Show that y^2-2yz+x-2y+3z=0 is a hyperbolic paraboloid

Show that $y^2-2yz+x-2y+3z=0$ is a hyperbolic paraboloid.

Explanation of Method

Showing that a second-degree equation describes a hyperbolic paraboloid.

Step 1: Completing the square

\[ \bigl(y-(z+1)\bigr)^2=(z+1)^2-3z-x=z^2-z+1-x=\left(z-\tfrac12\right)^2+\tfrac34-x . \]

Step 2: New coordinates

\[ u=y-z-1,\qquad v=z-\tfrac12,\qquad W=x-\tfrac34 . \]

This is an invertible affine change of coordinates, but not an orthogonal one. In the new coordinates: $W=v^2-u^2$.

This is the canonical form of a hyperbolic paraboloid.

Step 3: Orthogonal confirmation

The quadratic part $y^2-2yz$ has eigenvalues $\tfrac{1\pm\sqrt5}2$ and $0$, and the first two have opposite signs. The zero eigenvalue belongs to $\{1,0,0\}$, and the linear term $x$ has a nonzero component along it. So there is no center and the surface is a hyperbolic paraboloid. $\blacksquare$

Step 4: Sections

$z=c$ gives the parabola $x=-(y-c-1)^2+c^2-c+1$. $x=c$ gives the hyperbola

\[ u^2-v^2=\tfrac34-c , \] which becomes a pair of lines when $c=\tfrac34$.

$z=0$

Sections $z=c$ are parabolas $x=-(y-c-1)^2+c^2-c+1$ (blue; red for $z=0$). Their vertices rise and fall, giving the saddle shape.

Result

\[\boxed{\,W=v^2-u^2\,}\]

Remark

Cleanup: the source says the sections $x=c$ have two real $y$-values for every $z$ when $c\lt 1$. The minimum of $z^2-z+1$ is $\tfrac34$, so this holds only for $c\lt \tfrac34$. It also calls the change of variables a rotation; it is an affine map.