I am trying to find a canonical form for the quadric $x^2+y^2-3z^2-2xy-xz-6yz=0$ by completing squares. For example, if it were $x^2+y^2-3z^2-2xy-6xz-6yz=0$, I would write $$x^2+y^2-3z^2-2xy-6xz-6yz=(x-y-3z)^2-12(z+\frac12y)^2+3y^2=0$$ and letting $X:=x-y-3z, Z:=\sqrt{12}(z+\frac12y), Y:= \sqrt{3}y$. We get $Z^2=X^2+Y^2$ and so it is a cone.
I need to apply same method to above quadric but I could not.
Thanks for any help
On
If you really want the canonical form, caring about metric properties, orthogonally diagonalize: Your equation is approximately
$$-4.75182(0.158224x+ 0.478112y+ 0.863929z)^2 + 2.846(0.348633x-0.845649y+ 0.404145z)^2+0.905819(-0.923807x -0.237249y + 0.300488z)^2=0.$$
In this case, however, applying the standard dehomogenization, and looking at the parabola you get, there is a standard focus/directix form, that appeals to me:
$$2\cdot((x+\frac{47}{56}z)^2+(y-\frac{23}{56}z)^2-(x+y+\frac{61}{28}z)^2/2)=0$$ or $$\frac{(56x+47z)^2+(56y-23z)^2-2\cdot (28x+28y+61z)^2}{2^57^2}=0.$$
You can of course apply any metric properties preserving map (mapping any plane to $z=1$), dehomogenize and write the resulting conic section in focus/directrix form (for ellipses and hyperbolae there are two), homogenize again, and use the inverse map to get as many representations you want, few of which are nice.
You can do this: $$x^2+y^2-3z^2-2xy-xz-6yz=(x-y+3z)^2-7xz-12z^2= \\(x+y-3z)^2-12(z+\frac{7}{24}x)^2+\frac{49}{48}x^2$$