设P（x,y）,则|PA|^2=(x-m)^2+y^2=(x-m)^2+4x-4=[x-(m-2)]^2+4m-8∵x≥1,∴当m-2＜1,即m＜3时,x=1时有最小值（m-1）^2;当m-2≥1,即m≥3时,x=m-2时有最小值4m+8,综上,m＜3时,最小值|m-1|,m≥3时,最小值2√（m+2）....