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Take the parabolic element of PSL2(R)It fixes no point in $\mathbb{H}$. A Geodesic in the Poincare half-plane model is determined by its unordered pair of ideal endpoints: either two real numbers or one real number together with $\infty$. Translation adds one to every finite endpoint, so no such pair is invariant. Therefore this $A$ preserves neither a point nor a hyperbolic line in $\mathbb{H}$.

A Geodesic in the Poincare half-plane model is a vertical line or a semicircle orthogonal to the real axis. Given a hyperbolic line $\ell$, choose a real Möbius transformation sending its two ideal endpoints to $0$ and $\infty$. After multiplying its representing matrix by a scalar and, if necessary, changing its sign, it represents an element of $PS{L}_2(\mathbb{R})$ and sends $\ell$ to the positive imaginary axis ${\ell }_{+}$.

The chosen point $P\in \ell$ then maps to $iy$ for some $y>0$. The dilation $z\mapsto z/y$, represented byfixes ${\ell }_{+}$ and sends $iy$ to $i$. Thus every pointed line $(\ell ,P)$ can be sent to $({\ell }_{+},i)$, proving transitivity on pointed hyperbolic lines.

For a continuously differentiable curve $\gamma (t)=x(t)+iy(t)$ in the Poincare half-plane model, its hyperbolic length isThe hyperbolic lines are the geodesics: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.

The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane Möbius transformation to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $i{y}_0$ and $i{y}_1$. Every joining curve satisfiesEquality holds for the monotone vertical parametrizationThus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho (z,w)$.

Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the formThis is the normal form underlying the common perpendicular of ultraparallel hyperbolic lines. Put $z=e^{u+i\theta }$. Sincethe metric becomesEvery path from $l$ to $m$ therefore hasEquality requires $\theta =\pi /2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance

For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.

Forone has $L_y=0$ andA circle of radius $R$ centred at $(0,c)$ on the $y$-axis satisfiesOn any arc that does not cross $y=c$,which is constant. Hence $d{L}_{y^{\prime }}/dx=0=L_y$, so the Euler-Lagrange equation holds and $\delta I=0$. These circles are the coordinate-swapped form of a Geodesic in the Poincare half-plane model.

Here the integrandhas no explicit $x$-dependence. The Beltrami identity therefore givesfor a positive constant $R$. Thuswhose nonvertical solutions are the semicircles

The endpoint $(a,a)$ and the prescribed value at $x=b$ both lie onbecauseThe required positive arc is consequentlyThe condition $b<2a$ keeps the endpoint above the $x$-axis. The sketch is the descending part of the upper semicircle of radius $a$ centred at $(a,0)$, from $(a,a)$ toThis is precisely a Geodesic in the Poincare half-plane model.