) As shown in Figure 1, the parabola opening downwards is given by
$$
y = a x^{2} - 8 a x + 12 a
$$

It intersects the $x$-axis at points $A$ and $B$ (point $A$ is to the left of point $B$). There is another point $C$ on the parabola, located in the first quadrant, such that $\triangle O C A \sim \triangle O B C$.
(1) Find the length of $O C$ and the value of $\frac{B C}{A C}$.
(2) Suppose the line $B C$ intersects the $y$-axis at point $P$. When $C$ is the midpoint of $B P$, find the equations of the line $B P$ and the parabola.
(1) Given that $a<0$, and the equation $a x^{2}-8 a x+12 a=0$ has two roots $x_{1}=2, x_{2}=6$. Thus, $O A=2, O B=6$.
$\because \triangle O C A \backsim \triangle O B C$,
$$
\therefore O C^{2}=O A \cdot O B=12 \text {, }
$$

which means $O C=2 \sqrt{3}$, and
$$
\frac{B C^{2}}{A C^{2}}=\frac{S_{\triangle O O B}}{S_{\triangle O O A}}=\frac{O B}{O A}=\frac{6}{2}=3 .
$$

Therefore, $\frac{B C}{A C}=\sqrt{3}$.
(2) Since $C$ is the midpoint of $P B$, hence $O C=B C$. Thus, the coordinates of $C$ are $C(3, \sqrt{3})$.

Let the line $B P$ be $y=k x+b$. Since it passes through $B(6,0), C(3, \sqrt{3})$,
then $\left\{\begin{array}{l}0=6 k+b, \\ \sqrt{3}=3 k+b,\end{array}\right.$ which gives $\left\{\begin{array}{l}k=-\frac{\sqrt{3}}{3}, \\ b=2 \sqrt{3} .\end{array}\right.$
The line $B P$ is $y=-\frac{\sqrt{3}}{3} x+2 \sqrt{3}$.
Since $C$ lies on the parabola, then
$\sqrt{3}=a \times 3^{2}-8 a \times 3+12 a$. Solving this, we get $a=-\frac{\sqrt{3}}{3}$.
The equation of the parabola is
$$
y=-\frac{\sqrt{3}}{3} x^{2}+\frac{8}{3} \sqrt{3} x-4 \sqrt{3} .
$$