Finding the parametric equation of an ellipse in non-general form I have an equation of an ellipse in the form:
$(x-1)^2 + (y-1)^2 + 2axy=1$ where $-1<a<1$
I want to know how I could find a parametric equation of this ellipse. I have tried the old "complete the square" method but can't seem to get an answer.
Any help would be appreciated. Thanks!
 A: If polar coordinates is what you want:
$x^2+y^2+2axy-2x-2y+1=0$
$r^2+2ar^2\cos{\theta}\sin{\theta}-2r\cos{\theta}-2r\sin{\theta}+1=0$
$r^2(1+a\sin{2\theta})-2r(\cos{\theta}+\sin{\theta})+1=0$
$r=\frac{(\cos{\theta}+\sin{\theta})\pm\sqrt{(1-a)\sin{2\theta}}}{1+a\sin{2\theta}} $

A: The equations have non-zero cross terms (unless $a=0$) and are symmetric in $x$ and $y$, so the ellipses are rotated 45° from standard position. Their centers can be computed by taking partial derivatives and setting them to zero: $$\begin{align}2(x+ay-1)&=0\\2(ax+y-1)&=0\end{align}$$ with solution $\left(\frac1{1+a},\frac1{1+a}\right)$. You can then translate and rotate to compute the semi-axis lengths as ${\sqrt{1-a}\over1+a}$ and $\frac1{\sqrt{1+a}}$ for the axes $(1,1)$ and $(-1,1)$, respectively (or compute them directly from the eigenvalues and determinants of the associated matrix of the equation). A sense of symmetry prompts me to express the second semi-axis length as ${\sqrt{1+a}\over1+a}$.  
With this information you can construct a parameterization of these ellipses directly: $$p(t)=\left(\frac1{1+a},\frac1{1+a}\right)+{\sqrt{1-a}\over1+a}\cos t\,\left({1\over\sqrt2},{1\over\sqrt2}\right)+{\sqrt{1+a}\over1+a}\sin t\,\left(-{1\over\sqrt2},{1\over\sqrt2}\right)$$ or $$\begin{align}x &= {1\over\sqrt2(1+a)}\left(\sqrt2+\sqrt{1-a}\cos t-\sqrt{1+a}\sin t\right) \\ y &= {1\over\sqrt2(1+a)}\left(\sqrt2+\sqrt{1-a}\cos t+\sqrt{1+a}\sin t\right). \end{align}$$
