三角型ABC中,B(0,-2)C(0,2)其中周长等于12,求顶点A的轨迹方程

BC=|2-(-2)|=4
设A的坐标是(x,y)
则AB=√[(x-0)^2+(y+2)^2]
AC=√[(x-0)^2+(y-2)^2]
则有
AB+AC+BC=4+√[(x-0)^2+(y+2)^2]+√[(x-0)^2+(y-2)^2]=12
√[x^2+(y+2)^2]+√[x^2+(y-2)^2]=8
√[x^2+(y+2)^2]=8-√[x^2+(y-2)^2] 两边平方
x^2+(y+2)^2=64-16√[x^2+(y-2)^2]+x^2+(y-2)^2
64-8y=16√[x^2+(y-2)^2]
8-y=2√[x^2+(y-2)^2] 平方
64-16y+y^2=4x^2+4(y-2)^2=4x^2+4y^2-16y+16
4x^2+3y^2=48
x^2/12+y^2/16=1