Question

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,

Answer 1

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