1
$\begingroup$

On an old wikipedia page describing the ellipse, we have the following formula that parametrise an ellipse (semi-axis $a,b$, tilt $\phi$ and center $(r_0, \theta_0)$ inside the ellipse) :

$$r(\theta) = \frac{P(\theta)+Q(\theta)}{R(\theta)}$$ where

$$P(\theta) = r_0[(b^2-a^2)\cos(\theta+\theta_0-2\phi) + (a^2+b^2)\cos(\theta-\theta_0)]$$

$$R(\theta) = (b^2-a^2)\cos(2\theta-2\phi)+a^2+b^2$$

$$Q(\theta) =\sqrt{2}ab\sqrt{R(\theta)-2r_0^2 \sin^2(\theta-\theta_0)}$$

To what $P, Q$ and $R$ correspond geometrically and how to derive this expression?

$\endgroup$
2
  • $\begingroup$ It seems the expression is incorrect, I got $R(\theta)$ agree with it, but I got for $$P(\theta) = r_0[(b^2-a^2)\cos(\theta+\theta_0-\phi) + (a^2+b^2)\cos(\theta-\theta_0-\phi)]$$ and $$Q(\theta) =\sqrt{2}ab\sqrt{R(\theta)-2r_0^2 \sin^2(\theta-\theta_0-\phi)}$$ Can anyone check this result? $\endgroup$
    – MathFail
    Jun 2, 2023 at 10:50
  • $\begingroup$ @MathFail, interesting, how did you obtain those formulas though? $\endgroup$
    – edamondo
    Jun 2, 2023 at 14:52

1 Answer 1

1
+50
$\begingroup$

Let us first consider an ellipse $E$ whose center is the origin $O$.

enter image description here

Let $F_1,F_2$ be the foci. Let $P$ be a point on $E$. Let $\alpha=\angle{F_2F_1P}$.

Let $c=\sqrt{a^2-b^2}$. Applying the law of cosines to $\triangle{F_1F_2P}$ and using $F_1P+F_2P=2a$, we have $$(2a-F_1P)^2=F_1P^2+(2c)^2-2\cdot F_1P\cdot 2c\cos\alpha$$ i.e. $$F_1P=\frac{b^2}{a-c\cos\alpha}$$

Considering the rotation in complex plane with $F_1(-c\cos\phi,-c\sin\phi)$, $$P_x+iP_y-(-c\cos\phi-ci\sin\phi)=\bigg(0-(-c\cos\phi-ci\sin\phi)\bigg)(\cos\alpha+i\sin\alpha)\cdot\frac{F_1P}{F_1O}$$ gives $$P_x=\frac{b^2\cos(\phi+\alpha)}{a-c\cos\alpha}-c\cos\phi,\qquad P_y=\frac{b^2\sin(\phi+\alpha)}{a-c\cos\alpha}-c\sin\phi$$

Now, let us consider the ellipse $E'$ whose center is $(r_0\cos\theta_0,r_0\sin\theta_0)$.

A point on $E'$ is given by

$$r\cos\theta=P_x+r_0\cos\theta_0=\frac{b^2\cos(\phi+\alpha)}{a-c\cos\alpha}-c\cos\phi+r_0\cos\theta_0\tag1$$ $$r\sin\theta=P_y+r_0\sin\theta_0=\frac{b^2\sin(\phi+\alpha)}{a-c\cos\alpha}-c\sin\phi+r_0\sin\theta_0\tag2$$

Now, let us eliminate $\alpha$.

Let $$A=r\cos\theta+c\cos\phi-r_0\cos\theta_0$$ $$B=r\sin\theta+c\sin\phi-r_0\sin\theta_0$$

$(1)(2)$ can be written as

$$(-Ac-b^2\cos\phi)\cos\alpha+b^2\sin\phi\sin\alpha=-Aa\tag3$$

$$(-cB-b^2\sin\phi)\cos\alpha-b^2\cos\phi\sin\alpha=-aB\tag4$$

Solving $(3)(4)$ for $\cos\alpha,\sin\alpha$ gives

$$\cos\alpha=\frac{Aax+Bay}{Acx+cBy+b^2},\qquad \sin\alpha=\frac{Bax-Aay}{Acx+cBy+b^2}$$

So, we can eliminate $\alpha$ to have $$\bigg(\frac{Aax+Bay}{Acx+cBy+b^2}\bigg)^2+\bigg(\frac{Bax-Aay}{Acx+cBy+b^2}\bigg)^2=1$$ i.e. $$a^2(Ax+By)^2+a^2(Bx-Ay)^2-(Acx+cBy+b^2)^2=0$$ i.e. $$a^2(A^2+B^2)-(Acx+cBy+b^2)^2=0$$ which can be written as $$Cr^2+Dr+G=0\tag5$$ where

$$\begin{align}C&=a^2-c^2(x\cos\theta+y\sin\theta)^2 \\\\&=a^2-(a^2-b^2)\cos^2(\theta-\phi) \\\\&=\frac{b^2-a^2}{2}\cos(2\theta-2\phi)+\frac{a^2+b^2}{2}\end{align}$$ and $$\begin{align}D&=2a^2\bigg((c\cos\phi-r_0\cos\theta_0)\cos\theta+(c\sin\phi-r_0\sin\theta_0)\sin\theta\bigg) \\&\quad -2c\cos(\theta-\phi)\bigg(cx(c\cos\phi-r_0\cos\theta_0)+cy(c\sin\phi-r_0\sin\theta_0)+b^2\bigg) \\\\&=2a^2\bigg(c\cos(\theta-\phi)-r_0\cos(\theta-\theta_0)\bigg)-2c\cos(\theta-\phi)\bigg(a^2-cr_o\cos(\theta_0-\phi)\bigg) \\\\&=2r_0\bigg(-a^2\cos(\theta-\theta_0)+c^2\cos(\theta_0-\phi)\cos(\theta-\phi)\bigg) \\\\&=r_0\bigg(-2a^2\cos(\theta-\theta_0)+(a^2-b^2)\cos(\theta_0+\theta-2\phi)+(a^2-b^2)\cos(\theta-\theta_0)\bigg) \\\\&=-r_0\bigg((b^2-a^2)\cos(\theta+\theta_0-2\phi)+(a^2+b^2)\cos(\theta-\theta_0)\bigg)\end{align}$$

and

$$\begin{align}G&=a^2\bigg((c\cos\phi-r_0\cos\theta_0)^2+(c\sin\phi-r_0\sin\theta_0)^2\bigg) \\&\quad -(c^2\cos^2\phi-cr_0\cos\phi\cos\theta_0+c^2\sin^2\phi-cr_0\sin\phi\sin\theta_0+b^2)^2 \\\\&=a^2\bigg(c^2+r_0^2-2cr_0\cos(\theta_0-\phi)\bigg)-\bigg(a^2-cr_0\cos(\theta_0-\phi)\bigg)^2 \\\\&=b^2r_0^2-a^2b^2+c^2r_0^2\sin^2(\theta_0-\phi)\end{align}$$

Solving $(5)$ for $r$ and choosing $\color{red}+$, we get $$r=\frac{-D\color{red}+\sqrt{D^2-4CG}}{2C}$$ which can be written as $$r=\frac{P+Q}{R}$$ as the wikipedia page says.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .