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
Hint Let $O$ denote the centre of the ellipse.
Pick an arbitrary point $A$ on the ellipse. Draw a circle with the centre at $O$ and passing through $A$.
This circle will intersect the ellipse in 4 points* $A,B,C,D$. Show that $ABCD$ is a rectangle, with the edges parralel to the axes. Since the axes go through $)$ it is easy to construct them.
*If the circle intersects the ellipse in only two points, the two points are the intersection between the ellipse and one of the axes.
Of course, if the ellipse is degenerated (a circle) the construction above will fail, since there will be infinitely many points of intersection. But in that case, there is no axis to construct.
P.S. You don't even need the center to solve the problem. Given an ellipse, you can use the fact that the line passing through the midpoints of two parralel cords passes through the centre of the ellipse.
So given an ellipse without the centre, you start by doing the following:
Pick an arbitrary chord $EF$. Pick another point $G$ on the ellipse and draw the parralel through $G$ to $EF$. Let $H$ be the intersection between this parralel and the ellipse.
Pick another cord $IJ$ which is NOT parralel to $EF$. Draw another chord as above $KL \parallel EF$.
Now, if $M,N, P,Q$ are the midpoints of $EF, GH, IJ, KL$ then $MN \cap PQ= \{O\}$ is the centre of the ellipse.