Question

有一道数学题，急急急！在线等

已知抛物线Y方=2PX（P>0）的焦点为F，过点F做直线L与抛物线交 A B两点，抛物线的准线与X轴交于点C。 证明，<ACF = <BCF

Answer 1

y^2=2px
F(p/2,0)
过F弦AB ：y=k(x-p/2), A(x1,y1) B(x2,y2) (x1>x2)
k^2(x-p/2)^2=2px
l^2x^2-(pk^2+2p)x+p^2k^2/4=0
x1+x2=(pk^2+2p)/k^2=p+2p/k^2
x1x2=(p^2k^2/4) /k^2=p^2/4
C(-p/2,0)
tanACF=y1/(x1+p/2)=k(x1-p/2)/(x1+p/2)
tanBCF=-y2/(x2+p/2)=-k(x2-p/2)/(x2+p/2)
(x1-p/2)(x2+p/2)=x1x2-p^2/4+(x1-x2)p/2=(x1-x2)p/2
(p/2-x2)(x1+p/2)=p^2/4-x1x2+(x1-x2)p/2=(x1-x2)p/2
k(x1-p/2)(x2+p/2)=k(p/2-x2)(x1+p/2)
k(x1-p/2)/(x1+p/2)=k(p/2-x2)/(x2+p/2)
tanACF=tanBCF
角ACF=BCF

Answer 2

楼主，是不是缺了什么条件？