过点M（-2,0）的直线l与椭圆x^2+2y^2=2交于p1、p2两点,线段p1p2的中点为P,设直线l的斜率为k1,直

设P1（x1,y1）,p2(x2,y2),P(x0,y0)
则有x1^2+2y1^2=2
x2^2+2y2^2=2
则（x1^2-x2^2）+2（2y1^2-2y2^2）=0
（x1+x2）*（x1-x2）+2（y1+y2）*（y1-y2）=0 ①
又2x0=x1+x2,2y0=y1+y2
k1=(y1-y2)/(x1-x2)
即y1-y2=k1(x1-x2)
k2=(y0-0)/(x0-0)
即y0=k2x0
这样,代入①得
2x0(x1-x2)+2*2k2x0*k1(x1-x2)=0
1+2k1k2=0
k1k2=-1/2