Finding standard ellipse characteristics from specific ellipse parametrisation I have found the following ellipse representation $(x,y)=(x_0\cos(\theta+d/2),y_0\cos(\theta-d/2))$, $\theta \in [0,2\pi]$. This is a contour of bivariate normal distribution with uneven variances and correlation $\rho=\cos(d)$. I know that this is a rotated ellipse with centre $(0,0)$. How to find the lengths of major and minor axes and the angle between x-axis and major axis? 
 A: This formula rescales a standard ellipse $(\cos(\theta + d/2), \cos(\theta - d/2))$ (inscribed within the unit square) by the diagonal matrix $(x_0, y_0)$.  By symmetry, the values $\theta = 0$ and $\theta = \pi/2$ correspond to vertices of this standard ellipse, allowing us to find their coordinates (and thus the lengths of the semi-axes), whence we easily deduce its equation is $x^2 + y^2 - 2 \rho x y = 1 - \rho^2$.  Applying the diagonal matrix gives the conventional implicit form
$$\left(\frac{x}{x_0}\right)^2 + \left(\frac{y}{y_0}\right)^2 - 2 \rho \frac{x}{x_0} \frac{y}{y_0} = 1 - \rho^2 \text{.}$$
From here you can look up anything you want.
A: The semi-major axis will be the distance from the origin to the farthest point on the ellipse.  So $r^2=x_0^2\cos^2(\theta+d/2)+y_0^2\cos^2(\theta-d/2)$.  As $r^2$ is a monotonic function of $r$, you can maximize it instead, which simplifies things a bit.  The angle between the $X$ axis and the major axis will then be the arctangent of $y/x$ at this maximum point.  The semi-minor axis will be the distance from the origin to the closest point on the ellipse, and will be $\pi/2$ away from the semi-major axis.
