将空间曲线一般方程转化为参数方程x^2+y^2+z^2=3,x+y+z=1

Question

将空间曲线一般方程转化为参数方程
x^2+y^2+z^2=3,x+y+z=1

Answer 1

把z=1-x-y带入到x^2+y^2+z^2=3
得到x^2+y^2-x-y+xy=1
配方为(2x+y-1)^2+3(y-1/3)^2=16/3
令2x+y-1=4cost/√3
y-1/3=4sint/3
联立后解得
x=(2√3cost-2sint+1)/3
y=(1+4sint)/3
z=1-x-y=(1-2√3cost-2sint)/3
所以
x=(2√3cost-2sint+1)/3
y=(1+4sint)/3
z=(1-2√3cost-2sint)/3
即为参数方程