What is the parametrisation of $x^2-y^2+z^2=0$

Question

I need to be able to parametrise the surface $x^2-y^2+z^2=0$. I know it can be written as a cone in the form $y=\sqrt{x^2+z^2}$, but I don't know how to parametrise it and would appreciate any explanation of how you do it as well as a solution.

Answer 1

Rewrite as $$x^2+z^2=y^2$$ and notice that for a fixed value of $y$, you get the equation of a circle centered at $(0,0)$ (in the $xz$-plane) and with radius $|y|$. Let $y=r$ be a first parameter and then from the standard parametrization of a circle: $$\left\{ \begin{array}{rcl} x & = & r\cos t \\ y & = & r \\ z & = & r\sin t \end{array} \right.$$where $0 \le t \le 2\pi$ and $0 \le r < +\infty$ for a single cone, $-\infty < r < +\infty$ for the double cone.

Remark: since parametrizations aren't unique, there is no such thing as the parametrization, but rather a parametrization.