How can I convert $\dfrac{(x – 1)^{2}}{16} +\dfrac{(y + 2)^{2}}{9} = 1$ to polar form? How can I convert  $\frac{(x – 1)^{2}}{16} + \frac{(y + 2)^{2}}{9} = 1$to polar form?  This is an ellipse centered at $(1, -2)$.
I understand the general approach for converting between the coordinate systems.  However, in this case I got stuck on handling the terms $9x^2 + 16 y^2$.  Please provide sufficient details.  Thanks.
 A: Put $$x-1=4\cos\theta$$ and $$y+2=3\sin\theta$$
Hence the required form is $$(x,y)\equiv (1+4\cos\theta,-2+3\sin\theta)$$ Hope that answers your question.
A: The straightforward method of converting from rectangular to polar coordinates produces
$$ \ 9 \ (x-1)^2 \ + \ 16 \ (y+2)^2 \ = \ 144 \ \ \Rightarrow \ \ 9x^2 \ - \ 18x \ + \ 9 \ + \ 16y^2 \ + \ 64y \ + \ 64 \ = \ 144 $$
$$\Rightarrow \ \ 9x^2 \ - \ 18x \ + \ 16y^2 \ + \ 64y \  = \ 71 $$
[basically, putting the ellipse equation into "general form" (for a conic section) first]
$$\rightarrow \ \ 9r^2 \ \cos^2 \theta \ - \ 18r \ \cos \ \theta \ + \ 16r^2 \ \sin^2 \theta \ + \ 64r \ \sin \  \theta \  = \ 71 $$
$$\Rightarrow \ \ r^2 \ ( \ 9 \ \cos^2 \theta \ + \ 16 \ \sin^2 \theta \ ) \ +  \ r \ ( \  64 \ \sin \ \theta \ - \ 18 \ \cos \ \theta \ )  \  = \ 71 \ \ . $$
This is a perfectly good polar equation.  Expressing this as an explicit function $ \ r \ = \ f(\theta) \ $ is not so easy to manage, however.
Graphing the polar equation above does produce the desired ellipse:

