求到相距为2a的两定点A与B的距离之比为一常数b(b>0)的动点P的轨迹方程,并说明轨迹是何种曲线?

是圆.圆心在AB连线上
和A的距离是2ab^2/(b^2-1)
和B距离是2a/(b^2-1)
半径是2ab/(b^2-1)
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设AB都在X轴上 A(0,0) B(2a,0)
P坐标是(x,y) │PA│= b│PB│
√(X^2+Y^2)=b√[(X-2a)^2+Y^2]
X^2+Y^2=b^2(x^2-4ax+4a^2+Y^2)
(b^2-1)x^2-4ab^2x+4a^2b^2+(b^2-1)y^2=0
x^2 -4ab^2x/(b^2-1) +4a^2b^2/(b^2-1) +y^2=0
[x- 2ab^2/(b^2-1)]^2 +y^2=[2ab/(b^2-1)]^2
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