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直线 ${\displaystyle l:y=x-4}$ 交抛物线 ${\displaystyle \mathrm{\Gamma }:y^2=2px(p>0)}$ 于 ${\displaystyle A,B}$ 两点，${\displaystyle |AB|=6\sqrt{2}}$. ${\displaystyle C,D}$ 是位于 ${\displaystyle y}$ 轴和直线 ${\displaystyle l}$ 之间的抛物线 ${\displaystyle \mathrm{\Gamma }}$ 上两点，连接 ${\displaystyle BC,CD,AD}$.

1. 求抛物线 ${\displaystyle \mathrm{\Gamma }}$ 的标准方程;
2. 求四边形 ${\displaystyle ABCD}$ 的面积 ${\displaystyle S}$ 的最大值，以及 ${\displaystyle S}$ 取得最大值时直线 ${\displaystyle CD}$ 的方程.

1. 联立 ${\displaystyle l,\mathrm{\Gamma }}$: $$\begin{array}{r}{y}^2-2py-8p=0.\end{array}$$
   从而由韦达定理，$$\begin{array}{r}6\sqrt{2}=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{2}\sqrt{(y_1+y_2{)}^2-2y_1{y}_2}=\sqrt{2(4p^2+32p)}.\end{array}$$
   从而 ${\displaystyle p=1}$, ${\displaystyle \mathrm{\Gamma }:y^2=2x}$.
2. 固定 ${\displaystyle C}$,则要使 ${\displaystyle S}$ 最大，需要 ${\displaystyle S_{\mathrm{\Delta }ADC}}$ 最大，也即 ${\displaystyle D}$ 距离 ${\displaystyle AC}$ 最远，也即 ${\displaystyle D}$ 处抛物线的切线平行 ${\displaystyle AC}$. 根据地位的对称性，断言 ${\displaystyle S}$ 最大时 ${\displaystyle C}$ 处的切线也平行 ${\displaystyle BD}$. 将 ${\displaystyle p=1}$ 代入1中的联立式，解出 ${\displaystyle A(8,4),B(2,-2)}$. 设 ${\displaystyle C(2t_1^2,2t_1),D(2t_2^2,2t_2)}$, 则上述关系等价于$$\begin{array}{r}\{\begin{array}{rl}& (2t_1,1)//(2-2t_2^2,-2-2t_2)\Rightarrow 2t_1(-1-t_2)=1-t_2^2,\\ & (2t_2,1)//(8-2t_1^2,4-2t_1)\Rightarrow 2t_2(2-t_1)=4-t_1^2.\end{array}\end{array}$$
   当 ${\displaystyle t_2=-1}$, ${\displaystyle y_D=-2=y_B}$, 舍去；同理 ${\displaystyle t_1\ne 2}$. 故消去公因式后得 ${\displaystyle 2t_1=t_2-1}$, $$\begin{array}{r}\{\begin{array}{rl}& 2t_1=t_2-1,\\ & 2t_2=2+t_1\end{array}\Rightarrow \{\begin{array}{rl}& t_1=0,\\ & t_2=1.\end{array}\end{array}$$
   故 ${\displaystyle C(0,0),D(2,2)}$ , 也即此时 ${\displaystyle l_{CD}:y=x}$. 下面来计算 ${\displaystyle S}$ 的值. 由于 ${\displaystyle C(0,0),D(2,2),B(2,-2)}$, 故等腰直角三角形 ${\displaystyle CBD}$ 的面积是 ${\displaystyle 4}$. 而 ${\displaystyle D(2,2),B(2,-2),A(8,4),BD\perp x}$ 轴, 故三角形 ${\displaystyle ABD}$ 的面积是 ${\displaystyle 12}$, 从而 ${\displaystyle S_{max}=16}$.