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河南省2023-2024学年度高二年级12月八校联考数学.考卷答案试卷答案
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分析直线ax+by+2b-a=0化为a（x-1）+b（y+2）=0，令$\left\{\begin{array}{l}{x-1=0}\\{y+2=0}\end{array}\right.$，可得直线ax+by+2b-a=0过定点Q（1，-2）．可知：垂足N在以MQ为直径的圆上，圆心即相等MQ的中点C（0，-1）．|PN|的最大值为|PC|+r．
解答解：直线ax+by+2b-a=0化为a（x-1）+b（y+2）=0，令$\left\{\begin{array}{l}{x-1=0}\\{y+2=0}\end{array}\right.$，解得x=1，y=-2．
∴直线ax+by+2b-a=0过定点Q（1，-2）．
∴垂足N在以MQ为直径的圆上，
圆心即相等MQ的中点C（0，-1）．
其圆的方程为：x2+（y+1）2=2．
|PC|=$\sqrt{5}$．
∴|PN|的最大值为$\sqrt{5}+\sqrt{2}$．
故答案为：$\sqrt{5}+\sqrt{2}$．
点评本题考查了直线与圆的方程、两点之间的距离公式，考查了推理能力与计算能力，属于中档题．
