Yes, and it was given in any text book dealing with the conic sections that I have ever seen: If the equation of an ellipse centered at (0, 0) is [tex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/tex] then the same ellipse, centered at (a, b) has equation [tex]\frac{(x- a)^2}{a^2}+ \frac{(y-b)^2}{b^2}= 1[/tex]
Insert polar representations for "x" and "y", multiply out parentheses and simplify and redefine variables/constants. In particular, remember simplifying trig identities, such as, for example: [tex]2\sin^{2}\theta=1-\cos(2\theta)[/tex]
Now, HallsofIvy made out a special case, with the center with the same values as the lengths of the semi-axes. You shouldn't make that restriction here (call one of the (a,b)-pairs (c,d)-for example). To give you the first step on your way, multiplying up and out, we get (with (c,d) centre coordinates): [tex]b^{2}r^{2} \cos^{2}\theta+a^{2}r^{2} \sin^{2}\theta-2b^{2}cr \cos\theta-2a^{2} dr\sin\theta=a^{2}b^{2}-c^{2}-d^{2}[/tex] There would be various ways to simplify this expression further, and redefing independent constants. One very compact way of doing so would be to transform your equation into the following form: [tex]Ar^{2}\cos\gamma+Br\sin\phi=C[/tex] where the angle "phi" is a phase-shifted version of "gamma"/2 with a fourth constant D to be determined along with A, B and C (gamma being twice the value of "theta")