= Solution
Write $f(r)=1+r^2/a^2$. The geodesic <Lagrangian> is
$$
L=\frac12\left[-f\dot t^2+f^{-1}\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2\right].
$$
Its $\theta$ equation is
$$
\frac d{d\tau}(r^2\dot\theta)-r^2\sin\theta\cos\theta\dot\phi^2=0.
$$
Spherical symmetry lets us rotate the conserved angular-momentum plane into $\theta=\pi/2$. Equivalently, initial data $\theta=\pi/2$, $\dot\theta=0$ solve this equation for all $\tau$.
The cyclic coordinates give
$$
E=f\dot t,\qquad h=r^2\dot\phi.
$$
Using the timelike normalization $g_{\mu\nu}\dot x^\mu\dot x^\nu=-1$ gives
$$
\dot r^2+f\left(1+\frac{h^2}{r^2}\right)=E^2.
$$
After separating the constant terms,
$$
\boxed{
\frac12\dot r^2+V(r)=\frac12\left(E^2-1-\frac{h^2}{a^2}\right),
\qquad
V(r)=\frac12\left(\frac{r^2}{a^2}+\frac{h^2}{r^2}\right).
}
$$
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