How to obtain *A*, *B* and *C* from the equation of and oblique ellipse

Question

I have the equation of an ellipse centred at the origin and inclined to the coordinate axes: $$ \frac{(x\cos\theta + y\sin\theta)^2}{a^2} + \frac{(y\cos\theta - x\sin\theta)^2}{b^2} = 1 $$ In order to find the rotation angle I know that: $$ \tan(2\theta)=\frac B{A-C}\ $$ Now my problem is to obtain the forms of A, B and C for the latter equation, from the former. Any help will be very welcome.

Answer 1

All you need is to explicitly write your first equation by expanding the squares and grouping terms for $x'^2$, $y'^2$, and $x'y'$. Their coefficients are $A$, $C$, and $B$. You will be surprised to see $$\tan(2\theta)=\frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}$$ :)

Answer 2

In the world coordinates you have $$A x^2 + B y^2 + C x y - 1 = 0$$ and you want to find the parameters of the ellipse in a coordinate system that is aligned with the major/minor axis. We rotate the local coordinated by $\theta$ with:

$$ x = x' \cos\theta -y' \sin\theta \\ y = x' \sin\theta + y' \cos\theta $$

If you plug it in the ellipse equations you will find from the terms multiplying $x y$ that $$ \left. 2 C \cos^2 \theta + 2 (B-A) \sin \theta \cos\theta - C =0 \right\}\\ \left. C \cos(2 \theta)+(B-A) \sin(2 \theta) = 0 \right\}\\ \tan (2 \theta) = \frac{C}{A-B} $$

Use this angle to get the aligned ellipse

$$ \left( \tfrac{\sqrt{ (A-B)^2+C^2}+A+B}{2} \right) x'^2 + \left( \tfrac{-\sqrt{(A-B)^2+C^2}+A+B}{2} \right) y'^2 = 1 $$