How do I find where the major axis of an ellipse is when it is placed inside a quadrilateral and tangent to the four sides of that quadrilateral? I am an artist and I am inquiring of the math community for some help in understanding some concepts of geometry. The reason for this question stems from wanting a formulaic way of deriving where to place ellipses in perspective squares. This is applicable for perspective drawing. An ellipse is essentially a circle in perspective and a circle can fit inside of a square with 4 parts of the circle tangent to the 4 sides of the square. A square in perspective becomes a quadrilateral and the circle in perspective becomes an ellipse.
e.g. the image below depicts an ellipse inside a quadrilateral. In this case quadrilateral is symmetrical on either side (isosceles trapezoid) and the foci (points J, K) of the ellipse run on a line (the major axis of the ellipse) that is parallel to two sides of the bounding quadrilateral.

*

*note the line running through I is the perspective center (found by creating an x through the opposing sides of the quadrilateral) and the line running through F is the center found by dividing the height in half, which also seems to be the center of the ellipses.


As we add more ellipses to the example they are no longer bounded inside isosceles trapezoids. The major axis of the ellipses are no longer parallel with the sides of the quadrilaterals (Points P, Q, & V, W).

The ellipses in the examples were placed by 'eye-balling' but I would like a formulaic way to do this than adjusting the ellipses until they look correct.
How do I know where to place the major axis of the ellipses for a given quadrilateral?

If the quadrilateral $ABCD$ has two sides parallel to the horizon (as in your figure) then it's easy, because two tangency points of the ellipse are then the midpoints $E$, $F$ of the bases of the trapezoid, and their midpoint $O$ is the centre of the ellipse.
We can then find another tangency point $P$ on $AD$ as the vertex opposite to $E$ of a parallelogram having a diagonal on line $OD$, so that $EP$ is bisected by $OD$. Finally, we can construct diameter $LM$, conjugate to $EF$, taking $LM$ parallel to $AB$ and
$$
OL=OM={PH\cdot EO\over\sqrt{EO^2-HO^2}},
$$
where $H$ is the intersection of $EF$ with the line through $P$ parallel to $AB$.
Once you have constructed a pair of conjugate diameters, you can find the axes following the construction given here.

