= Solution
Multiplying the polar equation by its denominator gives $Ar-Bx=1$. Squaring and using $r^2=x^2+y^2$ yields
$$
(A^2-B^2)x^2+A^2y^2-2Bx=1.
$$
Take $C=\sqrt{A^2-B^2}>0$. <Completing the square> gives
$$
C^2\left(x-\frac{B}{C^2}\right)^2+A^2y^2
=1+\frac{B^2}{C^2}=\frac{A^2}{C^2}.
$$
The standard <ellipse> equation is consequently
$$
\boxed{\frac{(x-B/C^2)^2}{A^2/C^4}+\frac{y^2}{1/C^2}=1},
$$
whose centre is $\boxed{(B/C^2,0)}$. Squaring has introduced no unwanted branch: on this <ellipse> the minimum of $1+Bx$ is $A/(A+B)>0$, so $Ar=1+Bx$ follows with the correct positive sign.
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