$
\def \J {\Delta}
\def \half {\frac{1}{2}}
\def \kwart {\frac{1}{4}}
\def \hieruit {\quad \Longrightarrow \quad}
\def \slechts {\quad \Longleftrightarrow \quad}
\def \norm {\frac{1}{\sigma \sqrt{2\pi}} \; }
\def \EN {\quad \mbox{and} \quad}
\def \OF {\quad \mbox{or} \quad}
\def \wit {\quad \mbox{;} \quad}
\def \spekhaken {\iint}
\newcommand{\dq}[2]{\displaystyle \frac{\partial #1}{\partial #2}}
\newcommand{\oq}[2]{\partial #1 / \partial #2}
\newcommand{\qq}[3]{\frac{\partial^2 #1}{{\partial #2}{\partial #3}}}
\def \erf {\operatorname{Erf}}
$
Drawing an Ellipse
The following general expression will be proposed for the ellipse of inertia:
$$
\frac{\sigma_{yy} (x-\mu_x)^2 - 2 \sigma_{xy} (x-\mu_x) (y-\mu_y)
+ \sigma_{xx} (y-\mu_y)^2}{ \sigma_{xx} \sigma_{yy} - \sigma_{xy}^2 } = 1
$$
When making a drawing of such an ellipse, it would be more handsome to have it
in parametrized form. That is, we seek an equivalent like this:
\begin{cases}
x = \mu_x + a_x\cos(t) + a_y\sin(t) \\
y = \mu_y + b_x\cos(t) + b_y\sin(t)
\end{cases}
Multiply the first equation with $b_x$, the second with $a_x$ and substract:
$$
b_x (x-\mu_x) - a_x (y-\mu_y) = (a_y b_x - a_x b_y)\,\sin(t) \hieruit
$$ $$
\sin(t) = \frac{b_x (x-\mu_x) - a_x (y-\mu_y)}{a_y b_x - a_x b_y}
$$
Multiply the first equation with $b_y$, the second with $a_y$ and substract:
$$
b_y (x-\mu_x) - a_y (y-\mu_y) = (b_y a_x - b_x a_y)\,\cos(t) \hieruit
$$ $$
\cos(t) = \frac{b_y (x-\mu_x) - a_y (y-\mu_y)}{b_y a_x - b_x a_y}
$$
Now use the well known identity:
$$
\cos^2(t) + \sin^2(t) = 1
$$
Giving:
$$
\left(\frac{b_x x' - a_x y'}{b_y a_x - b_x a_y}\right)^2 +
\left(\frac{b_y x' - a_y y'}{a_y b_x - a_x b_y}\right)^2 = 1
$$
Where $x'= x-\mu_x$ and $y'= y-\mu_y$ . This result must be comparable with:
$$
\frac{\sigma_{yy} (x')^2 - 2 \sigma_{xy} x'y'
+ \sigma_{xx} (y')^2}{ \sigma_{xx} \sigma_{yy} - \sigma_{xy}^2 } = 1
$$
Drop the primes $'$ in both equations for the sake of simplicity. Look after:
$$
\left(\frac{b_x x - a_x y}{b_y a_x - b_x a_y}\right)^2 +
\left(\frac{b_y x - a_y y}{a_y b_x - a_x b_y}\right)^2 = 1
$$
And work out:
$$
\frac{(b_x^2 + b_y^2) \; x^2 - 2 (a_x b_x + a_y b_y) \; x y
+ (a_x^2 + a_y^2) \; y^2 }{ (b_y a_x - b_x a_y)^2 } = 1
$$
As compared with:
$$
\frac{\sigma_{yy} x^2 - 2 \sigma_{xy} x y
+ \sigma_{xx} y^2}{ \sigma_{xx} \sigma_{yy} - \sigma_{xy}^2 } = 1
$$
We thus find:
$$
a_x^2 + a_y^2 = \sigma_{xx} \EN
b_x^2 + b_y^2 = \sigma_{yy} \EN
a_x b_x + a_y b_y = \sigma_{xy}
$$
It also follows that the following must hold:
$$
\sigma_{xx} \sigma_{yy} - \sigma_{xy}^2 = (a_x b_y - b_x a_y)^2
$$
Check this:
$$
(a_x^2 + a_y^2)(b_x^2 + b_y^2) - (a_x b_x + a_y b_y)^2 =
$$ $$
a_x^2 b_x^2 + a_x^2 b_y^2 + a_y^2 b_x^2 + a_y^2 b_y^2
- (a_x^2 b_x^2 + a_y^2 b_y^2 + 2 a_x b_x a_y b_y) =
$$ $$
(a_x b_y)^2 + (b_x a_y)^2 - 2 (a_x b_y) (b_x a_y) = (a_x b_y - b_x a_y)^2
$$
Happy maybe that this is consistent, we are nevertheless faced with a little
problem: there are four unknowns $(a_x,a_y,b_x,b_y)$, but only three equations
to solve for them. This problem is readily tackled by the meaning of the
equations, though. The first equation means that the length of the vector
$\vec{a}$ is equal to $\sqrt{\sigma_{xx}}$. The second equation means that the
length of the vector $\vec{b}$ is equal to $\sqrt{\sigma_{yy}}$. The third
equation means that the cosine of the angle between $\vec{a}$ and $\vec{b}$ is
given by the inner product divided by these lengths:
$ \cos(\theta) = \sigma_{xy}/(\sqrt{\sigma_{xx}\sigma_{yy}}) $.
Now it is clear that these three facts are quite independent of an eventual
rotation of the coordinate system:
\begin{cases}
a'_x = + \cos(\tau) a_x + \sin(\tau) a_y \\
a'_y = - \sin(\tau) a_x + \cos(\tau) a_y
\end{cases}
When applied to the parametrized equations, for example the first of the two:
$$
x-\mu_x = a'_x\cos(t) + a'_y\sin(t) =
$$ $$
\left[\cos(\tau)\; a_x + \sin(\tau)\; a_y\right] \cos(t) +
\left[- \sin(\tau)\; a_x + \cos(\tau)\; a_y\right] \sin(t) =
$$ $$
\left[\cos(\tau) cos(t) - \sin(\tau) \sin(t)\right] a_x +
\left[\sin(\tau) cos(t) + \cos(\tau) \sin(t)\right] a_y =
$$ $$
\cos(t + \tau)\; a_x + \sin(t + \tau)\; a_y
$$
We conclude that a change of coordinate system is just the same as a different
offset for the running parameter. Nothing interesting actually. This the reason
why we are free to replace the four unknowns by only three. For example, choose
$\vec{a}$ in the same direction as the x-axis, meaning that $a_y = 0$.
Then the solution of our problem becomes:
$$
a_x^2 = \sigma_{xx} \hieruit a_x = \sqrt{\sigma_{xx}} \EN
a_x b_x = \sigma_{xy} \hieruit b_x = \sigma_{xy}/\sqrt{\sigma_{xx}}
$$ $$
b_x^2 + b_y^2 = \sigma_{yy} \hieruit
b_y = \sqrt{\sigma_{yy} - \sigma_{xy}^2/\sigma_{xx}}
= \sqrt{\frac{\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2}{\sigma_{xx}}}
$$
\begin{cases}
x = \mu_x + a_x\cos(t) + a_y\sin(t) \\
y = \mu_y + b_x\cos(t) + b_y\sin(t)
\end{cases}