问题
填空题
已知抛物线y2=2px(p>0),过点M(2p,0)的直线与抛物线相交于A,B,
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答案
设直线AB:x=ty+2p代入抛物线y2=2px消去x得,y2-2pty-4p2=0,
设A(x1,y1),B(x2,y2)
所以根据根与系数的关系可得:y1+y2=2pt,y1y2=-4p2
∴
•OA
=x1x2+y1y2=(ty1+2p)(ty2+2p)+y1y2OB
=t2y1y2+2pt(y1+y2)+4p2+y1y2
=-4p2t2+4p2t2+4p2-4p2=0.
故答案为:0.