Intersection of 2d plane and n-dimensional ellipsoid English is not my native language and I'm neither a mathematician nor a statistican by trade so please excuse me if I fail to explain my problem in a concise way.
I'm trying to visualize a multivariate reference region of normally distributed measurements. A multivariate reference region can mathematically be described as an $n$-dimensional ellipsoid of the form: 
$$
(x-cx)'A(x-cx)=1
$$ 
where $cx$ is a $n \times 1$ vector representing the center of the ellipsoid and where the eigenvectors and eigenvalues of $A$ ($n \times n$ pos definite matrix) contains information on the orientation and semiaxis of the ellipsoid (wiki). 
In order to present this I would like to plot the 2D intersection ellipse when $n-2$ positions of the vector $x$ is known. In the above equation $A$ and $cx$ will also be known. 
Example: Let's say I have an multivariate reference region ($A$ and $cx$ known) for age, systolic blood pressure and diastolic blood pressure ($n=3$). Then I would like to plot the ellipse describing what systolic and diastolic blood pressure could be expected in subjects with the exact age of $60$ for instance. Then the vector $x=(60, s, d)$ where $s$ and $d$ are the unknowns describing the ellipse I want to visualize. 
With the help of this article I've managed to find the orientation and dimensions of the 2D intersection ellipse of a 3-dimensional ellipsoid. It is, however, beyond me to generalize the solution into $n>3$ dimensions. One aspect that might simplify  the problem is that the intersection plane will be perpendicular to the axes of the coordinate system.
Suggestions? Solutions? Anyone?
 A: Without loss of generality suppose $x_3,\ldots,x_n$ are fixed. The resulting quadratic equation in $x_1$ and $x_2$ has a quadratic, a linear and a constant part. You can obtain those by separating $A,$ $c$ and $x$ into blocks separating the first two coordinates from the other ones.
The quadratic part is still given by the upper left $2\times2$ submatrix of the original matrix $A.$ That gives us the orientation of the axes of the intersection ellipse. The linear part is the sum of the original linear part (first two elements of the matrix product $Ac$) and the upper right $2\times(n-2)$ part of $A$ applied to the separate blocks $(x_1\ x_2)$ and $(x_3 \ldots x_n)$. That gives us the center of the intersection. For the actual size of the two axes we need to complete the squares in the quadratic and linear terms.
A: Assuming the first two components of $x$ are free we can have the quadratic form evaluated to
\begin{align}
1 
=& \,
((x_1, x_2, rx) - (cx_1, cx_2, rcx))^t A 
((x_1, x_2, rx) - (cx_1, cx_2, rcx)) \\
=& \,
a_{11} (x_1-cx_1)^2 + a_{22} (x_2 - cx_2)^2 +
(a_{12} + a_{21}) (x_1-cx_1)(x_2 - cx_2) + \\
& ((a_{13}+a_{31}) (x_3 - cx_3) + \ldots + 
   (a_{1n}+a_{n1}) (x_n-cx_n)) (x_1 - cx_1) + \\
& ((a_{23}+a_{32}) (x_3 - cx_3) + \ldots + 
   (a_{2n}+a_{n2}) (x_n-cx_n)) (x_2 - cx_2) + \\
& \sum_{i,j \ge 3} a_{ij} (x_i-cx_i)(x_j-cx_j) \\
=& \,
a (x_1-cx_1)^2 + b (x_2-cx_2)^2 + c (x_1-cx_1)(x_2-cx_2) +
d (x_1 - cx_1) + e (x_2 - cx_2) + (f + 1)
\end{align}
The coefficients $a, b, c, d, e, f$ contain only known values (the $x$ vector coordinates $x_3$ to $x_n$, the center vector coordinates $cx_i$ and matrix components $a_{ij}$).
Introducing $u = x_1 - cx_1$ and $v = x_2 - cx_2$ the above equation turns into
$$
a u^2 + b v^2 + c u v + d u + e v + f = 0
$$
which is the general form of a conic section in the $u, v$-plane, see conic sections.
If $b = a_{22} \ne 0$ we can try to solve for $v$:
\begin{align}
0 
&= v^2 + \left(\frac{c}{b} u + \frac{e}{b} \right) v + 
\frac{a}{b} u^2 + \frac{d}{b} u + \frac{f}{b} \\
&= \left( v + \frac{cu+e}{2b} \right)^2
- \left( \frac{cu+e}{2b} \right)^2 +
\frac{a}{b} u^2 + \frac{d}{b} u + \frac{f}{b}
\end{align}
which gives
$$
v = -\frac{cu+e}{2b}\pm\sqrt{\left( \frac{cu+e}{2b} \right)^2
- \frac{au^2 + du + f}{b}}
$$
