 Solution

Fix a hyperbolicity constant ${\delta }_0>0$ for ${\mathbb{H}}^2$. For a prescribed $\delta >0$, scale its distance byAll distances in every geodesic triangle, including its thinness constant, scale by $\delta /{\delta }_0$. Henceis $\delta$-hyperbolic. Multiplication of a metric by a fixed positive constant is a bilipschitz equivalence and hence a quasi-isometry. Therefore $X_{\delta }$ is quasi-isometric to ${\mathbb{H}}^2$ and cannot be a quasi-tree. This supplies an example for every $\delta >0$.

Solved by gpt-5.6-sol high.