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I am trying to understand how the Poincaré disk is constructed using the stereographic projection for the hyperboloid $x^2+y^2-z^2=-1$.

So I want to project a line from the fixed point $(0,0,-1)$ to a point on the hyperboloid.

The parametric equation of a line between $(0,0,-1)$ and a point $(x,y,z)$ is given by $\vec{r}(t)=t(x,y,z)+(1-t)(0,0,-1)=(tx,ty,tz+t-1)$, where $t \in [0,1]$.

Now I want to find where this line intersects the hyperboloid. So

$(tx)^2+(ty)^2-(tz+t-1)^2=-1 \iff$ $(x^2+y^2-z^2-2z-1)t^2+(2z+2)t=0$

Now $x^2+y^2-z^2=-1$ because I want to find the intersection. So I get the equation

$-(2z+2)t^2+(2z+2)t=0,$

whose only solutions are $t=0$ or $t=1$.

Shouldn't I obtain an expression for $t$ in terms of $x,y,z$?

Bernard
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wwinters57
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1 Answers1

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You should be projecting from the hyperboloid (i.e., a point such that $x^2+y^2-z^2=-1$) to the plane $z=0$. What you are doing now is you are projecting the hyperboloid ($x^2+y^2-z^2=-1$) to itself ($x^2+y^2-z^2=-1$), so that is why nothing changes (for $t=1$).

Now I want to find where this line intersects the hyperboloid => Now I want to find where this line intersects the plane $z=0$.

The correct $t$ is given by $tz+t-1=0$.

Zeno Rogue
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