= Solution
Use the <Poincaré half-plane model>, $\mathbb H=\{x+iy:y>0\}$ with metric $ds^2=(dx^2+dy^2)/y^2$. Its <hyperbolic lines> are vertical lines and semicircles centred on the real boundary, hence orthogonal to that boundary. Each has two <ideal endpoints> in $\mathbb R\cup\{\infty\}$. Distinct <parallel hyperbolic lines> are disjoint inside $\mathbb H$ and share one ideal endpoint; distinct <ultraparallel hyperbolic lines> are disjoint and share none. Intersecting lines are neither. Because the metric is conformal to the Euclidean metric, the two notions of angle agree.
Apply a hyperbolic <isometry> taking one line $l$ to the imaginary axis, whose ideal endpoints are $0,\infty$. If the other line $l'$ is ultraparallel, its endpoints are two finite nonzero real numbers of the same sign. Reflect if necessary to write them as $0<a<b$. Then $l'$ is the semicircle with centre $c=(a+b)/2$ and radius $r=(b-a)/2$.
Every <hyperbolic line> perpendicular to the imaginary axis must be a semicircle centred at zero: at the intersection its tangent has to be horizontal, forcing its real centre to coincide with zero. Let its radius be $R$. Orthogonality to the circle representing $l'$ requires $c^2=R^2+r^2$, since the radii to the intersection form a right triangle. Consequently
$$
\boxed{R^2=c^2-r^2=ab.}
$$
This has exactly one positive solution. The circles intersect above the real axis: $c-r=a<\sqrt{ab}<b=c+r$. Hence the required <common perpendicular of ultraparallel hyperbolic lines> exists and is unique.
Conversely, if $l'$ intersects the imaginary axis, its endpoints straddle zero and the same equation would require $R^2=ab<0$, which is impossible. If $l'$ has endpoint zero, it gives $R^2=0$, not a line. If its shared endpoint is infinity, $l'$ is another vertical line; no circle centred at zero meets a distinct vertical line with horizontal tangent. Thus parallel lines have no common perpendicular either. These cases exhaust distinct lines, proving \b[a unique common perpendicular exists exactly for ultraparallel lines].
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