解:焦点(p/2,0) 设过焦点的 直线方程为:y/(x-p/2)=1/n
x= ny+p/2代入抛物线方程y^2=2p(ny+p/2)
y^2-2pny-p^2=0
根据伟达定理;y1y2=-p^2
y1+y2=2pn (2)因为C在准线上且AC平行X轴,所以AC垂直准线且C为垂足,设F为焦点,AC=AF又设A(x1,y1)
B(x2,y2)F(+p/2,0)
C(-p/2,y1)原点O(0,0),只要证明CO的斜率与BO的斜率相等,即证明CO与BO共线 CO的斜率K=(y1)/(-p/2)
BO的斜率K°=y2/x2 x2=ny2+p/2k°-k=y2/(ny2... |