Ellipse projection knowing semiaxes vectors I'd like to know the semi-projection of a tilted ellipse on $x$ and $y$ axes, called as $O_\parallel$ and $O_\perp$, knowing the vectors $\vec{\zeta}=(\zeta_x,\zeta_y)$  and $\vec{ \nu}=(\nu_x,\nu_y)$ and that, as obvious $\hat{x}=(1,0) , \hat{y}=(0,1)$.
I know that's fairly easy in 2D by looking at the parametric equation and by imposing some condition on the derivative 
but, because I need a feasible method that works in many dimension, i'd like to find a direct expression from the vector components.
 A: This problem has a nice solution. There's probably a simpler way than what I did below, but I took an optimization approach. 
Suppose we are given $n$ orthogonal vectors $w_1,\dots,w_n$ in $\mathbb{R}^n$, which define the axes of an ellipse. Then the ellipse is given by
$$
E=\{x\in\mathbb{R}^n\ |\ x^\text{T}Ax\leq1\},
$$
where the matrix $A$ is given by $A=VDV^\text{T}$, where the $i^\text{th}$ column of $V$ is $w_i\big/\|w_i\|$, and $D$ is diagonal, with $d_{ii}=1\big/\|w_i\|^2$. So what you would like to do, is for each coordinate $i=1,\dots,n$, solve the convex optimization problem
$$
\begin{array}{rl}
\max\ & x_i \\
\text{s.t.}\ & x^\text{T}Ax\leq1
\end{array}
$$
Using the KKT conditions (I'm omitting the details), we can then derive the optimal solution to this optimization problem for any coordinate $i$. The solution for direction $i$ is
$$
z^*=\sqrt{V_i^\text{T}D^{-1}V_i^{}},
$$
where $V_i$ is the $i^\text{th}$ row of $V$. This can be simplified even further. Let $W$ denote the matrix with $w_i$ as its columns, and let $W_i$ denote the $i^\text{th}$ row of $W$. Then the optimal solution is
$$
z^*=\|W_i\|_2.
$$
Here's a picture computed using this method with $w_1=(3,\ 2)$ and $w_2=(-1,\ 3/2)$.

A: The extreme values of any particular variable $x_i$ on an ellipsoid are given by the intersections of the tangent hyperplanes to the ellipsoid that are normal to the $x_i$-axis with that axis. In homogeneous coordinates, these hyperplanes will have the form $(\mathbf e_i^T; -\tau)^T$, where $\tau$ is one of those extreme values. If the ellipsoid is given by the matrix $Q$, then these hyperplanes satisfy the dual quadric equation $$(\mathbf e_i^T;-\tau)Q^{-1}(\mathbf e_i^T;-\tau)^T=0.$$ This equation reduces to $$\pmatrix{1&-\tau}\pmatrix{[Q^{-1}]_{(i,i)} & [Q^{-1}]_{(i,n+1)} \\ [Q^{-1}]_{(n+1,i)} & [Q^{-1}]_{(n+1,n+1)}}\pmatrix{1\\-\tau} = 0,\tag{*}$$ which is a simple quadratic equation in $\tau$.  
If the principal half-axes of the ellipse are given by the vectors $V=(\mathbf v_1,\dots,\mathbf v_n)$ and its center is at $\mathbf c$, the matrix $$M = \pmatrix{V & \mathbf c \\ \mathbf 0^T & 1}$$ is the affine transformation that maps the unit hypersphere onto the ellipoid. We then have $Q=M^{-T}\operatorname{diag}(1,\dots,1,-1)M^{-1}$ and so $$Q^{-1} = M\operatorname{diag}(1,\dots,1,-1)M^T = \pmatrix{VV^T-\mathbf c\mathbf c^T & -\mathbf c \\ -\mathbf c^T & -1}.$$ Substituting back into equation (*) and expanding by coordinates, the quadratic equation for $\tau$ becomes $$\pmatrix{1&-\tau} \pmatrix{[VV^T-\mathbf cc^T]_{(i,i)} & -c_i \\ -c_i & -1} \pmatrix{1\\-\tau} = \sum_{k=1}^n [v_k]_i^2-c_i^2+2c_i\tau-\tau^2=0.$$
