4. Given a parabola $y=a x^{2}+b x+c$ opening upwards passing through the points $(4,-6)$ and $(-2,0)$, intersecting the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. Then the minimum value of the area $S$ of $\triangle A B C$ is ( ).
(A) $10+4 \sqrt{6}$
(B) $10-4 \sqrt{6}$
(C) $10+4 \sqrt{5}$
(D) $10-4 \sqrt{5}$
4. A.

From the problem, we have
$$
\left\{\begin{array}{l}
16 a+4 b+c=-6, \\
4 a-2 b+c=0 .
\end{array}\right.
$$
(1) - (2) gives $12 a+6 b=-6$, which simplifies to
$$
b=-2 a-1 \text {. }
$$

Substituting equation (3) into equation (2) gives $c=-8 a-2$.
Thus, $y=a x^{2}-(2 a+1) x-8 a-2$
$$
=(a x-4 a-1)(x+2) \text {. }
$$

When $y=0$, $(a x-4 a-1)(x+2)=0$.
Solving this, we get $x_{1}=4+\frac{1}{a}, x_{2}=-2$.
Therefore, $A B=\left|\left(4+\frac{1}{a}\right)-(-2)\right|=\left|6+\frac{1}{a}\right|$.
When $x=0$, $y=-8 a-2$.
Thus, $O C=|-8 a-2|=|8 a+2|$.
Since the parabola opens upwards, we have $a>0$.
Hence, $A B=6+\frac{1}{a}, O C=8 a+2$.
Therefore, $S=\frac{1}{2} A B \cdot O C$
$$
\begin{array}{l}
=\frac{1}{2}\left(6+\frac{1}{a}\right)(8 a+2)=10+24 a+\frac{1}{a} \\
\geqslant 10+2 \sqrt{24}=10+4 \sqrt{6} .
\end{array}
$$

Thus, the minimum value of $S$ is $10+4 \sqrt{6}$.