Question

这个轨迹方程的参数怎么消?X=m^2/(1+m^2)(^2是平方的意思）Y=(m^2-m+1)/(1+m^2)可以的话,

Answer 1

x=m^2/(1+m^2)

x+xm^2=m^2

m^2=x/(1-x)

y=(m^2-m+1)/(1+m^2)

y+ym^2=m^2-m+1

(y-1)m^2+y-1=-m

(y-1)(m^2+1)=-m

m+1/m=1/(1-y)

两边平方

m^2+2+1/m^2=1/(1-y)^2

x/(1-x)+2+(1-x)/x=1/(1-y)^2

1/[x(1-x)]=1/(1-y)^2

x(1-x)=(1-y)^2