已知抛物线的顶点在原点，焦点在 $x$ 轴上，$\triangle ABC$ 三个顶点都在抛物线上，且 $\triangle ABC$ 的重心为抛物线的焦点，若 $BC$ 边所在直线的方程为 $4x + y - 20 = 0$，则抛物线方程为 \((\qquad)\)
【难度】
【出处】
2011年清华大学夏令营试题
【标注】
【答案】
A
【解析】
设抛物线方程为$${y^2} = 2px,p > 0.$$设 $BC$ 的中点为 $M$，连接 $AM$，则$$\overrightarrow {AF} = 2\overrightarrow {FM} .$$设 $B\left( {{x_1},{y_1}} \right)$，$C\left( {{x_2},{y_2}} \right)$，$A\left( {{x_0},{y_0}} \right)$，$M\left( {{x_M} ,{y_M}} \right)$，则 ${y_1}^2 = 2p{x_1}$，${y_2}^2 = 2p{x_2}$，${y_0}^2 = 2p{x_0}$，${y_M} = - \dfrac{p}{4}$，${x_M} = 5 - \dfrac{{{y_M}}}{4} = 5 + \dfrac{p}{{16}}$．
因为\[\begin{split} &\overrightarrow {AF} = 2 \overrightarrow {FM} \\ \Leftrightarrow &\left( {\dfrac{p}{2} - {x_0} ,- {y_0}} \right) = 2\left( {5 + \dfrac{p}{{16}} - \dfrac{p}{2} ,- \dfrac{p}{4}} \right)\\ \Leftrightarrow &\begin{cases}
\dfrac{p}{2} - {x_0} = 10 - \dfrac{{7p}}{8} ,\\
- {y_0} = - \dfrac{p}{2} ,
\end{cases}\end{split}\]所以$${x_0} = \dfrac{{11p}}{8} - 10,{y_0} = \dfrac{p}{2}.$$而 ${y_0}^2 = 2p{x_0}$，所以 $\dfrac{{{p^2}}}{4} = 2p \cdot \left( {\dfrac{{11p}}{8} - 10} \right)$，解得 $p = 8$．所以 ${y^2} = 16x$ 为所求抛物线方程．
因为\[\begin{split} &\overrightarrow {AF} = 2 \overrightarrow {FM} \\ \Leftrightarrow &\left( {\dfrac{p}{2} - {x_0} ,- {y_0}} \right) = 2\left( {5 + \dfrac{p}{{16}} - \dfrac{p}{2} ,- \dfrac{p}{4}} \right)\\ \Leftrightarrow &\begin{cases}
\dfrac{p}{2} - {x_0} = 10 - \dfrac{{7p}}{8} ,\\
- {y_0} = - \dfrac{p}{2} ,
\end{cases}\end{split}\]所以$${x_0} = \dfrac{{11p}}{8} - 10,{y_0} = \dfrac{p}{2}.$$而 ${y_0}^2 = 2p{x_0}$，所以 $\dfrac{{{p^2}}}{4} = 2p \cdot \left( {\dfrac{{11p}}{8} - 10} \right)$，解得 $p = 8$．所以 ${y^2} = 16x$ 为所求抛物线方程．
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