Jacobi equation and conjugate points on solution curves of the Van der Pol vector field

Question

Dear  colleagues

I had already asked this question in MO and I present it here at  physicsoverflow

https://mathoverflow.net/questions/439319/jacobi-equation-and-conjugate-points-on-solution-curves-of-the-van-der-pol-vecto

Let $X$ be  a  geodesible non vanishing  vector field on a manifold $M$. Namely there is a Riemannian  structure $(M,g)$  such  that all integral  curves of  $M$  are unparametrized geodesics of the  metric $g$.  [This  is  a  beautiful non example](https://mathoverflow.net/a/274981/36688)

> Is the index of an orbit or a  closed orbit (i.e the index of a   geodesic or  a  closed geodesics ) encoded in the vector  field $X$? Namely can we  compute the number of  conjugate points on a  (closed) orbits of $X$ with information just from the vector field and nothing  else?

This  question could play a crucial role in investigation of the following post about a [negatively curved  structure  on the punctured plane for which the solution curves of the Van der Pol equation would be geodesics](https://mathoverflow.net/q/160945/36688).The post search  for  a negative curvature metric on the punctured plane for which the  Van der Pol foliation is a  foliation by geodesics.

 Because if the answer to this  current question is affirmative(or there are some modified way to compute the index of the closed orbits of the underling vector field as closed  geodesics of the corresponding compatible metric) and we  get a non zero index, then this  would implies that **there is  no any   metric with negative curvature on the punctured plane making all  solutions of the  Van der Pol vector field into geodesics since negative curvature implies non existence of  conjugate points**. 

>The  next question:  Regardless of  the sign of the  curvature, is there a  Riemannian metric on the  punctured plane  such that  all solution curves of the  Vander pol equation are geodesics and there is  no any conjugate points at all?