已知一条曲线C在y轴右边,C上每一点到点F(1,0)的距离减去它到y轴的差都是1.(1)求曲线的方程； (2)是否存在正

按抛物线定义：此题变为到点F(1,0)的距离与到x=-1距离相等.∴(1)y^2=4x；(2)设直线方程为y=k(x-m) [x为R] 设A(x1,y1),B(x2,y2)则FA=（x1-1,y1),FB=(x2-1,y2),y1y2=k^2(x1-m)(x2-m)∴FA.FB=x1x2-(x1+x2)+1+y1y2=(k^2+1)x1x2-(mk^2+1)(x1+x2)+1+(km)^2联解抛物线与直线得：k^2x^2-(4+2mk^2)x+(km)^2=0(注意k=0舍去)∴x1+x2=（4+2mk^2)/k^2,x1x2=(km)^2/k^2=m^2带入得FA.FB=(k^2+1)x1x2-(mk^2+1)(x1+x2)+1+(km)^2=(k^2+1)m^2-(mk^2+1)[（4+2mk^2)/k^2]+(km)^2+1=[(m^2-6m+1)k^2-4]/k^2>0左边是二次函数要大于0恒成立则：m^2-6m+1>0且△