Question

设抛物线Y平方=2PX(P大于0)的焦点为F.其准线与X轴交与点C.过点F作它的炫长AB.若角CBF=90度，则AF-BF为多少

高手进，非诚勿扰

Answer 1

设AB方程为:y=k(x-p/2)(假设k存在)
联立得k^2(x^2-px+p^2/4)=2px
(k^2)x^2-(k^2+2)px+(kp)^2/4=0
设两交点为A(x1,y1),B(x2,y2),
∠CBF=90°即(x1-p/2)(x1+p/2)+y1^2=0
x1^2+y1^2=p^2/4
x1^2+2px1-p^2/4=0
(x1+p)^2=(5/4)p^2
x1=(-2+√5)p/2或(-2-√5)p/2(舍)
∴A((-2+√5)p/2,√(-2+√5)p)
|AC|=√{[(1+√5)/2]^2+(-2+√5)}p=√[(-1+√5)/2]p
|AF|=√{[(-3+√5)/2]^2+(-2+√5)}p=√[(3-√5)/2]p==(-1+√5)p/2
∵ΔCAF∽ΔBAC,故|AB|/|AC|=|AC|/|AF|
∴|AB|=|AC|^2/|AF|=p
∴|BF|=|AB|-|AF|=(3-√5)p/2
|AF|-|BF|=4p/2
AF-BF=2P