Analytical proof for Kepler's first law using cartesian co-ordinates Many detailed proofs are available for Kepler's first law, like this one.  But all of them use polar co-ordinates. There is one which doesn't even use differential equations. I'm looking for a proof which is done using cartesian co-ordinates.
For simplicity lets assume the Sun is at origin and the planet is located on the x-axis and is having a suitable velocity which is parallel to positive y-axis.
Background: I'm a high school physics teacher. My students are not familiar with polar co-ordinates, however they are comfortable with calculus
 A: Take the Sun(mass M) and the planet(mass m) to be point masses. Let the Sun be fixed at the origin and the planet be moving in the x-y plane, initial velocity of the planet be $v_o\hat{j}$ and initial position of the planet be $r_o\hat{i}$. At any given instant, let the planet's position, velocity and acceleration be $\vec{r}, \vec{v}, \vec{a}$ respectively. Let $\theta$ be the angle subtended between $\vec{r}$ and positive x-axis.
The gravitation force is always acting towards the origin, hence torque won't be generated on the planet, about the origin.
Therefore, angular momentum of the planet should be conserved about the origin.
$$\vec{r}\times\vec{p}=r_omv_o\hat{k}$$
$$\vec{r}\times\vec{v}=r_ov_o....(1)$$
This can also be written as 
$$I\vec{\omega}=r_omv_o\hat{k}$$
$$mr^2{\frac{d\vec{\theta}}{dt}}=r_omv_o\hat{k}$$
$$r^2\frac{d{\theta}}{dt}\hat{k}=r_ov_o\hat{k}$$
$$r^2\frac{d{\theta}}{dt}=r_ov_o....(2)$$ 
From Newton's law of Gravitation,
$$\vec{F}=\frac{-GMm}{r^3}\vec{r}$$
From Newton's second law of motion,
$$\vec{F}=m\vec{a}$$
$$m\vec{a}=\frac{-GMm}{r^3}\vec{r}$$
$$\frac{d\vec{v}}{dt}=\frac{-GM}{r^3}\vec{r}$$
Multiplying and diving Left Hand Side by $d\theta$ and substituting
$$\vec{r}=r cos\theta\hat{i}+r sin\theta\hat{j}~(where~r^2=x^2+y^2,cos\theta=\frac{x}{r}~and~sin\theta=\frac{y}{r})$$
$$\frac{d\vec{v}}{d\theta}.\frac{d\theta}{dt}= -\frac{-GM(r cos\theta\hat{i}+r sin\theta\hat{j})}{r^3}$$
$$\frac{d\vec{v}}{d\theta}.r^2\frac{d\theta}{dt}=-GM(cos\theta\hat{i}+sin\theta\hat{j})$$
From (2), 
$$\frac{d\vec{v}}{d\theta}.r_ov_o=-GM(cos\theta\hat{i}+sin\theta\hat{j})$$
$$Let ~~~\alpha=\frac{GM}{r_ov_o}$$
$$\int_{\vec{v_o}}^{\vec{v}}\vec{dv}=-\alpha(\int_{0}^{\theta}cos\theta.d\theta.\hat{i}+\int_{0}^{\theta}sin\theta.d\theta.\hat{j})$$
$$\vec{v}-\vec{v_o}=-\alpha (sin\theta\hat{i}-cos\theta\hat{j})\Big|_0^{\theta}$$
$$\vec{v}-\vec{v_o}=-\alpha (sin\theta\hat{i}-(1-cos\theta\hat{j}))$$
$$\vec{v}=-\alpha sin\theta\hat{i}+( \alpha cos\theta-\alpha+v_o)\hat{j}$$
Now,
$$\vec{r}\times\vec{v}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\rcos\theta&rsin\theta&0\\-\alpha sin \theta & \alpha cos\theta+v_o-\alpha&0\end{vmatrix}$$
$$=(rcos\theta)(\alpha cos\theta+v_o-\alpha)-(-\alpha sin\theta)(sin\theta)\hat{k}$$
$$=(r)(\alpha cos^2\theta+\alpha sin^2\theta-\alpha cos\theta+v_ocos\theta)\hat{k}$$
$$=(r)(\alpha+v_ocos\theta-\alpha cos\theta)$$
From (1),
$$r_ov_o\hat{k}=(r)(\alpha+(v_o-\alpha)cos\theta)$$
$$r=\frac{r_ov_o}{\alpha(1+(\frac{v_o-\alpha}{\alpha})cos\theta)}$$
$$Let ~\frac{r_0v_0}{\alpha}=h~~~and ~~~\frac{v_o-\alpha}{\alpha}=p$$
$$Then~~~r=\frac{h}{1+pcos\theta}$$
$$Substituting~~~cos\theta=\frac{x}{r},$$
$$r(1+p\frac{x}{r})=h$$
$$r+px=h$${\tiny }
$$r^2=(h-px)^2$$
$$x^2+y^2=h^2+p^2x^2-2hpx$$
$$x^2(1-p^2)+2hpx+y^2=h^2$$
$$When~~~~~(1-p^2)\neq0,$$
$$x^2+\frac{y^2}{1-p^2} +\frac{2hpx}{1-p^2}=\frac{h^2}{1-p^2}$$
$$Adding ~~~\frac{h^2}{(1-p^2)^2} ~~~on ~both~ sides,$$
$$x^2+\frac{h^2p^2}{(1-p^2)^2}+\frac{y^2}{1-p^2} +\frac{2hpx}{1-p^2}=\frac{h^2}{1-p^2}+\frac{h^2p^2}{(1-p^2)^2}$$
$$(x+\frac{hp}{1-p^2})^2+\frac{y^2}{1-p^2}=\frac{h^2}{(1-p^2)^2}$$
This takes the form
$$\frac{(x+\frac{hp}{1-p^2})^2}{\frac{h^2}{(1-p^2)^2}}+\frac{y^2}{\frac{h^2}{(1-p^2)}}=1$$
If$(1-p^2)>0$, then the equation will take the form of a shifted ellipse. $\frac{(x+x_o)^2}{a^2}+\frac{y^2}{b^2}=1$
$$1-p^2>0$$
$$p^2-1<0$$
$$(p-1)(p+1)<0$$
$$-1<p<1$$
$$-1<\frac{v_o^2}{(\frac{GM}{r_o})}-1<1$$
$$0<\frac{v_o^2}{(\frac{GM}{r_o})}<2$$
$$0<v_o^2<\frac{2GM}{r_o}$$
$$0<v_o<\sqrt{\frac{2GM}{r_o}}$$
Hence for a suitable velocity, the planet will be orbiting Sun in an elliptical path.
A: Indeed, as already suggested by other answers, it is straightforward to transform the ellipse from Cartesian co-ordinates to polar co-ordinates even for students not familiar with polar co-ordinates.
Once you have expressed the ellipse in terms of r and , the attached figure proves, using only the most elementary calculus in Cartesian coordinates, that the ellipse is a solution. The only assumptions are Newtonian (square law) gravity and conservation of angular momentum.

