First we  explain our  Motivation:

>First note that there is  no  a  Riemannian metric  on an open set  of  the  plane  which possess two  nested closed geodesics $\gamma_1, \gamma_2$ and the  curvature in the  annular region made by theses two closed geodesics has constant sign(either positive or  negative).

 >But the same  is  true if we keep the  closed geodesic $\gamma_1$  and replace the  smooth closed geodesic  $\gamma_2$ by a  non smooth one  denoted again by  $\gamma_2$. In fact $\gamma_2$ is the  union of two pieces of  geodesics whose union make a  closed curve which can be  counted as a 2-polygon(with two vertex), whose sum angles is $2\pi$, look at the following    figure:**



>In the  above  figure the  (horizontal) $x$-axis is assumed to be a geodesic and we assume that all geodesics in the figure which transversally intersect the  x axis, they intersect with the same  angle.

>Of course the  sum angle of  the $2$-polygon in the figure is equal to $2\pi$ iff the $x$-axis is  an isocline.

>On the other  hand, [in this   question, Limit Cycles  as  closed Geodesics(2),]( we  had intention to  consider limit cycles  of  a  quadratic  vector  field as closed geodesic. Our  approach is  based on usage of Gauss  Bonnete theorem so the  sign of  Gauss  curvature  play a  crucial role. So we observe that certain "Isocline -type  property" of  foliation by  geodesics could  be  an  obstruction for Looking at limit cycles as  closed geodesics in a  negatively curved space.  We explain this  "Isocline- type property" in the  main part of  our  question 

[[A Picture explaining the question][2]][2]

**Main part of our  question:**

Let $M$  be  a  Riemannian surface with  a  foliation $\mathcal{F}$ whose  leaves  are  geodesics. We say that $\mathcal{F}$ has the local isocline property if for  every $x\in M$ there exist  a  geodesic $\alpha$ passing $x$ which is locally transverse to the foliation and intersects the leaves of the foliation with a  fix angel. We call such a  geodesic  $\alpha$ an isocline.

Let  $X=P\partial_x+Q\partial_y$ be a quadratic polynomial vector  field on $\mathbb{R}^2$.

Define the algebraic curve $C=\left\{(x,y)\mid yP-xQ=0\right\}$. Let $W$ be a vector field proportional to $x\partial_x+y\partial_y$.

Associated to the above quadratic vector field, we consider the  Riemannian metric on $M=\mathbb{R}^2\setminus C$ whose orthonormal frame is the following: 
$$\left\{\frac{x^2+y^2}{yP(x,y)-xQ(x,y)}V,\ W\right\}.$$

Then all orbit  of the  vector  field  are  geodesic of  the  metric  with the above frame. So  we have  a  foliation by geodesics  $\mathcal{F}$ on the  Riemannian surface $M$.

>Question: Is there an example  of  such situation with the  above Local isocline property? As a more general question: Is it true to say that every  foliation associated to a  quadratic  vector  field necessarily satisfies the  above  local property? (A possible positive  answer to the latter two questions is  an obstruction for  consideration of  limit  cycles  as  closed geodesics in negatively or  positively curved space.)   But if the  answer is  "No", can we choose  an  appropriate $W$ in the  above  frame  such that this local property fail at all point  $x\in M= \mathbb{R}^2\setminus C$?

**Note 1:**  The  above  definition of "Isocline  Local property" is  a  corrected version of  the  following post:

**Note 2:** With the same argument as the materials of this  post one can prove the following obvious observation:

**Obvious Fact:**

There is  no a  conformal Riemannian metric on the  punctured plane with non vanishing curvature with the following two properties:

1) The  orbits  of the  solutions  of the  Van der Pol equation $$\begin{cases}x'=y-(x^3-x)\\ y'=-x \end{cases}$$ are unparametrized geodesics.(Foliation by geodesics)

2)The positive $y$-axis  is a  geodesic.

Please  see this  MO  question as  a  related post: