I have a short section of an ellipse (red) and I wish to find the focal points of that ellipse. I can find the normals at 3 (or more) points on the section.
Given that the normals subdivide the angles between the points and the focal points; can I work backwards to find those focal points?
Update: I can certainly pick 6 points and find their normals and tangents, but my ellipse section will always be (less than) 1 quadrant of the ellipse; and I can't see how to construct the self-crossing hexagon required for Pascal's theorom?
Here's a geometric construction of the ellipse, given three points $P$, $Q$, $R$ on it and two tangents $RT$, $QT$ (see figure below). I won't prove all the details, please ask to know more.
Construct the midpoint $V$ of $QR$: line $TV$ is then a locus of center $C$ of the ellipse. From $P$ construct a line parallel to $QR$, meeting the tangents at $L$ and $L_1$. Construct point $K$ on $LL_1$ such that $LK=\sqrt{LP\cdot L_1P}$; line $RK$ intersects line $VT$ at $D$, which is a point on the ellipse. Center $C$ is that point on $DT$ satisfying: $$ DC:TC=DV:TD $$ (a possible construction of $C$ is shown in the figure: construct $Dv=DV$ parallel to $Td=TD$ and intersect $dv$ with $DT$).
Finally, draw through $C$ a line parallel to $QR$, intersecting tangent $TQ$ at $t$. Let $n$ be the intersection between $Ct$ and the parallel to $TD$ throgh $Q$. Construct $A$ on $Ct$ such that $CA=\sqrt{Cn\cdot Ct}$: point $A$ is then on the ellipse and $CA$, $CD$ are a pair of conjugate semidiameters.
If you want to construct the axes of the ellipse, follow the procedure outlined here.