解析几何中有这样一个结论,即命题1在抛物线y2=2px(p0)中,过顶点O作互相垂直的两直线交抛物线于A,B两点,连A,B交x轴于E点,则E为定点.图1证设A(x1,y1),B(x2,y2),直线AB:x=ky+m,代入y2=2px,得y2-2pky-2pm=0.故y1y2=-2pm.又OA⊥OB,得x1x2+y1y2=0,(1)21y22故y4p2+y1y2=0,m2-2pm=0, There is a conclusion in analytical geometry that proposition 1 is in the parabola y2=2px(p0). Two vertices perpendicular to the vertex O are intersected by two straight lines at A and B. Even A and B cross the x axis at point E. E is a fixed point. Figure 1 proves that A (x1, y1), B (x2, y2), straight line AB: x = ky + m, into y2 = 2px, get y2-2pky-2pm = 0. So y1y2 = - 2pm. Also OA⊥OB, get x1x2+y1y2=0, (1)21y22 so y4p2+y1y2=0, m2-2pm=0,