Analytic Geometry: One sheeted hyperboloid

Question

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!

Answer 1

This Wolfram article explains how to categorize a quadratic surface: http://mathworld.wolfram.com/QuadraticSurface.html

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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.

Answer 2

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.