Binnet equation, $$\frac{d^2}{d\theta^2}\left(\frac{1}{r}\right)+\frac 1r=-\frac{m}{L^2}F(r)r^2,$$ where $m$ and $L^2$ are positive constants, can be used to give the orbit $r(\theta)$ of a particle in a central force field. The condition $L^2\neq 0$ implies that the orbit is a strictly increasing function.
Let $r_1$ be a maximum or a minimum value of $r(\theta)$ such that $r(\theta_1)=r_1$. This point, $(r_1,\theta_1)$, is called an apsis. Then I read the following sentence from a book:
From Binnet equation's form it follows that the orbit is symmetric in $\theta$ about the segment $OA_1$ going from the origin to the apsis $A_1$.
I understand the meaning of this sentence but I don't understand why it is true. What specifically in Binnet equation leads to this symmetry about $OA_1$? I see that Binnet equation is invariant under $\theta\rightarrow -\theta$ and $\theta\rightarrow \theta+\theta_0$. I still don't see how these could explain the sentence above. I would appreciate very much a non mathematicians' answer.
I think I got the answer.
Binet equation is invariant under $\theta\rightarrow -\theta$. But only by defining $\theta=0$ at a turning point $A_1=(r_1,\theta_1=0)$ the solution can be reflected about the segments $OA_1$. This is so because the initial conditions $$r(\theta_1=0)=r_1,\quad r'(\theta_1=0)=0,$$ are also invariant under the transformation. The corresponding about any other segment $OA$, which is not an apsis, will not happen in general, because $$r'(\theta=0)=v_\theta\neq 0,$$ may not be invariant. Thanks for @AloneAndConfused for pointing out that the answer was indeed int the reflection symmetry of the equation.