√[(x+1)²+y²] / √[(x-1)²+y²] = a
(x+1)²+y² = a² [(x-1)²+y²]
(a² -1)x² - 2(a²+1)x + (a²-1)y² = -(a²-1) ,a>0
1)
如果a=1,则a² -1=0
上式化为:x=0,轨迹是一条直线(在Y轴上)
2)
如果a≠1,则a² -1≠0
上式化为:x² - 2(a²+1)/(a² -1)x +1+ y² =0
化简 [x - (a²+1)/(a² -1)]² + y² = [(a²+1)/(a² -1)]² -1
轨迹是圆