Question:
You're given the parallelogram $ABCD$ shown below, with vertices at $A(2,3)$, $B(12, 9)$, $C(14, 17)$, $D(4,11)$. And you want to determine the ellipse that is tangent to all four sides of the parallelogram, such that its major or minor axis is at an angle $+30^\circ$ with positive $x$ axis. To determine the ellipse, find its algebraic equation, or find its major and minor semi-axes, and their inclination angle from the positive $x$ axis.
Remarks:
From symmetry, it follows that the center of the ellipse is the center of the parallelogram.
The gradient of the ellipse (the normal vector) is pointing in the outward direction at the points of contact with the sides of the parallelogram.
I'll propose instead a ruler-and-compass construction, because it's fairly easy. Let then $ABCD$ be the given parallelogram, $O$ its center, and $OE$ the given major axis, intersecting side $BC$ at $E$ (see figure below).
Construct the bisector of $\angle BCD$, intersecting the major axis at $H$, and line $CI\perp CH$, intersecting the major axis at $I$. Draw circle $HCI$ and construct from $O$ a tangent $OK$ to this circle. Construct then on the major axis points $M$ and $N$ such that $OM=ON=OK$. Points $M$ and $N$ are the foci of the ellipse.
To find one of the contact points we can then construct $M'$ reflecting $M$ about line $CD$: the intersection $P$ between $M'N$ and $CD$ is a contact point. Having the foci and a point, the ellipse is completely determined and it's easy to find its vertices, if wanted.
I have no time now to explain why this construction works, I'll probably add something later on.