直角坐标平面xoy中,若定点A(1,2)与动点P(x,y)满足OP•OA=4,则点P的轨迹方程是 _.

问题描述:

直角坐标平面xoy中,若定点A(1,2)与动点P(x,y)满足

OP
OA
=4,则点P的轨迹方程是 ______.

设点P(x,y),则

OP
=(x,y)
因为A(1,2)
所以
OA
=(1,2)
因为
OP
OA
=4

所以(x,y)•(1,2)=4
即x+2y=4,
即x+2y-4=0
故答案为:x+2y-4=0