设P坐标为(x,y)
PO=√(x^2+y^2)
PA=√[(x-c)^2+y^2]
|P0^2-PA^2|=|x^2+y^2-[(x-c)^2+y^2]|
=|x^2-(x-c)^2|=C
两边平方得
x^2-(x-C)^2=C^2
x^2-(x^2-2CX+C^2)=C^2
x^2-x^2+2Cx-C^2=C^2
2CX=2C^2
x=C