Determining the major/minor axes of an ellipse from general form
If you are looking for the length of the minor and major axis you can calculate $ r_{min} $ and $ r_{max} $ (see formulae below).
If you are trying to determine the bounding box, you can calculate the left-most, right-most, top-most and bottom-most points.
As far as the angle of rotation is concerned, I use the algorithm and formulae below.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form: $ a x^2 + b x y + c y^2 + d x + e y + f = 0 $.
The equation represents an ellipse if $ b^2 - 4 a c < 0 $ , or similarly, $ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
For convenience, I'll use the form with double terms,
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0.$$
First you need to center the ellipse with
$$A(x-x_c)^2+2B(x-x_c)(y-y_c)+C(y-y_c)^2+2D(x-x_c)+2E(y-y_c)+F=0.$$
The center is found by canceling the linear terms, wich gives
$$Ax_c+By_c=D,\\Bx_c+Cy_c=E.$$
The equation then reduces to
$$Ax^2+2Bxy+Cy^2=Ax_c^2+2Bx_cy_c+Cy_c^2-F=G.$$
In polar coordinates,
$$(A\cos^2(t)+2B\cos(t)\sin(t)+C\sin^2(t))r^2=G.$$
We want to find the extrema of $r$. They are also the extrema of the trigonometric factor and we determine them by cancelling the derivative.
$$-A\cos(t)\sin(t)+B(\cos^2(t)-\sin^2(t))+C\sin(t)\cos(t)=0$$ which we rewrite, by the double angle formulas
$$(C-A)\sin(2t)+2B\cos(2t)=0$$
which gives
$$\tan(2t)=\frac{2B}{A-C}.$$
The solution is $$t=\frac12\arctan\left(\frac{2B}{A-C}\right)+k\frac\pi2,$$
(giving the directions of the axis, which are orthogonal), and the semi-axis lengths are
$$r=\sqrt{\frac G{A\cos^2(t)+2B\cos(t)\sin(t)+C\sin^2(t)}}.$$
One can notice that the ellipse parameters are directly related to the Eigenvalues and Eigenvectors of the matrix
$$\left(\begin{matrix}A&B\\B&C\end{matrix}\right).$$
After diagonalization of the matrix, the equation of the ellipse becomes
$$\lambda x^2+\mu y^2=G,$$ which is of the well-known form
$$\left(\frac x{\sqrt{\frac G\lambda}}\right)^2+\left(\frac y{\sqrt{\frac G\mu}}\right)^2=1.$$