过点Q(2,-4)做圆O:x2+y2=9的割线,交圆O于A,B求AB中点P的轨迹方程.AB中点P(x,y) 2x=xA+xB 2y=yA+yB 过Q(2,-4)作圆O:x^2+y^2=9的割线L:y+

题目

过点Q(2,-4)做圆O:x2+y2=9的割线,交圆O于A,B求AB中点P的轨迹方程.AB中点P(x,y) 2x=xA+xB 2y=yA+yB 过Q(2,-4)作圆O:x^2+y^2=9的割线L:y+4=k(x-2) k=(y+4)/(x-2)…… ……AB中点P的轨迹方程:(x-1)^2+(y+2)^2=5 满足(x^2+y^2

答案

P(x,y)
k(OP)=y/x
k(AB)=(y+4)/(x-2)
AB垂直OP,k(OP)*k(AB)=-1
(y/x)*[(y+4)/(x-2)]=-1`
(x-1)^2+(y+2)^2=5
因为P(x,y)满足题目的条件,所以就是点P的轨迹方程.