Analytic Geometry: One sheeted hyperboloid Good afternoon!
I have a question about analytic geometry. I don't actually know if the answer is quite simple, and I missed something while revising, or if it is actually more complicated than I originally thought.
The canonical expression for a one sheeted hyperboloid is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$$
which is fairly easy to draw. But this expression orients the hyperboloid centered at the z axis. When the hyperboloid is centered at any random vector, the expression changes. I don't want to know particularly how to get that expression, as it is not my objective. I want to know how do I realize at plain sight that, for example, the expression $$x^2-yx=1$$ is also an hyperboloid, but centered at other axis. I don't know if my question is unclear or too vague, if so, please ask me to clarify it. Thank you! 
 A: This Wolfram article explains how to categorize a quadratic surface:
http://mathworld.wolfram.com/QuadraticSurface.html

Edit: More specifically, here's a another method:
$$x^2-yx=1$$
Complete the square:
$$(x - \frac 12 y)^2-\frac 14 y^2=1$$
Let's pick the different coordinates:
\begin{cases}x'=x-\frac 12 y \\ y'=\frac 1 2 y\end{cases}
Then the equation becomes:
$$(x')^2-(y')^2=1$$
This is the equation of a hyperbole. Or in 3 dimensions it is the equation of a cylindrical hyperbole.
A: You don't need to actually find eigenvectors for the Gram/Hessian matrix; it is enough to find the count of positive eigenvalues and the count of negative. If any are zero, the story changes. 
http://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
So, your orginal (diagonal) example has two + and one - eigenvalue, then the target value $1$ is positive. So, you get one sheet. One sheet if the sign of the target number agrees with the repeated eigenvalue.
You next $x^2 - y z,$ the Hessian matrix is
$$ H \; = \;  
 \left(  \begin{array}{rrr}
  2  &  0  &  0 \\
  0   &  0  &  1 \\
  0  &  1   &  0  
\end{array} 
  \right)  ,
  $$
eigenvalues $2,1,-1.$ 
Again, two + eigenvalues, and the target 1 is also +, so one sheet.  Look at another way, rotated basis gives $y = \frac{u+v}{\sqrt 2},z = \frac{u-v}{\sqrt 2}, $ the end result being $x^2 - \frac{u^2}{2} + \frac{v^2}{2}. $ Oh, if the target were $0$ you would get a (double-nappe) cone; this would be called the light cone in Lorentz/Minkowski 2+1 space. 
Cautions: if you throw in linear terms, $px+qy+rz,$ it throws off everything. Also if you have three positive or three negative eigenvalues, you have a definite form, and the result is an ellipsoid or a single point or empty.  
