I am trying a derivation of the polar form of an ellipse using vector notation. Beginning with a definition of an ellipse as the set of points in $\vec{R}^2$ for which the sum of the distances from two points is constant, I have
$|\vec{r_1}|+|\vec{r_2}| = c$
Thus, $|\vec{r_1}|^2+|\vec{r_1}||\vec{r_2}|=c|\vec{r_1}|$
ellipse diagram, Inductiveload on Wikimedia
Choosing a coordinate system (similar to the one in the diagram), such that
$\vec{r_1}=\vec{r}=x\vec{i}+y\vec{j}$, and
$\vec{r_2}=\vec{F_2 P}=(x-a)\vec{i}+y\vec{j}$
Noticing that $|\vec{r_1}||\vec{r_2}|=\frac{\vec{r_1}\cdot\vec{r_2}}{\cos(\theta)}=\frac{x^2-ax+y^2}{\cos(\theta)}=\frac{r^2-ax}{\cos(\theta)}$
and using $x=r\cos(\theta)$, we get
$r^2+\frac{r^2}{\cos(\theta)}-ar=cr$, thus I get a polar equation for an ellipse of
$r=\frac{a+c}{1+\sec(\theta)}$
whereas I expect $r=\frac{a(1-e^2)}{1+e\cos(\theta)}$. Please tell me where this argument takes the wrong turn!
Best Answer
I'm not sure yet what exactly you are doing yet, but here at least is an approach which works.
I think the issue is that you claim that $x=r \cos(\theta)$ and $\vec{r}_1 \cdot \vec{r}_2=r_1 r_2 \cos(\theta)$. It is not obvious to me that these angles would be the same.
Label the foci as $F$ and $F'$. Let $\vec{r}$ be the vector which points from $F$ to a point $p$ ont he ellipse. Let $\vec{r}'$ be the vector which points from $F'$ to $p$. Let $d$ be the distance between $F$ and $F'$. Let the $a$ represent the semi-major axis of this ellipse. By considering one of the points of the ellipse which lies on the major axis we can conclude that,
$$ r+r'=2a.$$
If $\theta$ is the angle between $\vec{r}$ and the line connecting $F$ to $F$' then we can use the law of cosines to write,
$$ r^2 + d^2+2rd\cos(\theta) = r'^2$$
$$ r^2 + d^2+2rd\cos(\theta) = (r-2a)^2$$
$$ \color{blue}{ r^2} + d^2+2rd\cos(\theta) = \color{blue}{ r^2}-4ar+4a^2$$
$$ d^2+2rd\cos(\theta) = -4ar+4a^2$$
$$ 2rd\cos(\theta) +4ar= 4a^2-d^2$$
$$ 2r(d\cos(\theta) +2a)= 4a^2-d^2$$
$$ r= \frac{a(1-e^2)}{e\cos(\theta) +1}$$