Let $(\mathcal{M},\mathbf{g})$ be the globally hyperbolic Lorentzian manifold corresponding to the exterior region of the Schwarzschild solution, i.e.
$\mathcal{M}=\mathbb{R}\times[\mathbb{R}^3\backslash\overline{B_{2m}(0)}]$
and
$\mathbf{g}=-\left(1-\frac{2m}{r}\right)\mathbf{d}t^2+\left(1-\frac{2m}{r}\right)^{-1}\mathbf{d}r^2+r^2(\mathbf{d}\theta^2+\sin^2\theta\mathbf{d}\varphi^2)$.
How can I argue that the point $q=\left(\pi\sqrt{\frac{r_0^3}{m}},r_0,\frac{\pi}{2},\pi\right)$ is conjugate to the point $p=\left(0,r_0,\frac{\pi}{2},0\right)$, where $r_0=3m$, along the curve
$c(t)=\left(t,r_0,\frac{\pi}{2},\sqrt{\frac{m}{r_0^3}}t\right)$
without solving the Jacobi equation? Does it suffice to show that there is a family of geodesics connecting $p$ to $q$?