已知点A(0,-2),B(0,4),动点P(x,y)满足PA•PB=y2-8,则动点P的轨迹方程是 _ .

问题描述:

已知点A(0,-2),B(0,4),动点P(x,y)满足

PA
PB
=y2-8,则动点P的轨迹方程是 ___ .

∵点A(0,-2),B(0,4),动点P(x,y)满足

PA
PB
=y2-8,
则有(-x,-y-2)•(-x,4-y)=y2-8,即 x2+y2-2y-8=y2-8,
化简可得x2=2y,
故答案为:x2=2y.