Suppose you have an ellipsoid with known semi-axes, say $a,b,c$. The ellipsoid is then rotated so that the three axes of the ellipsoid are aligned with the three columns of a known rotation matrix $R$. Keeping this orientation, you place the ellipsoid on the $xy$ plane, in the first octant (where $ x \ge 0, y \ge 0, z \ge 0$). Next, you drag the ellipsoid, while maintaining its spatial orientation, along the $xy$ plane, in a direction parallel to the $x$ axis, till it touches the $yz$ plane, then you drag it parallel to the $y$ axis, till it touches the $xz$ plane. Now the ellipsoid is tangent to the $3$ coordinate planes. The question is find the tangency points of the ellipsoid with the three coordinate planes.
As for context, this question can be considered a extension of $2D$ ellipses in general orientation, to the $3D$ case involving an ellipsoid instead of an ellipse.
My Progress:
The algebraic equation of an ellipsoid is
$ (r - r_0)^T Q (r - r_0) = 1 $
where $ r = [x , y, z ]^T $ and $ r_0 = [r_{0x}, r_{0y}, r_{0z}]^T $ is the center of the ellipsoid (unknown). And $Q$ is known and is of the form
$ Q = R D R^T $
with $ D = \text{diag}\bigg(\ \dfrac{1}{a^2},\ \dfrac{1}{b^2}, \ \dfrac{1}{c^2} \bigg) $, and $R $ is a rotation matrix, whose columns (mutually orthogonal unit vectors) give the direction of the three axes of the ellipsoid.
From tangency to the $xy$ plane, let $r_1$ be the tangency point, then
$ (r_1 - r_0)^T Q (r_1 - r_0) = 1 $
and the gradient of the ellipsoid (which is a vector normal to its surface) points in a direction of $- \mathbf{k} $ (where $\mathbf{k} $ is the unit vector pointing in the positive $z$-direction). Hence,
$ Q (r_1 - r_0) = -\alpha \mathbf{k} $
for some $\alpha \gt 0 $
Similar equations can written for $r_2$ and $r_3$, the tangency points with the $xz$ plane and the $ y z $ plane respectively.
Continuing with $r_1$, and from the above equation,
$ r_1 - r_0 = - \alpha Q^{-1} \mathbf{k} $
Plugging this into the equation of the ellipsoid, yields
$ \alpha = \dfrac{1}{\sqrt{ \mathbf{k}^T Q^{-1} \mathbf{k} } } $
Everything on the right hand side is known, so $\alpha$ is now known.
Recall that $r_1$ is the tangency point with $xy$ plane, so its $z$-coordinate is $0$. Premultiplying the above equation for $r_1 - r_0$ by $ \mathbf{k}^T$ therefore, will give
$ r_{0z} = \sqrt{ \mathbf{k}^T Q^{-1} \mathbf{k} } $
Similarly, we will get
$ r_{0y} = \sqrt{ \mathbf{j}^T Q^{-1} \mathbf{j} } $
and
$ r_{0x} = \sqrt{ \mathbf{i}^T Q^{-1} \mathbf{i} } $
where $\mathbf{i}$ and $\mathbf{j} $ are the unit vectors in the positive $ x $ and $y$ directions, respectively.
Having found $r_0$ completely (the center of the ellipsoid). Then now we can use the equations above for $(r_1 - r_0)$ , $(r_2 - r_0)$ , $(r_3 - r_0) $ to solve for $r_1, r_2, r_3$.
Here is another point of view.
You may know in plane geometry the fact that the locus of points from which one can draw an orthogonal pair of tangents to the ellipse with equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ is a circle called the director circle of the ellipse with equation $$x^2+y^2=a^2+b^2.$$
This property has a straightforward equivalent in space geometry : The locus of points of intersection of 3 mutually perpendicular tangent planes to the ellipsoid with equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ is a sphere with equation :
$$x^2+y^2+z^2=a^2+b^2+c^2$$
This sphere is called the director-sphere of the given ellipsoid.
I have found a reference (and a proof) of this property on p. 72 of "Elementary Treatise on Solid Geometry" by Charles Smith, Macmillan, London, 1907 (!) that you can find here.
Do you see you this property can be exploited for your issue ? (in fact, you should find back some parts of your calulations)