Subfield: Geometry


Question: 已知抛物线 $y^{2}=6 x$ 上的两个动点 $A\left(x_{1}, y_{1}\right)$ 和 $B\left(x_{2}, y_{2}\right)$, 其中 $x_{1} \neq x_{2}$ 且 $x_{1}+x_{2}=4$. 线段 $A B$ 的垂直平分线与 $x$ 轴交于点 $C$, 求 $\triangle A B C$ 面积的最大值.

Solution:
设线段 $A B$ 的中点为 $M\left(x_{0}, y_{0}\right)$, 则

$$
\begin{aligned}
& x_{0}=\frac{x_{1}+x_{2}}{2}=2, y_{0}=\frac{y_{1}+y_{2}}{2}, \\
& k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{6}-\frac{y_{1}^{2}}{6}}=\frac{6}{y_{2}+y_{1}}=\frac{3}{y_{0}} .
\end{aligned}
$$

线段 $A B$ 的垂直平分线的方程是

$y-y_{0}=-\frac{y_{0}}{3}(x-2) \quad \quad (1)$.

易知 $x=5, y=0$ 是 (1) 的一个解, 所以线段 $A B$ 的垂直平分线与 $x$ 轴的交点 $C$ 为定点, 且点$C$的坐标为 $(5,0)$.

由 (1) 知直线 $A B$ 的方程为

$$
y-y_{0}=\frac{3}{y_{0}}(x-2) \text {, 即 } x=\frac{y_{0}}{3}\left(y-y_{0}\right)+2 \text {. } \quad \quad (2)
$$

(2) 代入 $y^{2}=6 x$ 得

$$
y^{2}=2 y_{0}\left(y-y_{0}\right)+12 \text {, 即 } y^{2}-2 y_{0} y+2 y_{0}^{2}-12=0 \text {. } \quad \quad (3)
$$

依题意, $y_{1}, y_{2}$ 是方程 (3) 的两个实根, 且 $y_{1} \neq y_{2}$, 所以

$$
\begin{aligned}
& \Delta=4 y_{0}^{2}-4\left(2 y_{0}^{2}-12\right)=-4 y_{0}^{2}+48>0, \\
& -2 \sqrt{3}<y_{0}<2 \sqrt{3} . \\
& |A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}
\end{aligned}
$$

$$
\begin{aligned}
& =\sqrt{\left(1+\left(\frac{y_{0}}{3}\right)^{2}\right)\left(y_{1}-y_{2}\right)^{2}} \\
& =\sqrt{\left(1+\frac{y_{0}^{2}}{9}\right)\left[\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}\right]} \\
& =\sqrt{\left(1+\frac{y_{0}^{2}}{9}\right)\left(4 y_{0}^{2}-4\left(2 y_{0}^{2}-12\right)\right)} \\
& =\frac{2}{3} \sqrt{\left(9+y_{0}^{2}\right)\left(12-y_{0}^{2}\right)} .
\end{aligned}
$$

定点 $C(5,0)$ 到线段 $A B$ 的距离

$$
\begin{aligned}
h= & |C M|=\sqrt{(5-2)^{2}+\left(0-y_{0}\right)^{2}}=\sqrt{9+y_{0}^{2}} . \\
S_{\triangle A B C} & =\frac{1}{2}|A B| \cdot h=\frac{1}{3} \sqrt{\left(9+y_{0}^{2}\right)\left(12-y_{0}^{2}\right)} \cdot \sqrt{9+y_{0}^{2}} \\
& =\frac{1}{3} \sqrt{\frac{1}{2}\left(9+y_{0}^{2}\right)\left(24-2 y_{0}^{2}\right)\left(9+y_{0}^{2}\right)} \\
& \leq \frac{1}{3} \sqrt{\frac{1}{2}\left(\frac{9+y_{0}^{2}+24-2 y_{0}^{2}+9+y_{0}^{2}}{3}\right)^{3}} \\
& =\frac{14}{3} \sqrt{7} .
\end{aligned}
$$

当且仅当$9+y_{0}^{2}=24-2 y_{0}^{2}$，即$y_{0}=\pm \sqrt{5}$，$A \left(\frac{6+\sqrt{35}}{3}, \sqrt{5}+\sqrt{7} \right)$,
$B \left(\frac{6-\sqrt{35}}{3}, \sqrt{5}-\sqrt{7} \right)$或
$A\left(\frac{6+\sqrt{35}}{3},-(\sqrt{5}+\sqrt{7})\right), B\left(\frac{6-\sqrt{35}}{3},-\sqrt{5}+\sqrt{7}\right)$ 时等号成立.

所以 $\triangle A B C$ 面积的最大值为 $\frac{14}{3} \sqrt{7}$.
设线段 $A B$ 的中点为 $M\left(x_{0}, y_{0}\right)$, 则

$$
\begin{aligned}
& x_{0}=\frac{x_{1}+x_{2}}{2}=2, y_{0}=\frac{y_{1}+y_{2}}{2}, \\
& k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{6}-\frac{y_{1}^{2}}{6}}=\frac{6}{y_{2}+y_{1}}=\frac{3}{y_{0}} .
\end{aligned}
$$

线段 $A B$ 的垂直平分线的方程是

$y-y_{0}=-\frac{y_{0}}{3}(x-2) \quad \quad (1)$.

易知 $x=5, y=0$ 是 (1) 的一个解, 所以线段 $A B$ 的垂直平分线与 $x$ 轴的交点 $C$ 为定点, 且点$C$的坐标为 $(5,0)$.

设 $x_{1}=t_{1}^{2}, x_{2}=t_{2}^{2}, t_{1}>t_{2}, t_{1}^{2}+t_{2}^{2}=4$, 则

$$
\begin{aligned}
& S_{\triangle A B C}=\frac{1}{2}\left|\begin{array}{ccc}
5 & 0 & 1 \\
t_{1}^{2} & \sqrt{6} t_{1} & 1 \\
t_{2}^{2} & \sqrt{6} t_{2} & 1
\end{array}\right| \text { 的绝对值, } \\
& S_{\triangle A B C}^{2}=\left(\frac{1}{2}\left(5 \sqrt{6} t_{1}+\sqrt{6} t_{1}^{2} t_{2}-\sqrt{6} t_{1} t_{2}^{2}-5 \sqrt{6} t_{2}\right)\right)^{2}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{3}{2}\left(t_{1}-t_{2}\right)^{2}\left(t_{1} t_{2}+5\right)^{2} \\
& =\frac{3}{2}\left(4-2 t_{1} t_{2}\right)\left(t_{1} t_{2}+5\right)\left(t_{1} t_{2}+5\right) \\
& \leq \frac{3}{2}\left(\frac{14}{3}\right)^{3}, \\
S_{\triangle A B C} & \leq \frac{14}{3} \sqrt{7},
\end{aligned}
$$

当且仅当 $\left(t_{1}-t_{2}\right)^{2}=t_{1} t_{2}+5$ 且 $t_{1}^{2}+t_{2}^{2}=4$,

即 $t_{1}=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{6}}, t_{2}=-\frac{\sqrt{7}+\sqrt{5}}{\sqrt{6}}, A\left(\frac{6+\sqrt{35}}{3}, \sqrt{5}+\sqrt{7}\right), B\left(\frac{6-\sqrt{35}}{3}, \sqrt{5}-\sqrt{7}\right)$ 或

$A\left(\frac{6+\sqrt{35}}{3},-(\sqrt{5}+\sqrt{7})\right), B\left(\frac{6-\sqrt{35}}{3},-\sqrt{5}+\sqrt{7}\right)$ 时等号成立.

所以 $\triangle A B C$ 面积的最大值是 $\frac{14}{3} \sqrt{7}$.

Final answer: $\frac{14}{3}, \sqrt{7}$