Take, for example, the hyperboloid $x^2-y^2-z^2=1$ and the open disk (equipped with the Poincare metric) $x=1, \, y^2+z^2<1$. If we multiply a solution $(1,y,z)$ to the second set of equations by some scalar $s$ to get a solution to the first equation $(s, sy, sz)$, is the resulting map an isometry?
It's intuitive to me that if, given a two-sheeted hyperboloid, you center an open disk equipped with the Poincare metric at one of the hyperboloids vertices (is that what they're called?) such that the radius of the disk is the radius of the asymptotic cone at that point along its axis of symmetry, then performing the projection I described is an isometry of the hyperbolic plane. I haven't thought much about how to prove this, I feel as though I would need to know how to calculate distance on the hyperboloid, and trying to do so with an integral gives an equation that I'm pretty sure isn't solvable analytically.
Is my intuition correct? If so, How would you go about proving it? If not, there is an isometry between these spaces, because they're both models for the hyperbolic plane, right?