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?

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