分析:设点M的坐标为(x,y),点P的坐标为(x0,y0),由题意可得{■(2x=x_0+3"," @2y=y_0 "," )┤

　　所以{■(x_0=2x"-" 3"," @y_0=2y"," )┤再代入圆的方程即可.

解:由题意设点M(x,y),P(x0,y0),则{■(2x=x_0+3"," @2y=y_0 "," )┤

　　所以{■(x_0=2x"-" 3"," @y_0=2y"." )┤

　　又因为点P(x0,y0)在圆x2+y2=1上,

　　所以(2x-3)2+4y2=1,

　　所以(x"-" 3/2)^2+y2=1/4.

　　故点M的轨迹方程为(x"-" 3/2)^2+y2=1/4.

★10.若直线x+y-m=0被曲线y=x2所截得的线段长为3√2,求m的值.

分析:直线与曲线交于两点,可设出这两点的坐标,然后灵活应用根与系数的关系求解.

解:设直线x+y-m=0与曲线y=x2相交于A(x1,y1),B(x2,y2)两点,联立直线与曲线方程,得{■(x+y"-" m=0",①" @y=x^2 ".②" )┤

　　将②代入①,得x2+x-m=0,

　　所以{■(x_1+x_2="-" 1"," @x_1 x_2="-" m"." )┤

　　所以|AB|=√("(" x_1 "-" x_2 ")" ^2+"(" y_1 "-" y_2 ")" ^2 )=√(1+"(-" 1")" ^2 )·|x1-x2|

　　=√2·√(〖"(" x_1+x_2 ")" 〗^2 "-" 4x_1 x_2 )=√2·√(1+4m)=3√2,

　　所以√(1+4m)=3,所以m的值为2.