Problem 860

Question 166, Solution: Let the equation of the parabola be $y^{2}=2 p x$, then $F\left(\frac{p}{2}, 0\right)$. Solving the system of equations $\left\{\begin{array}{c}y^{2}=2 p x \\ 4 x+y-20=0\end{array}\right.$, we get $(20-4 x)^{2}=2 p x \Rightarrow 8 x^{2}-(80+p) x+200=0$. According to Vieta's formulas, we have $\left\{\begin{array}{c}x_{B}+x_{c}=\frac{p}{8}+10 \\ y_{B}+y_{C}=-\frac{p}{2}\end{array}\right.$. Noting that the centroid of $\triangle A B C$ is $F\left(\frac{p}{2}, 0\right)$, hence
$$\left\{\begin{array} { l } { x _ { A } + x _ { B } + x _ { c } = \frac { 3 p } { 2 } } \\ { y _ { A } + y _ { B } + y _ { C } = 0 } \end{array} \Rightarrow \left\{\begin{array}{c} x_{A}=\frac{11 p}{8}-10 \\ y_{A}=\frac{p}{2} \end{array}\right.\right.$$

Since point $A$ lies on the parabola, we have $\left(\frac{p}{2}\right)^{2}=2 p \cdot\left(\frac{11 p}{8}-10\right)$, solving this gives $p=8$. In conclusion, the equation of the parabola is $y^{2}=16 x$.