Let $(M, \partial M)$ be a Riemannian manifold with sectional curvature $\geq 0$ and convex boundary $\partial M$ of sectional curvature $\geq 0$. Then it is well known that the function $\operatorname{dist}(\partial M , x) : M \setminus \partial M \rightarrow \mathbb{R}$ is concave.
Assume $M=\mathbb{H}^2$ i.e. the hyperbolic plane. In the upper half plane model the right quadrant is a convex set $C$. What can be said about the eigenvalues of the hessian of $\operatorname{dist}(\partial C , x) : M \setminus \partial M \rightarrow \mathbb{R}$ at a point $p \in C$. Clearly one eigenvalue will be $0$ what is the second?
Is there a quick reference?
Edit: Short answer is $\tanh(d(p,\partial C))$ (see answer below).
As you observed, the Hessian always has a null direction along the geodesic joining $p$ to the closest point of $\partial C$. Since the Hessian is symmetric, the remaining eigenspace will be orthogonal to this geodesic, meaning it will coincide with the tangent space to the level set of the distance function $r = \operatorname{dist}(\partial C,\cdot).$ The restriction of the Hessian to this tangent space is just the second fundamental form of the level curve $\{ r = r(x) \}$, so the corresponding eigenvalue is just the geodesic curvature of this curve.
It's well known that in the half-plane model (with your $C$) these curves (known as hypercycles) are just the Euclidean rays from the origin. You can deduce this fairly simply: the geodesic joining $\{r=r(x)\}$ to $\partial C$ must be perpendicular to both $\partial C$ and $\{r = r(x)\}.$ The former implies it's an arc of a circle centred at the origin, and thus the latter means the tangent to $\{r = r(x)\}$ is radial.
Thus the eigenvalue you're looking for is just the geodesic curvature at $x$ of the ray from the origin to $x$. A routine calculation (just parametrise this curve by arclength and compute its covariant acceleration) should result in the answer $\tanh(r(x)).$