Question

若点P(x,y)满足√x^2+(y-2)^2+√x^2+(y+2)^2=8,则点p的轨迹方程

求过程

Answer 1

若点P(x,y)满足√x^2+(y-2)^2+√x^2+(y+2)^2=8,
即点P到点A(0,2) B(0,-2)距离之和=8
根据椭圆定义可知
点P的轨迹为椭圆 焦点在y轴上
c=2 2a=8 a=4
b^2=a^2-c^2=12
点p的轨迹方程
y^2/16+x^2/4=1