設(shè)A(X1,Y1),B(X2,Y2)則 y1^2=2px1,y2^2=2px2
∠AOB=90
(y1*y2)/(x1*x2)=-1 即y1*y2=-4P^2
由直線AB得:y-y1=(y1-y2)/(x1-x2)*(x-x1)
因?yàn)?y1^2=2px1,y2^2=2px2兩式相減
y1^2-y^2=2p(x1-x2)
(y1+y2)(y1-y2)=2p(x1-x2)
(y1-y2)/(x1-x2)=2p/(y1+y2)
故y-y1=2p/(y1+y2)*(x-x1)
又y1*y2=-4P^2,y1^2=2px1,y2^2=2px2
(y-y1)(y1+y2)=2p*(x-x1)
yy1+yy2-y1^2-y1y2=2px-2px1
yy1+yy2-2px1+4p^2=2px-2px1
yy1+yy2=2px-4p^2
故(y2+y1)*y=2p*(x-2p)
x=2p時(shí),y恒為0
所以直線AB過(guò)定點(diǎn)(2p,0)
A,B是拋物線y^2=2px(p>0)上的兩點(diǎn),滿(mǎn)足OA垂直O(jiān)B,求證直線AB恒過(guò)一定點(diǎn)
A,B是拋物線y^2=2px(p>0)上的兩點(diǎn),滿(mǎn)足OA垂直O(jiān)B,求證直線AB恒過(guò)一定點(diǎn)
這個(gè)答案最后一步怎么得到的不理解啊
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這個(gè)答案最后一步怎么得到的不理解啊
(y1+y2)*y=2p(x-2p)怎么求?
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數(shù)學(xué)人氣:273 ℃時(shí)間:2019-10-17 00:37:11
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