Rotated Ellipse It is well known that the equation
$$\frac {(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac {(x\sin\alpha-y\cos\alpha)^2}{b^2}=1\tag{1}$$
(where $\beta\neq\alpha$) represents an ellipse centred at the origin with semimajor/minor axes $a,b$, and rotated by $\alpha$. 

Question
The equation 
  $$\frac {(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac {(x\sin\beta-y\cos\beta)^2}{b^2}=1\tag{2}$$
  represents a rotated ellipse centred at the origin, but its semimajor/minor axes are no longer $a,b$. 
How can this be transformed into a form similar to $(1)$, such that the semimajor/minor axes and angle of rotation can be easily determined?

 A: If you rewrite your equation as
$$
ux^2 + vy^2 +2wxy = 1
$$
then the squared reciprocal lengths of the semimajor/semiminor axes of the ellipse are the eigenvalues of the matrix
$$
\begin{pmatrix} u & w \\ w & v \end{pmatrix}
$$
You get
$$
\begin{eqnarray*}
u & = & \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \beta}{b^2} \\
v & = & \frac{\sin^2 \alpha}{a^2} + \frac{\cos^2 \beta}{b^2} \\
w & = & \frac{\cos\alpha\sin\alpha}{a^2} - \frac{\cos\beta\sin\beta}{b^2}
\end{eqnarray*}
$$
The eigenvalues of the matrix are
$$
\lambda_{1,2} = \frac{u+v}{2} \pm \sqrt{\left(\frac{u-v}{2}\right)^2+w^2}
$$
Using 
$$
\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \\
\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \\
\cos^2\alpha+\sin^2\alpha = 1
$$
this simplifies a lot:
$$
\lambda_{1,2} = \frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)
\pm\frac{1}{2}\sqrt{\frac{1}{a^4} - \frac{2\cos(2\alpha-2\beta)}{a^2b^2}
+\frac{1}{b^4}}
$$
The length of the semimajor axis is $1/\sqrt{\lambda_2},$ and the length of the semiminor axis is $1/\sqrt{\lambda_1}.$
