1638

已知直线 ${\displaystyle l:y=x-2}$ 与 ${\displaystyle x}$ 轴、${\displaystyle y}$ 轴分别交于点 ${\displaystyle M,N}$, 点 ${\displaystyle A}$ 在曲线 ${\displaystyle y=x^2-\ln x}$ 上, 点 ${\displaystyle B}$ 在 ${\displaystyle l}$ 上, 点 ${\displaystyle P}$ 满足 ${\displaystyle \overrightarrow{AP}=3\overrightarrow{PB}}$, 则 ${\displaystyle \overrightarrow{PM}\cdot \overrightarrow{PN}}$ 的最小值为_____.

设 ${\displaystyle A(x_1,y_1)}$, ${\displaystyle B(x_2,y_2)}$, 则 ${\displaystyle P(\frac{x_1+3x_2}{4},\frac{y_1+3y_2}{4})}$. 这样$$\begin{array}{rl}\overrightarrow{PM}\cdot \overrightarrow{PN}& =(x_p-2,y_p)\cdot (x_p,y_p+2)=(x_p-1{)}^2+(y_p+1{)}^2-2\\ & \ge \frac{1}{2}(x_p-y_p-2{)}^2-2=\frac{1}{32}(x_1-y_1+3(x_2-y_2)-8{)}^2-2\\ & =\frac{1}{32}(x_1^2-\ln x_1-x_1+2{)}^2-2.\end{array}$$ 记 ${\displaystyle f(x)=x^2-\ln x-x+2}$, 则 ${\displaystyle f^{\prime }(x)=\frac{(2x+1)(x-1)}{x}}$, 所以 ${\displaystyle f(x)\ge f(1)=2}$. ^[1] 所以 ${\displaystyle \overrightarrow{PM}\cdot \overrightarrow{PN}}$ 的最小值为 ${\displaystyle \frac{2^2}{32}-2=-\frac{15}{8}}$.