= Solution
Parametrize a nonconstant <hyperbolic geodesic> by unit speed. Its energy is $(\dot x^2+\dot y^2)/(2y^2)$, and the <Euler-Lagrange equation> in the <cyclic coordinate> $x$ gives $\dot x/y^2=c$. Unit speed gives $\dot x^2+\dot y^2=y^2$. If $c=0$ the curve is vertical. Otherwise, on either side of a possible highest point,
$$
\frac{dx}{dy}=\frac{\pm cy}{\sqrt{1-c^2y^2}},\qquad (x-a)^2+y^2=c^{-2}.
$$
The two branches join smoothly at the top. Conversely these curves with their unit-speed parametrizations satisfy the energy equations. Thus \b[the unparametrized complete <geodesics> are vertical half-lines and upper semicircles centered on the real axis].
Let the <tractrix> point have positive height $r(\theta)$. Its tangent segment to the axis has length one and makes angle $\theta$, so $r=\sin\theta$. Since its slope is $\tan\theta$, $h'(\theta)=\cos^2\theta/\sin\theta$ for its axial coordinate. Rotating $X(\theta,v)=(h(\theta),\sin\theta\cos v,\sin\theta\sin v)$ gives orthogonal coordinate vectors with squared lengths $\cot^2\theta$ and $\sin^2\theta$. Hence the stated <first fundamental form> follows.
Now put \b[$x=v$, $y=\csc\theta$]. Direct substitution gives
$$
\frac{dx^2+dy^2}{y^2}=\sin^2\theta\,dv^2+\cot^2\theta\,d\theta^2.
$$
On the specified intervals this coordinate map is a <diffeomorphism> onto the stated rectangle, so is an <isometry> there and a <local isometry> on the <tractoid>.
Set $b=\csc\beta$, $a=\csc\alpha$. In the unwrapped coordinate, connect $(x_0,b)$ to $(x_0+2\pi,b)$ along the semicircle centered at $(x_0+\pi,0)$ of radius $\sqrt{b^2+\pi^2}$. Its height lies between $b$ and that radius. The hypothesis $a^2-b^2>\pi^2$ puts the whole arc in the closed height strip $b\leq y\leq a$. Projecting the angular coordinate modulo $2\pi$ therefore gives \b[a <geodesic> segment in the closed <tractoid> annulus which winds once and returns to its starting point]. Taking $x_0$ to represent any chosen angular position proves the assertion for every point of its $\theta=\beta$ boundary. The returned tangent need not equal the starting tangent, so this proves the claimed returning segment without asserting a periodically smooth <closed geodesic>.
\Image[/past-exam-of-the-geometry-course-of-the-university-of-oxford/2007-tractoid-return.png]
{title=A returning <geodesic> in an unwrapped <tractoid> annulus}
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