帮帮忙，抛物线问题

由点斜式可得AB直线方程:Y=X-P/2 将此代人抛物线方程得:
X^-PX+P^/4=2PX (^表示平方)
X^-3PX+P^/4=0
X=3/2P±√2P
∵A点在X轴上方 ∴A(3/2P+√2P,ya);B(3/2P-√2P,yb)
∴│AF│=√2(3/2P+√2P-P/2)=√2P+2P
│BF│=√2[P/2-(3/2P-√2P)]=2P-√2P
∴│AF│:│BF│=3+2√2

p>0
y^2=2px的焦点F(0.5p,0)
过F做倾斜角为45度的直线L:y=x-0.5p,x=y+0.5p
y^2=2px=2p(y+0.5p)
y^2-2py-p^2=0
yA+yB=2p,yA*yB=-p^2
xA+xB=(yA+yB)+p=3p
(xA-xB)^2=(yA-yB)^2=(yA+yB)^2-4yA*yB=4p^2+4p^2=8p^2
AB^2=(xA-xB)^2+(yA-yB)^2=16p^2
|AB|=4p
|AF|+|BF|=4p......(1)
AF^2=(xA-0.5p)^2+yA^2
BF^2=(xB-0.5p)^2+yB^2
AF^2-BF^2=(xA+xB)*(xA-xB)-p(xA-xB)+(yA+yB)*(yA-yB)
=3p*2√2p-p*2√2p+2p*2√2p
=8√2p^2
(|AF|+|BF|)*(|AF|-|BF|)=8√2p^2
4p*(|AF|-|BF|)=8√2p^2
|AF|-|BF|=2√2p......(2)
(1)+(2):
|AF|=(2+√2)p
(1)-(2):
|BF|=(2-√2)p
|AF|/|BF|=(2+√2)p/[(2-√2)p]=3+2√2