The projective model

The hyperboloid sheet $\mathcal{H}$ is topologically the same as a circular disk. We can make this equivalence precise by considering the ``projection'' along straight lines from the origin. Given any point $(x,y,z)$ with $z^2>x^2+y^2$ (the union of the two solid cones), there is precisely one value of $t$ such that $t^2 (z^2-x^2+y^2) = 1$ and $tz>0$. That is, there is exactly one point on the line through $(0,0,0)$ and $(x,y,z)$ lying on $\mathcal{H}$.

Now let $\mathbf{x}=(x,y,z)$ be on $\mathcal{H}$ and consider the point $(\frac{x}{z},\frac{y}{z},1)$ also on the line through the origin and $\mathbf{x}$. Note that

$$\left(\frac{x}{z}\right)^2 +\left(\frac{y}{z}\right)^2 +\frac{1}{z^2} = 1$$

Thus, the point $(u,v)= (\frac{x}{z},\frac{y}{z})$ is a point in the interior of the unit disk. This defines a bijection from the hyperboloid $\mathcal{H}$ to the open unit disk.

This bijection carries plane cross-sections in $\mathbb{R}^3$ defined by $ax+by+cz=0$ to lines $au+bv+c=0$ in $\mathbb{R}^2$. The disk equipped with the straight lines as geodesics is often called the projective or Klein model of hyperbolic geometry.