設(shè)AB方程為y=k(x-p/2) A(x1,y1)B(x2,y2)
與y^2=2px聯(lián)立得 k^2(x-p/2)^2=2px
k^2x^2-(k^2+2)px+k^2p^2/4=0
x1+x2=(1+2/k^2)p x1x2=p^2/4
對(duì)y^2=2px 兩邊求導(dǎo)得 2y*y`=2p
在A點(diǎn)切線斜率為y`=2p/2y1=p/y1 方程為y-y1=p/y1(x-x1)
在B點(diǎn)切線斜率為y`=p/y2 方程為y-y2=p/y2(x-x2)
兩式聯(lián)立得 y=p(x1-x2)/(y1-y2)=(y1+y2)/2
則x=y1y2/2p
故CF^2=(x+p/2)^2+y^2=y1^2y2^2/4p^2+p^2/4+(y1^2+y2^2)/4
=x1x2+p^2/4+p(x1+x2)/2
又AF=a=x1+p/2 BF=x2+p/2=b
則AF*BF=x1x2+(x1+x2)p/2+p^2/4=ab=CF^2
故CF=根號(hào)ab