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I have an ellipse with polar form equation: $$r=\frac{a\left(1-e^{2}\right)}{1+e\cos\left(\theta\right)}$$ where $e$ is eccentricity, and $a$ is semi-major axis. How do I convert this from polar form to canonical form $\frac{x^2}{b^2}+\frac{y^2}{c^2}=1?$

I tried to use $b$ = $a(1-e^2)$ and $c = 1+e \cos{\theta}$, but that didn't work, because I don't know what $\theta$ is. Is there a solution to this?

Blue
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WarpPrime
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  • Do you know the standard polar-to-cartesian conversion rules? eg, $r=\sqrt{x^2+y^2}$, $\cos\theta =\cdots$? – Blue Oct 14 '20 at 02:08
  • @Blue I am not familiar with these rules. – WarpPrime Oct 14 '20 at 02:15
  • See here, for instance. ... How have you come upon this particular problem? Not knowing the rules suggests that this isn't a textbook exercise, and/or that you may not be at all familiar with polar coordinates. Saying more about your own skill level here will help people tailor their responses to best serve you, while avoiding wasting time (theirs or yours) explaining things you do know or using techniques that are currently beyond your level. – Blue Oct 14 '20 at 02:24

1 Answers1

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

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

$$r^2=(a\left(1-e^{2}\right)-ex)^2$$

$$x^2+y^2=(a\left(1-e^{2}\right)-ex)^2$$

$$y^2=-(1-e^2)((x+ae)^2-a^2)$$

$$a^2(1-e^2)(\frac{(x+a e)^2}{a^2}+\frac{y^2}{a^2(1-e^2)}-1)=0.$$

Jan-Magnus Økland
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