# How can we show polar coordinates r(theta) is an ellipse?

$r(θ) = a(1 − β^2)/(1 + β \cos θ)$

and I want to show this $r(θ)$ is an ellipse described by

$\dfrac{(x+\sqrt{a^2 − b^2})^2}{a^2}+\dfrac{y^2}{b^2}= 1$, when $0<β<1$.

How can we show this?

## 1 Answer

With the relations $$r=\sqrt{x^2+y^2}, r\cos\theta=x$$ we can rewrite the equation as $$r=\frac{ar(1-\beta^2)}{r+\beta x}$$ or, equivalently (disregarding $r=0$ that's not a solution), $$r+\beta x=a(1-\beta^2)$$ that becomes $r=a(1-\beta^2)-\beta x$; now square and get $$x^2+y^2=a^2(1-\beta^2)^2-2a(1-\beta^2)\beta x+\beta^2x^2$$ Reorder: $$x^2(1-\beta^2)+2a(1-\beta^2)\beta x+y^2=a^2(1-\beta^2)^2$$ Divide everything by $1-\beta^2$: $$x^2+2a\beta x+\frac{y^2}{1-\beta^2}=a^2(1-\beta^2)$$ Complete the square: $$x^2+2a\beta x+a^2\beta^2+\frac{y^2}{1-\beta^2}=a^2$$ Set $c=a\beta$ and $a^2(1-\beta^2)=b^2$: $$(x+c)^2+\frac{a^2}{b^2}y^2=a^2$$ Divide by $a^2$: $$\frac{(x+c)^2}{a^2}+\frac{y^2}{b^2}=1$$ Note that $c=\sqrt{a^2-b^2}$.

• I think you are wrong my friend. Your last c doesn't match with first one! Read my answer. Your relation between polar and Cartesian parameters are wrong. – A.F.23 Oct 10 '15 at 14:03
• @A.F.23 There was a wrong division. The relation between polar and cartesian coordinates is of course correct. – egreg Oct 10 '15 at 14:16
• Yes, I thought two equations are in two coordinate systems and we should use $x=r\cos\theta-f$. Sorry. – A.F.23 Oct 10 '15 at 18:47