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!


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,

$$ r^2 + d^2+2rd\cos(\theta) = r'^2$$

$$ r^2 + d^2+2rd\cos(\theta) = (r-2a)^2$$

$$ \color{blue}{ r^2} + d^2+2rd\cos(\theta) = \color{blue}{ r^2}-4ar+4a^2$$

$$ d^2+2rd\cos(\theta) = -4ar+4a^2$$

$$ 2rd\cos(\theta) +4ar= 4a^2-d^2$$

$$ 2r(d\cos(\theta) +2a)= 4a^2-d^2$$

$$ r= \frac{a(1-e^2)}{e\cos(\theta) +1}$$

  • $\begingroup$ Yes you're quite right I have two different angles with the same name! $\endgroup$
    – Arty
    Feb 19 '17 at 15:55
  • $\begingroup$ How did you get the equality where r is highlighted. shouldn't the right hand side be $r^2-4ar+4a^2$ $\endgroup$
    – excalibirr
    Dec 1 '18 at 17:46
  • $\begingroup$ @exodius, I think you are correct that I made a typo. Unfortunately I don't have the time to make the appropriate edit right now. Thank you for catching the error. $\endgroup$
    – Spencer
    Dec 3 '18 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.