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} $
Best Answer
The general equation of an ellipse is: $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ if: $$4AC - B^2 > 0$$ The trick is to eliminate B so that the xy term vanishes.
If $B<>0$ then the ellipse is rotated and the angle of rotation is obtained from: $$tan(2 \theta) = \frac {B}{A-C}$$ $0 < \theta < \frac {\pi}{4}$
$$\cos {\theta} = \sqrt{\frac{1 + \cos{2 \theta}}{2}}$$ $$\sin{\theta} = \sqrt{\frac{1 - cos{2 \theta}}{2}}$$ Determine the new equation of the ellipse by calculating the following coefficients:
$A' = A \cos^2{\theta} + B \cos{\theta} \sin{\theta} + C \sin^2{\theta}$
$B' = 0$
$C' = A \sin^2{\theta} - B \cos{\theta} \sin{\theta} + C \cos^2{\theta}$
$D' = D \cos{\theta} + E \sin{\theta}$
$E' = -D \sin{\theta} + E \cos{\theta}$
$F' = F$
The resulting equation: $A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0$
After writing this equation in the form: $$ \frac{(x'-x'_0)^2}{a^2} + \frac{(y'-y'_0)^2}{b^2} = 1$$
We get: $$x'_0 = \frac{-D'}{2A'}$$ $$y'_0 = \frac{-E'}{2C'}$$
$$a^2 = \frac{-4F'A'C'+C'D'^2+A'E'^2}{4A'^2C'}$$ $$b^2 = \frac{-4F'A'C'+C'D'^2+A'E'^2}{4A'C'^2}$$
The coordinates of the centerpoint are found by rotating back about angle $\theta$
$x_0 = x'_0 \cos{\theta} - y'_0 \sin{\theta}$
$y_0 = x'_0 \sin{\theta} + y'_0 \cos{\theta}$
Cheers!