已知A(-1,0),B(1,0),点C、点D满足|
|
设C、D点的坐标分别为C(x0,y0),D(x,y),
则
=(x0+1,y0),AC
=(2,0),AB
则
+AB
=(x0+3,y0),AC
故
=AD
(1 2
+AB
)=(AC
,x0+3 2
)y0 2
∵
=(x+1,y).AD
∴x0=2x-1 y0=2y
代入|
|=AC
=4,整理得x2+y2=4,(x0+1)2+y02
点D的轨迹方程为x2+y2=4
∵x= x0+1 2 y= y0 2
代入x2+y2=4得(x+1)2+y2=16
∴点D的轨迹方程为(x+1)2+y2=16
故答案为(x+1)2+y2=16,x2+y2=4,