已知抛物线方程 y²=4x ,若过焦点F且倾斜角为60°的直线m交抛物线与A,B两点,

设x=y√3/3+1
x=y√3/3+1
y²=4x 
y²-4√3y/3-4=0
设A(x1,y1)B(x2,y2)
y1+y2=4√3/3
y1y2=-4
设M(-1,m)
kMF=m/-2
kMA=(y1-m)/(x1+1)
kMB=(y2-m)/(x2+1)
kMA+kMB=(y1-m)/(x1+1)+(y2-m)/(x2+1)
=[y1+y2-2m+y2x1+y1x2-m(x1+x2)]/(x1x2+x1+x2+1)
={y1+y2-2m+y1y2(y1+y2)/4-m[(y1+y2)²-2y1y2]/4}/{y1²y2²+[(y1+y2)²-2y1y2]/4+1}
={4√3/3-4√3/3-[(4√3/3)²+8]m/4-2m}/{1+[(4√3/3)²+8]m/4+1}
=-m=2kMF