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)

设AO y=kx联立求得A(2P/k2，2P/k)
BO y=-1/kx得B(2Pk²，-2Pk)
AB方程点斜式化简后得(1/k-k)y+2P=x
y=0时x=2P故AB过定点C(2P，0)
而OD⊥DC，OC为定点，
故D为以OC为直径的圆上一点，
方程为(x-P)²+y²=P²