解:设直线l方程为y=k(x+p),代入y2=4px. 得k2x2+(2k2p-4p)x+k2p2=0.Δ=4(k2p-2p)2-4k2·k2p2>0,得0<k2<1. 令A(x1,y1)、B(x2,y2),则x1+x2=-,y1+y2=k(x1+x2+2p)=, AB中点坐标为(,).AB垂直平分线为y-=-(x-). 令y=0,得x0==p+.由上可知0<k2<1,∴x0>p+2p=3p.∴x0>3p. (2)解:∵l的斜率依次为p,p2,p3,…时,AB中垂线与x轴交点依次为N1,N2,N3,…. ∴点Nn的坐标为(p+,0). |NnNn+1|=|(p+)-(p+)|=,=, 所求的值为[p3+p4+…+p21]=,因为0<k2<1,所以0<P<1 |