For example,
- Circle ($\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$) : $$ \left\{ \begin{array}{c} x\to a\cos \theta \\ y\to a \sin \theta \\ \end{array} \right. $$
- Oval ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$) : $$ \left\{ \begin{array}{c} x\to a\cos \theta \\ y\to b \sin \theta \\ \end{array} \right.$$
Any other equations containing $x$,$y$,$x^2$, and $y^2$ can be substituted to trigonometric equations like this, with some translations.
However, I can't find out any way to do this on equations containing $xy$ terms.
For example, $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$, where $C \ne 0.$ How can I do that? Thanks!
I will use the slightly more standard $ax^2+2bxy+cy^2+2dx+2ey+f=0,$ assumed to be an ellipse. Its center is $(\frac{cd-eb}{b^2-ac},\frac{ae-bd}{b^2-ac})$ and translating this to the origin we get
$$ax'^2+2\,bx'y'+cy'^2={\frac {-cd^2+acf-ae^2-fb^2+2deb}{{b}^{2}-ac}}$$ Rotating through the angle $\theta'=\arctan{(\frac{-c+a+\sqrt{c^2-2ac+a^2+4b^2}}{2b})}$ we get $${\frac {-4\,c{b}^{2}-2\,{b}^{2}\sqrt {{c}^{ 2}-2\,ac+{a}^{2}+4\,{b}^{2}}-{c}^{3}+2\,a{c}^{2}-{c}^{2}\sqrt {{c}^{2} -2\,ac+{a}^{2}+4\,{b}^{2}}-c{a}^{2}+ac\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\, {b}^{2}}}{-4\,{b}^{2}-{c}^{2}+2\,ac-c\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{ b}^{2}}-{a}^{2}+a\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2}}}}x''^2+{\frac {4\,c{b}^{2}-2\,{b}^{2}\sqrt {{c}^{2}-2 \,ac+{a}^{2}+4\,{b}^{2}}+{c}^{3}-2\,a{c}^{2}-{c}^{2}\sqrt {{c}^{2}-2\, ac+{a}^{2}+4\,{b}^{2}}+c{a}^{2}+ac\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^ {2}}}{4\,{b}^{2}+{c}^{2}-2\,ac-c\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2 }}+{a}^{2}+a\sqrt {{c}^{2}-2\,ac+{a}^{2}+4\,{b}^{2}}}}y''^2 -{\frac {-cd^2+acf-ae^2-fb^2+2deb}{{b}^{2}-ac}}=0$$ This can now be put into the form $\frac{x''^2}{a''^2}+\frac{y''^2}{b''^2}=1.$
In an example: $-5x^2+4xy+7x-4y^2-4y+24=0$ is an ellipse with center $(\frac58,-\frac3{16}).$ The method yields $-{\frac {16}{425}}\,{\frac {{x''}^{2} \left( 68-4\,\sqrt {17} \right) }{ -17-\sqrt {17}}}-{\frac {16}{425}}\,{\frac {{y''}^{2} \left( -68-4\, \sqrt {17} \right) }{17-\sqrt {17}}}-1$ or $a''={\frac {85}{32}}+{\frac {5}{32}}\,\sqrt {17}$ $b''={\frac {85}{32}}-{\frac {5}{32}}\,\sqrt {17}$.
Going back through the substitutions, $x''=\cos{\theta'}x'+\sin{\theta'}y',y''=-\sin{\theta'}x'+\cos{\theta'}y'$ and $x'=x-\frac58$,$y'=y+\frac3{16}$ we get $$\begin{array}{cc} x=&\frac58-{\frac {5}{16}}\,\sqrt {2}\sin \left( t \right) \sqrt {17-\sqrt { 17}}+{\frac {5}{16}}\,\cos \left( t \right) \sqrt {34+2\,\sqrt {17}} \\y=& -\frac3{16}+{\frac {5}{1088}}\,\sqrt {17} \left( -\sqrt {2}\sqrt {17}\sin \left( t \right) \sqrt {17-\sqrt {17}}+17\,\sqrt {2}\sin \left( t \right) \sqrt {17-\sqrt {17}}+\cos \left( t \right) \sqrt {34+2\, \sqrt {17}}\sqrt {17}+17\,\cos \left( t \right) \sqrt {34+2\,\sqrt {17 }} \right). \end{array}$$