25~26高三上·安徽合肥一模·第14题

$-\dfrac{15}{8}$

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

【答案】$-\dfrac{15}{8}$

【解析】当直线$l$平移到与曲线$y = x^2 – \ln x$相切于点$A$，此时切点$A$是曲线上的点到直线$l$的距离最小的点.

由$y = x^2 – \ln x$，得$y’ = 2x – \dfrac{1}{x}$，
因为直线$l$的斜率为$1$，所以令$2x – \dfrac{1}{x} = 1$，整理得$2x^2 – x – 1 = 0$，
解得$x = -\dfrac{1}{2}$（舍去）或$x = 1$。又$y = 1^2 – \ln 1 = 1$，故此时切点$A(1,1)$，
且此时$A$到直线$l:y = x – 2$的距离为$d = \dfrac{|1 – 1 – 2|}{\sqrt{1 + 1}} = \sqrt{2}$.

又$\overrightarrow{AP} = 3\overrightarrow{PB}$，故此时$P$到直线的距离为$\dfrac{1}{4} \times \sqrt{2} = \dfrac{\sqrt{2}}{4}$.

取$MN$的中点为$F$，$PF \perp l$时，$|PF|$的长取得最小值$\dfrac{\sqrt{2}}{4}$，如图所示：

由直线$l:y = x – 2$，可得$M(2,0)$，$N(0,-2)$，
所以$F(1,-1)$，所以$|FM| = \sqrt{(1 – 2)^2 + (-1 – 0)^2} = \sqrt{2}$.

又$\overrightarrow{PM} \cdot \overrightarrow{PN} = (\overrightarrow{PF} + \overrightarrow{FM}) \cdot (\overrightarrow{PF} + \overrightarrow{FN})$
$= (\overrightarrow{PF} + \overrightarrow{FM}) \cdot (\overrightarrow{PF} – \overrightarrow{FM})$
$=|\overrightarrow{PF}|^2 – |\overrightarrow{FM}|^2 = |\overrightarrow{PF}|^2 – 2$.

故$|\overrightarrow{PF}|$最小时，$\overrightarrow{PM} \cdot \overrightarrow{PN}$的值最小，且最小值为$\left(\dfrac{\sqrt{2}}{4}\right)^2 – 2 = -\dfrac{15}{8}$.

故答案为：$-\dfrac{15}{8}$.