Intersection of an ellipsoid and plane If a non-axis-aligned ellipsoid is defined by
$$\tag{1} \bf x^T A x=1$$
where $\bf x$ $=[x, y, z]^T$, $\bf A$ $=\begin{bmatrix}a_{xx}&a_{xy}&a_{xz}\\a_{yx}&a_{yy}&a_{yz}\\a_{zx}&a_{zy}&a_{zz} \end{bmatrix}$, and $a_{ij}=a_{ji}$,
then Eqn(1) is the following when expanded:
$$\tag{2} a_{xx}x^2+a_{yy}y^2+a_{zz}z^2+2a_{xy}xy+2a_{xz}xz+2a_{yz}yz=1$$
at $z=0$ the coordinates of the ellipsoid is:
$$\tag{3} a_{xx}x^2+a_{yy}y^2+2a_{xy}xy=1$$
which I think is also an equation of an ellipse since its the intersection of the ellipsoid and the x-y-plane.
Equation 3 is in implicit form.  How can its parametric equations be found?
 A: Diagonalize the top-left $2\times2$ matrix, and you will reduce to a form
$$\lambda u^2+\mu v^2=1$$ where $\lambda,\mu$ are the two Eigenvalues, which are positive. Then
$$u=\frac{\cos\theta}{\sqrt\lambda},v=\frac{\sin\theta}{\sqrt{\mu}}$$ and $x,y$ are the linear combinations of $u,v$ given by the Eigenvectors.
A: It suffices to transform in a first step
$$a_{xx}x^2+a_{yy}y^2+2a_{xy}xy=1\tag{0}$$
into a polar equation using ... polar coordinates 
$$x=r \cos \theta, \ y=r \sin \theta, \tag{1}$$
giving :
$$r^2(a_{xx}\cos^2 \theta +a_{yy}\sin^2 \theta+a_{xy}(2 \sin \theta \cos \theta))=1.\tag{2}$$
Using formulas 
$$\begin{cases}\sin 2 \theta &=& 2 \sin \theta \cos \theta \\ \cos^2 \theta&=&\tfrac12(1+\cos 2 \theta) \\ \sin^2 \theta&=&\tfrac12(1-\cos 2 \theta)\end{cases}$$ 
one obtains from (2) the following polar equation : 
$$r=\dfrac{1}{\sqrt{A \cos(2 \theta)+B+a_{xy} \sin (2 \theta)}}\tag{3}$$
with $A:=\tfrac12(a_{xx}-a_{yy})$ and $B:=\tfrac12(a_{xx}+a_{yy})$.
(See remark below).
Having polar equation (3), it is now straightforward to convert it into a parametric equation by using again formulas (1) but with the expression of $r$ given by (3).
Remark : instead of (3), we could have taken as well 
$$r=\dfrac{-1}{\sqrt{...}}\tag{4}.$$
Indeed, equation (0) is homogeneous with degree 2, meaning that it is invariant when $(x,y)$ is replaced by its opposite $(-x,-y)$ ; as a consequence, (4) generates the very same curve.
