Recall first a few basic constructions.
The current point is defined by the draggable cyan point m on the circular directrix.
The given line is defined by the green draggable point d.
Knowing these basic constructions, that we won't detail anymore, let's search the intersection of our
two given conic sections.
As they have the same center, from symmetry, the intersection points build a parallelogram (for the diagonals intersect in their midpoint O)
The sides of this parallelogram are conjugate directions for both the conic sections.
Let (d) any diameter and (d') the conjugate diameter of (d) with respect to the first conic section Γ, then (d") the conjugate diameter of (d') with respect to the second conic section Γb.
We thus define (step 1) a projective homography (d) → (d") in the pencil of lines O* : green line (d) → orange line (d")
The directions of the parallelogram sides are then the fixed directions in this homography.
They are constructed by projecting (d) → (d") onto any circle going through O, the homography axis intersects the circle in fixed points.
For doing that (step 2) we chose any three points a,d,c and their transformed a",d",c", and we use the axis definition :
locus of the intersection points of ad" and a"d.
The chosen points are (a) focal axis of Γ, the conjugate is the perpendicular axis and (a") is the conjugate of this perpendicular axis with respect to Γb
We do the same with (c) the perpendicular axis of Γ and (c") the conjugate of the focal axis of Γ with respect to Γb
Finally (d) and (d") are a current direction, defined by draggable point d.
We thus get the directions (magenta) Ou, Ov of the parallelogram sides.
But we have now to construct the sides themselves.
Let's chose (step 3) any line (L) intersecting the conic sections in (M, M') and (Mb, M'b)
The pencil of conic sections defined by Γ and Γb defines on (L) a Desargues involution fully determined by M ↔ M' and Mb ↔ M'b
If this involution is hyperbolic, it has two real fixed points I and J.
For that, just chose (L) so that one chord of MM' or MbM'b is strictly inside the other.
Also to get an effective construction, these fixed points shouldn't be at infinity, hence (L) shouldn't be parallel to the previously constructed conjugate diameters. This allows however a wide choice for (L).
In the applet (L) is defined by the two "current" points M and Mb of the two conic sections (driven by the cyan points on the circular directices).
Let's construct then (step 4) the fixed points of this involution, that we obtain by projecting it on any circle, for instance the one previously chosen.
The axis of the involution on the circle is constructed : line through the intersection points of mmb with m'm'b, and of mm'b with m'mb
The intersections of this axis with the circle give the fixed points I and J after projecting back to (L).
The involution M ↔ M' on (L) is then characterized by (M,M',I,J) = -1, harmonic range.
Let Q the intersection of (L) with diameter Ou and E,E' the intersections of (L) with the searched degenerated conic section, made of the two common chords parallel to Ou.
Q is obviously the midpoint of EE' and we can then construct (step 5) E and E' through QE² = QE'² = QI.QJ (classical).
We get then the common chords as the parallels to Ou through E and E', then the intersections of these lines with Γ ( or with the other degenerated conic section parallel to Ov, constructed in a similar way).
Important note :
The choice of the "any points" impacts the construction.
If we change the conic sections,we might also sometimes change (manually) these construction points ("ctrl pnt" button) :
The construction even completely fails if the two conic sections have the same axes :
We should then chose a second arbitrary line in step 1.
However in this case the construction is simpler : the fixed directions are just the axes !
An automatic choice of these control points would lead to an irrelevant figure, most of the points being outside the drawing and would make the construction even more complicated (165 constructed geometric elements in the applet)
Even the simple construction of an intersection or the conjugate requires several intermediate elements as shown in the preliminary constructions.
Even just project a direction (d) on a circle requires two elements ! Just intersecting (d) with the circle doesn't work : we must grant that it is the other intersection point than O etc.
Although it is obvious manually, it is not so simple inside an applet, and it is often needed to use artificial constructions, for instance in this case it is the reflection of O through the perpendicular to (d) from center ω.
Here also, the key is to construct a degenerated conic section in the pencil, and the Desargues involution.
Details (these applets are too big to fit in the same page)