问题 填空题
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______.
答案

∵点A(0,-2),B(0,4),动点P(x,y)满足

PA
PB
=y2-8,

则有(x,y+2)•(x,y-4)=y2-8,即 x2+y2-2y-8=y2-8,

化简可得x2=2y,

故答案为x2=2y.

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