设A(x1,y1) B(x2,y2) M(x0,y0)
M1(-p/2,y0) F(p/2,0)
点差法计算AB斜率:
A,B满足抛物线方程y1^2=2px1 y2^2=2px2
两式相减y1^2-y2^2=2px1-2px2
K(AB)=(y1-y2)/(x1-x2)=2p/(y1+y2)=p/y0
两点计算斜率K(M1F)=(y0-0)/(-p/2-p/2)=-y0/p
K(AB)*K(M1F)=-1= =>FM1垂直于AB
即有角AFM1=角AA1M1=90
又有AA1=AF,AM1=AM1
故有三角形AA1M1全等于三角形AM1F