We had this problem once before. Check out
and scan for "ellipse". I think you'll want to skip over the responses
by someone named "rusin"; the others are much clearer.
In the last sentence of that document you wrote:
> The more general problem, of constructing the foci from 5 _arbitrary_
> points on the ellipse, seems even more formidable.
If I remember well this was a standard problem in a course about
projective geometry that I took some (well, many) years ago. The
solution was more or less as follows:
1) Given the points M, N, P, Q and R use Pascal's theorem (limit case)
to construct the tangent m to the ellipse at M (the rule suffices).
2) Draw a circle c tangent to m at M. This circle is homologous to the
ellipse, with M as center of homology (this last word is used here with
the meaning it has in the context of projective geometry, of course).
3) Let N', P' and Q' be the (second) points where MN, MP and MQ meet
the circle c, respectively. The axis e of the homology is determined by
the points NP.N'P' and PQ.P'Q'.
4) Let S be the point where P'Q' intersect a parallel to PQ by M. The
parallel i' to the axis e by S' is the "limit line", i.e. the homologous
of the line i at infinity. Let X' be the intersection of m and i'. With
center X' and radius X'M intersect i' at points Y' and Z'. Since the angle
Y'MZ' is right, MY' and MZ' give the directions of the principal diameters.
5) Let O' be the pole of the line i' with respect to c. Let the line O'Y'
meet c at points A' and B', and let O'Z' meet c at C'and D'. The points A,
B, C and D homologous of A', B', C' and D' are the vertices of the ellipse.
The point O = AB.CD (homologous of O') is the center. The foci are easily
found: if AB is the major axis draw a circle with center C and radius OA
and intersect it with AB.
In other post cited in the same document you say:
>Back in the century in which emperors and presidents gave mathematical
>proofs there was so much collective understanding of Euclidean constructions
>that someone would have leapt in to demonstrate the elegant geometrical
It seems that I'm older than I thought :-(
José H. Nieto
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Practically do the following:
Centre: Draw two arbitrary parallel chords of
the given ellipse.
Find the midpoints of the two chords.
Draw the chord of the ellipse passing
through the two midpoints.This chord
is a diameter of the ellipse.
Find the midpoint of the last chord
drawn,this is the centre of the ellipse.
Axes: Draw a circumference centred on the
centre of the ellipse so as to
intersect it at four points.
Choose three of these points and draw
the two mutually perpendicular chords
passing through them.
Draw two chords passing through the
centre of the ellipse and
parallel to the above two chords.
They are the major and minor axes of
Foci: Draw a circumference centering on one
of the vertices of
the minor axis and using half lenght
of the major axis as the radius.
The two intersections with the major
axis are the foci.