How to find the vertices of a particular ellipse with straightedge and compass?

Question

In order to provide and alternative solution to a well-known problem $^{(*)}$ I would like to solve the following sub-problem in the most effective way (i.e. in the least number of steps).

$A,B,C,T$ are four distinct points in the plane such that $TA=TB$ and $T,C$ lie on opposite sides of the $AB$-line. Find with straightedge and compass the vertices of the ellipse through $A,B,C$ such that $TA,TB$ are the tangents to the ellipse in $A,B$.

Exploting the usual projective properties of a conic it is not difficult to draw the tangent at $C$ in a few steps. But what is the most effective way to find the vertices of an ellipse, given three tangents?

$^{(*)}$^{The mentioned well-known problem is the following: $U,V,W$ are three distinct points inside a circle $\Gamma$. Find a triangle $ABC$, inscribed in $\Gamma$, such that $U\in BC,V\in AC,W\in AB$. This is usually solved through Pascal's theorem.}

Answer 1

The line through $T$ and the midpoint of $\overline{AB}$ passes through the center of the ellipse. (In fact, it's one of the axes, which will be important later.) Moreover, if the tangents at $B$ and $C$ meet at, say, $U$, then the line through $U$ and the midpoint of $\overline{BC}$ also passes through the center. With two lines pinpointing its location, we may take the center of the ellipse as known.

Consequently, we can start our investigation from here (with considerable changes of notation):

Given point $P$ on an ellipse with center $O$, and a point $X$ on one of the ellipse's (extended) axes such that $\overleftrightarrow{PX}$ is tangent to the ellipse, how do we construct the endpoints of the given axis?

If our ellipse were a circle with radius $r$, then we'd know that $$|\overline{OM}|\;|\overline{OX}| = r^2$$ Since an ellipse is a circle deformed by scaling in the directions of its axes, and since $\overleftrightarrow{OX}$ is assumed to be the direction of an axis (I told you that would be important!), the relation is preserved in this form: $$|\overline{OM}|\;|\overline{OX}| = a^2$$ where $a$ is the "radius" for that axis.

This says exactly that $a$ is the geometric mean of $|\overline{OM}|$ and $|\overline{OX}|$, making it an easily-constructible length. We simply build a semi-circle with diameter $\overline{OX}$; the point $A$ at which the perpendicular at $M$ meets the semi-circle is such that $\overline{OA}$ is a leg of a right triangle with hypotenuse $\overline{OX}$; moreover, thanks to similar triangles $\triangle AOM$ and $\triangle XOA$, we have that $|\overline{OA}|$ is the geometric mean we seek. Therefore, the circle about $O$ through $A$ has radius $a$, and where this circle meets $\overleftrightarrow{OX}$ are the endpoints of corresponding axis of the ellipse. $\square$

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So, once you have the center of the ellipse, construction involves three steps:

• Construct the midpoint of $\overline{OX}$ (not shown).
• Construct the semicircle about that midpoint, through $O$ and $X$.
• Construct the circle about $O$, through the intersection $A$ of the semicircle and $\overleftrightarrow{PM}$.

The endpoints of the other axis can be determined in the same way, based on the point $Y$ where tangent at $P$ meets the perpendicular through $O$:

Note that this construction is not at all helpful for determining non-axis "diameters".

Answer 2

My solution, for now:

1. We take the line $l$ through $BT\cap AC,AT\cap BC$, $D=l\cap AB$, then $DC$ is tangent to the ellipse at $C$;
2. Let $E=AT\cap DC$ and $\Gamma$ be the circle tangent to $AT,BT$ at $A,B$;
3. Let $F=CT\cap\Gamma$ (one of the two intersections);
4. Let $G$ be the intersection between $AT$ and the tangent to $\Gamma$ at $F$;
5. Let $H$ be one intersection between $\Gamma$ and the internal angle bisector of $\widehat{ATB}$;
6. Let $I$ be the intersection between $AT$ and the perpendicular to $TH$ through $H$;
7. Let $J$ be the intersection between $DG$ and the parallel to $IH$ through $E$;
8. Let $K$ be $TJ\cap DI$;
9. Then the parallel to $IH$ through $K$ meet the angle bisector of $\widehat{ATB}$ in a vertex of the ellipse;
10. With three extra lines we may find the opposite vertex;
11. We have an axis of the ellipse, hence we may exploit the fact that an ellipse is the image of a circle under a dilation, find the dilation ratio through the position of $A$, then the other two vertices.

This is a solution with $\color{red}{2}$ ausiliary circles and $\color{red}{22}$ ausiliary lines, a bit too many, I bet you can do better. To translate an algebraic solution in terms of a straightedge-and-compass construction may provide a huge improvement, probably. Here I just exploited the fact that there is a projective map, having $AT$ and $BT$ as fixed lines, that sends $\Gamma$ into our ellipse, then considered polar lines and applied the Desargue's theorem.