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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?

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2 Answers

up vote 6 down vote accepted

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.

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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.

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+1, also good answer, but you need to know about the parametrisation of the ellipse inscribed in unit square. Now I know a lot about it :) What had me confused though it is obvious that $\theta=0$ and $\theta=\pi/2$ will not give the major and minor axes vertices if $x_0\neq y_0$. This will not teach me again that it is dangerous to do math when tired. –  mpiktas Dec 24 '10 at 11:57
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@mpiktas The symmetry of the unit square (under transposing $x$ and $y$) implies the equation of an inscribed ellipse must be in the form $x^2 + y^2 + 2 B x y = F$. Setting $\theta = d/2$ locates $(\rho, 1)$ on the ellipse and setting $\theta = 0$ locates $(\cos(d/2), \cos(d/2))$ on it. These give two linear equations for $F$ and $B$, etc. ... I do appreciate the difficulties of doing math when one is tired... –  whuber Dec 24 '10 at 17:56
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