把下列曲面化成标准方程,并写出相应的坐标变换公式x^2+y^2+5z^2-6xy+2xz-2yz-4x+8y-12z+14=0

Question

把下列曲面化成标准方程,并写出相应的坐标变换公式x^2+y^2+5z^2-6xy+2xz-2yz-4x+8y-12z+14=0

Answer 1

先对含有x的项配方,然后对余下的多项式对含y的配方,最后对z配方：
0=x^2+y^2+5z^2-6xy+2xz-2yz-4x+8y-12z+14
=(x^2-6xy+2xz-4x)+y^2+5z^2-2yz+8y-12z+14
=(x-3y+z-2)^2-9y^2-z^2-4+6yz-12y+4z+y^2+5z^2-2yz+8y-12z+14
=(x-3y+z-2)^2-8y^2+4z^2+4yz-4y-8z+10
=(x-3y+z-2)^2-8(y^2-yz/2+y/2)+4z^2-8z+10
=(x-3y+z-2)^2-8(y-z/4-1/4)^2+z^2/2+z+1/2+4z^2-8z+10
=(x-3y+z-2)^2-8(y-z/4-1/4)^2+9/2*(z-7/9)^2+70/9
=u^2-8v^2+9w^2/2+70/9
坐标变换公式为：u=x-3y+z-2,v=y-z/4-1/4,w=z-7/9.
当然也可以用二次型的正交变换来做.