Point on plane that touches ellipsoid Consider a plane described by:
$\overrightarrow{x} \cdot \overrightarrow{w}=c$.
Consider also the ellipsoid described by:
$(x_{1}/s_{1})^{2} + ... + (x_{N}/s_{N})^{2} = d$.
Here, d is such that the ellipsoid just touches the plane. How do I find the point where they touch?
If I'm thinking correctly, in the case of $s_{i}$ being the same for all i, the ellipsoid is just an N-dimensional sphere, and the point is simply where the plane intersects with its own plane vector:
Plane: $\overrightarrow{x} \cdot \overrightarrow{w}=c$.
Line: $\overrightarrow{x} = b \cdot \overrightarrow{w}$
Resulting in
$b = c / (\overrightarrow{w} \cdot \overrightarrow{w})$
$\overrightarrow{x} = \overrightarrow{w} \cdot c / (\overrightarrow{w} \cdot \overrightarrow{w})$
Gut feeling says that if $s_{i}$ are not all the same value, then this solution should be scaled by $\overrightarrow{s}$ somehow.
 A: You are right. In the rescaled coordinates $z_i=x_i/s_i$,
$$z_1^2+z_2^2+\cdots z_n^2=d$$
and the transformed plane
$$(s_1w_1)z_1+(z_2w_2)z_2+\cdots(s_nw_n)z_n=c$$ is tangent to that sphere.
Hence the solution vector is parallel to the normal to the plane, $$z_i=\lambda s_iw_i$$ or $$x_i=\lambda s_i^2w_i$$ where
$$\lambda=\frac{c}{\sum (s_iw_i)^2}.$$
A: When working with properties such as tangency that are invariant under affine transformations, scaling the coordinate system so that the ellipsoid you’re working with becomes a sphere is a good general way to simply those problems. Yves Daoust’s answer picks up where you left off with that approach. Since we’re looking at tangents to quadrics, another approach that can often lead to a simple solution is to use the properties of polar points and hyperplanes.  
In block form, the homogeneous covector that represents the given plane is $\pi=[\mathbf w^T,-c]$ and the matrix of the ellipsoid is $$Q=\begin{bmatrix}S^{-2}&0\\0&-d\end{bmatrix},$$ where $S=\operatorname{diag}(s_1,\dots,s_N)$. Tangent planes to the ellipse satisfy the dual quadric equation $$\pi Q^{-1}\pi^T=\mathbf w^TS^2\mathbf w-\frac{c^2}d=0$$ from which $$d={c^2\over\mathbf w^TS^2\mathbf w}.\tag1$$ The polar point of a hyperplane tangent to a quadric is its only intersection with the quadric. The polar point of $\pi$ is $\mathbf p=Q^{-1}\pi^T=[\mathbf w^TS^2,c/d]^T$, which in Cartesian coordinates is $\frac dcS^2\mathbf w$. (If $c=0$, then $\pi$ passes through the origin and the ellipsoid is degenerate.) Substituting the value of $d$ from (1) and expanding gives for the point of tangency $${c\over\mathbf w^TS^2\mathbf w}S^2\mathbf w={c\over\sum_{i=1}^Ns_i^2w_i^2}(s_1^2w_1,\dots,s_N^2w_N).\tag2$$
