过抛物线y^2=2px(p>0)的焦点F(p/2,0)、斜率为k(k>0)的直线方程为y=k(x-p/2),设该直线与抛物线的交点为A(u,k(u-p/2))、B(v,k(v-p/2)),于是u、v应该是方程组y^2=2px与y=k(x-p/2)的解,即u、v满足方程2px=(k^2)(x-p/2)^2,即x^2-(1+2/k^2)px+p^2/4=0,因此2pu=(k^2)(u-p/2)^2,2pv=(k^2)(v-p/2)^2,u+v=(1+2/k^2)p,uv=p^2/4
又|AF|^2=(u-p/2)^2+(k^2)(u-p/2)^2=(1+k^2)(u-p/2)^2；|BF|^2=(v-p/2)^2+(k^2)(v-p/2)^2=(1+k^2)(v-p/2)^2,故由|AF|=2|BF|得：(1+k^2)(u-p/2)^2=4(1+k^2)(v-p/2)^2
将2pu=(k^2)(u-p/2)^2,2pv=(k^2)(v-p/2)^2代入上式中即得u=4v
再将u=4v分别代入u+v=(1+2/k^2)p,uv=p^2/4中并解所得方程组,得k=2倍根2