[help-3dldf] Re: Constructing ellipse from 4 points

[Laurence Finston]

Just in case anyone's interested, I've found a solution
to the problem of finding the ellipse determined by 6 (not 4)
non-colinear points.  It's called the "Braikenridge-Maclaurin
Construction" and is the converse of Pascal's Theorem.  I found it here:
http://www.maths.gla.ac.uk/~wws/cabripages/trace.html
and
http://www.maths.gla.ac.uk/~wws/cabripages/conpascal.html

It seems to require 6 to start with, but then you use the
other 5 to find as many more as you want.

I've attached an image and the 3DLDF code that generates it.
Since it's 2D, I could have done it with MetaPost, but converting
it should be straightforward, if anyone is so inclined.

The text on the first website states that the ellipse is traced by F when
the line XYZ is rotated about X.  I believe this to be an error.  At least,
I couldn't get it to work.  Instead, I shifted Z' along the line AE and
took Y' as the intersection point of XZ' and BE.

Since I can generate as many points as I want this way, I believe it
should be possible to find a pair of parallel secants, or a pair of
conjugate axes.  There are then easy ways to find the main axes,
center, and foci.  I think I will do this another day, though.

For the intersection of a plane with an ellipsoid,
I'll have to transform my points to a major plane and then transform the
ellipse using the inverse of the first transformation, but I do this often
anyway.  So, with a bit of luck, I should be be able to solve the problem
now.  Then I can start working on the intersection of a line with an
ellipsoid.

Laurence

ellpcn01.ps.gz
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ellpcn01.ldf
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