Construct focii of ellipse given center and four tangent lines

We are given 4 distinct lines $a,b,c$ and $d$ which are said to be tangents to an ellipse. Let's consider that the 4 meetings of these lines form an convex quadrilateral $ABCD.$

There is a theorem which garantees that the center $E$ of ellipse inscribed in $ABCD$ must lie in the line joining the midpoints of $AC$ and $BD$ (diagonals of $ABCD).$

I've been wondering how can we construct such an ellipse (namely the focii or the axis of the ellipse) with the four lines and its center $E.$

EDIT: in case of people wondering: yes, any convex quad admits at least one (but usually infinite) inscribed ellipse

EDIT2: I forgot to mention that I'm looking for an straightedge and compass construction. Sorry.

The equation of an ellipse in the plane is given by

$ (r - E)^T Q (r - E) = 1 \hspace{20pt} (1)$

where $E$ is the center, and $Q$ is a $2 \times 2$ symmetric matrix.

At any point on the ellipse the gradient vector which is the normal (perpendicular) vector to the ellipse curve is given by

$ g = 2 Q (r - E) \hspace{20pt} (2) $

If the tangency point on $AB$ is $r_1$ then the gradient vector at $r_1$ will be parallel to the normal to the line segment $AB$, that is,

$ Q (r_1 - E) = \alpha n_1 \hspace{20pt} (3)$

where $n_1$ is known (it is the normal vector to $AB$). From this it follows that

$ r_1 - E = \alpha Q^{-1} n_1 \hspace{20pt} (4)$

Since $r_1$ is on the ellipse we can plug this expression into the equation of the ellipse, to deduce that

$\alpha = \dfrac{1}{\sqrt{n_1^T Q^{-1} n_1 } } \hspace{20pt} (5)$

On the other hand, premultiplying $(4)$ by $n_1^T$ and using $(5)$

$n_1^T (r_1 - E) = \sqrt{n_1^T Q^{-1} n_1 } $

The equation of the line segment $AB$ is $n_1^T (r - A) = 0 $, and we have

$n_1^T (r_1 - E) = n_1^T (r_1 - A + A - E) = n_1^T (A - E )$

because $n_1^T (r_1 - A) = 0$. Thus we have

$n_1^T (A - E) = \sqrt{ n_1^T Q^{-1} n_1 } \hspace{20pt} (6.1)$

Similar equations can be written for the other three line segments $BC, CD, DA$, whose normals are $n_2, n_3, n_4$:

$n_2^T (B - E) = \sqrt{ n_2^T Q^{-1} n_2 } \hspace{20pt}(6.2) $

$n_3^T (C - E) = \sqrt{n_3^T Q^{-1} n_3 } \hspace{20pt} (6.3) $

$n_4^T (D - E) = \sqrt{n_4 ^T Q^{-1} n_4 } \hspace{20pt} (6.4)$

Only $3$ equations out of $4$ are needed to find the matrix $Q^{-1}$. Using the theorem mentioned in the question, a matrix $Q$ satisfying any three of the equations $(6.1) - (6.4)$ will satisfy the fourth equation if the center $E$ is on the line segment between the midpoint of $AC$ and the midpoint of $BD$. ( I don't have a proof of that but I verified it numerically).

Solving any three of equations $(6.1) - (6.4)$ can be done quite easily. Let matrix

$M = Q^{-1} = \begin{bmatrix} M_{11} && M_{12} \\ M_{12} && M_{22} \end{bmatrix} $

Then from $(6.1)$,

$n_1^T Q^{-1} n_1 = \left( n_1^T (A - E) \right)^2 \hspace{20pt} (7) $

The right hand side of $(7)$ is known, while the left hand side is linear in the entries of $Q^{-1} $, namely,

$n_1^T Q^{-1} n_1 = M_{11} n_{1x}^2 + M_{22} n_{1y}^2 + 2 M_{12} n_{1x} n_{1y} \hspace{20pt} (8) $

Two more equations can be written creating a $3 \times 3$ in the unknowns $M_{11}, M_{22}, M_{12} $, and solved.

Once $Q^{-1}$ is found, the ellipse specification is complete. Finding all the parameters of the ellipse follows directly by diagonalizing matrix $Q$ as follows:

$Q = R D R^T \hspace{20pt} (9)$

It can be assumed that $D_{11} \le D_{22}$. The semi-major axis $a = \dfrac{1}{\sqrt{D_{11}}}$ and the semi-minor axis $b = \dfrac{1}{\sqrt{D_{22}}} $

The focii are along the major axis, and are given by

$F_1 = E + a e R_1 $ and $F_2 = E - a e R_1 \hspace{20pt} (10) $

where $R_1$ is the first column vector of the matrix $R$ specified in $(9)$, and $e$ is the eccentricy,

$e = \sqrt{ 1 - \left( \dfrac{b}{a} \right)^2 }$

The figure below shows an example with the above method.

Let $F$ and $G$ be the intersection points of the opposite sides of quadrilateral $ABCD$, as shown in figure below.

The intersection point $O$ of the diagonals lies on the segment joining tangency points $M\in CD$ and $N\in AB$, as explained here. Moreover, it is a general property of the ellipse that line $EF$ bisects $MN$.

To find $M$ and $N$ we must then construct a chord of $\angle BFC$ which is bisected by $EF$. A possible construction is shown in figure below. Draw chord $HK$ of that angle passing through $E$ and perpendicular to $EF$. Draw $FH'=EH$ and $FK'=EK$ both parallel to $HK$. Join $EK'$ and draw the line parallel to it passing through $H'$ and meeting $EF$ at $J$. If $L$ is the midpoint of $EJ$, all chords of $\angle BFC$ parallel to $HL$ will be bisected by line $EF$.

We can then find $M$ and $N$ drawing through $O$ the parallel to $HL$. The other two tangency points $P$ and $Q$ can be easily constructed as explained here. Finally, we can construct a pair of conjugate diameters and the axes of the ellipse as explained here.

To construct one of the ellipses inscribed in the given convex quadrilateral, let us use Brianchon's hexagram theorem (see the book cited bellow, pp. 261-265). I used Corollary 1 (p.263) in my construction: An auxiliary green line cuts the quadrilateral to give a pentagone. The ellipse inscribed in this pentagone is unique. We then construct one point of the ellipse on each side. GeoGebra finishes the work.
To check the construction, have a look here. It is possible to move the big blue points.

The orange lines are used for construction of the center by hand.

Doerrie, Heinrich, 100 great problems of elementary mathematics. their history and solution. Transl. from the German by David Antin, New York: Dover Publications, Inc. X, 393 p. (1965).