圆x^2+y^2=1,求点[(x(x+y),y(x+y)]的轨迹方程
问题描述：
圆x^2+y^2=1,求点[(x(x+y),y(x+y)]的轨迹方程
答
回答即可得2分经验值，回答被选为满意回答可同步增加经验x^2+y^2-x-y=0值和财富值
答
设a=x(x+y),b=y(x+y),
有a+b=(x+y)^2,
a-b=(x+y)(x-y),
则(a-b)^2=(x+y)^2*(x-y)^2=(a+b)*(x-y)^2,
有(x-y)^2=(a-b)^2/(a+b)
于是a+b+(a-b)^2/(a+b)=(x+y)^2+(x-y)^2=2
化简得a^2+b^2=a+b;
答
令m=x(x+y),n=y(x+y)
则m-n=x^2-y^2,
于是m-n+1=2x^2,1-m+n=2y^2
(m+n-1)^2=4x^2y^2=(m-n+1)(1-m+n)
整理得：
m^2+n^2-m-n=0
于是点[(x(x+y),y(x+y)]的轨迹方程为：x^2+y^2-x-y=0