Kepler's first law Consider a particular solution curve of the differential equation. We have two constants of the motion for this system, namely, the angular momentum $\ell$ and total energy $E$. We assume $\ell \neq 0$. We will show that in polar coordinates in configuration space, a solution with nonzero angular momentum lies on a curve given by $r(1+ \epsilon\cos(\theta))=κ$, where $\epsilon$ and $κ$ are constants. This equation defines a conic section, as can be seen by rewriting this equation in Cartesian coordinates. This fact is known as Kepler’s first law.
my question is as follows:
To prove this, recall that $\ell = mr^{2}\theta^{\prime}$ is constant and nonzero.
(1) Why the sign of $\theta$ remains constant along each solution curve?
From this, it is clear that  $\theta$ is always increasing or always decreasing in time.
(2) Why we may also regard $r$ as a function of $\theta$ along the curve?
 A: $$\vec{M}=\vec{r}\times\vec{p}\Rightarrow \vec{M}\perp \vec{r},\vec{M}={\rm \ const} 
\Rightarrow \vec{r} {\rm\ lies\ in\ plane}$$
$$x=r\cos\varphi,\,y=r\sin\varphi$$
$$M=M_z=(\vec{r}\times\vec{p})_z=xp_y-yp_x=m(x\dot{y}-y\dot{x})=
mr^2\dot{\varphi}$$
$$E=\frac{m(\dot{x}^2+\dot{y}^2)}{2}+U(r)=\frac{m}{2}\left(\dot{r}^2+r^2\dot{\varphi}^2\right)+U(r)$$
$$E=\frac{m\dot{r}^2}{2}+\frac{M^2}{2mr^2}+U(r)$$
$$U=-\frac{\alpha}{r}, \alpha > 0$$
$$E=\frac{m\dot{r}^2}{2}+\frac{M^2}{2mr^2}-\frac{\alpha}{r}$$
$$\dot{r}^2=\frac{2}{m}\left(E+\frac{\alpha}{r}-\frac{M^2}{2mr^2}\right)$$
$$\frac{d\varphi}{dr}=\frac{\dot{\varphi}}{\dot{r}}=\frac{\frac{M}{mr^2}}{\sqrt{\frac{2}{m}\left(E+\frac{\alpha}{r}-\frac{M^2}{2mr^2}\right)}}$$
$$\varphi=\varphi_0+\frac{M}{\sqrt{2m}}
\int_{r(\varphi_0)}^{r(\varphi)}
\frac{dr}{r^2\sqrt{E+\frac{\alpha}{r}-\frac{M^2}{2mr^2}}}$$
$$p=\frac{M^2}{m\alpha}, e=\sqrt{1+\frac{2EM^2}{m\alpha^2}}$$
$$\varphi=\varphi_0+\int_{r(\varphi_0)}^{r( 
\varphi)}\frac{dr}{r\sqrt{\left(\frac{e 
r}{p}\right)^2-\left(1-\frac{r}{p}\right)^2}}= 
\varphi_0+\arccos\frac{p-r}{er}-\arccos\frac{p-r_0}{er_0}$$
$${\rm Let\ choose\ polar\ axis\ direction\ such\ that\ }\varphi_0-\arccos\frac{p-r_0}{er_0}=0$$
$$r=\frac{p}{1+e\cos\varphi}$$
