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
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![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](/uploads/image/z/8666000-8-0.jpg?t=%E5%B7%B2%E7%9F%A5%E7%9B%B4%E7%BA%BFl%E8%BF%87%E6%8A%9B%E7%89%A9%E7%BA%BFy%5E2%3D2px%28p%3E0%29%E4%BA%A4%E4%BA%8EAB%E4%B8%A4%E7%82%B9%E4%B8%94OA%E2%8A%A5OB%2COD%E5%9E%82%E7%9B%B4AB%E4%BA%A4AB%E4%BA%8ED%2C%E6%B1%82%E8%AF%81%E7%9B%B4%E7%BA%BFl%E8%BF%87%E5%AE%9A%E7%82%B9%E5%B9%B6%E6%B1%82%E5%87%BA%E5%AE%9A%E7%82%B9)
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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
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
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