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

Question

$r(θ) = a(1 − β^2)/(1 + β \cos θ)$

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?

Answer 1

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 $c=\sqrt{a^2-b^2}$.