This is a problem I came up with recently. You have the ellipsoid
$$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1 $$
where $a = b = 1$ and $c = 0.8$, and $x, y, z \ge 0 $. Inscribe a cube in the region bounded by the ellipsoid and the three coordinate planes in the first octant, such that four of its vertices lie on the ellipsoid surface, two vertices lie on the $xy$ plane, one vertex on the $xz$ plane and one vertex on the $yz$ plane.
This is shown in the figure below.
My question is: Identify this cube, that is, find its center and orientation, as well as its side length.
My attempt:
My attempt is included in my answer that follows.
From the symmetry resulting from $a = b = 1$, the cube faces are either parallel to the plane $x = y$ or are perpendicular to it, and the center lies in this plane. Let the center be $C = (x_1, x_1, z_1) $ and let the tilt angle (angle between the base of the cube and the $xy$ plane) be $\theta$, then we can define a reference frame attached to the cube with its origin at the center of the cube, and whose $x',y',z'$ axes are normal to the faces of the cube. Then we can define the following unit vectors along these axes
$\hat{z'} = ( c_1 \cos \theta, c_1 \cos \theta, \sin \theta) $
$\hat{x'} = (- c_1, c_1, 0 ) $
$\hat{y'} = (-c_1 \sin \theta, -c_1 \sin \theta, \cos \theta) $
where $c_1 = \dfrac{1}{\sqrt{2}} $
Define the rotation matrix $R = [ \hat{x'}, \hat{y'} , \hat{z'} ] $
If the side length of the cube is $s = 2 a$ then the local coordinates of the vertices are $a (\pm 1, \pm 1, \pm 1)$
The local coordinates of one of the two vertices that lies on the $xy$ plane is $P_1 = a (1, -1, -1) $, and that of the vertex that lies on the $yz$ plane is $P_2 = a (1, 1, -1) $ , and the coordinates of two vertices of the cube that lie on the surface of the ellipsoid are
$P_3 = a (1, 1, 1)$ and $P_4 = (1, -1, 1) $
The (world) coordinates of $P_i$ is given by $Q_i = C + R P_i$. Using $P_1 $ and $P_2$ and imposing $z = 0$ for the first and $x=0$ for the second, one can relate the center coordinates $x_1, z_1$ in terms of $a $ and $\theta$. Next, taking $P_3$ and $P_4$ and substituting the corresponding $Q_3$ and $Q_4$ into the equation of the ellipsoid, we get an equation in the angle $\theta$ that is of the form
$ A \cos(\theta) + B \sin(\theta) + C \cos(2 \theta) + D \sin(2 \theta) + E = 0 $
for known constants $A,B,C,D,E$. Solving, we get the value of $\theta$, and substituting that, we find the side length $s$. I got $\theta \approx 47.37359^\circ$ and $ s \approx 0.458567 $.