已知直线l过抛物线y^2=2px(p>0)交于AB两点且OA⊥OB,OD垂直AB交AB于D,求证直线l过定点并求出定点

设A(a²/(2p),a),B(b²/(2p),b)
OA的斜率=a/[a²/(2p)] = 2p/a
OB的斜率=b/[b²/(2p)] = 2p/b
OA⊥OB:(2p/a)(2p/b) = -1,ab = -4p²
AB的方程:(y - b)/(a - b) = [x - b²/(2p)]/[a²/(2p) - b²/(2p)]
y - b = (2px - b²)/(a +b)
(a + b)y - ab - b² = 2px - b²
y = 2px/(a + b) + ab
y = 2px/(a + b) - 4p²
直线l过定点(0,-4p²)