A Construction for Arcs on a conic:
Here comes a non-trivial Solution to a non-trivial problem.
How to get an arc on a conic.
(Perhaps I will implement a method ArcOnConic based on this if I have more time).
In fact it is a nice exercise in projective geometry.
So assume you want to have an arc on a conic C through three points a,b,c on C.
You need a circle K and three points a' b' c' on the circle and furthermore (and this is the Problem)
a projective transformation that maps
a' --> a
b' --> b
c' --> c
and
K --> C.
You can specify such a Transformation by the image of four points. So you need one more point pair d, d' with
d' --> d
such that also K --> C.
Now you need a property of conics and crossratios.
If you have four points fixed a,b,c,d on a fixed conic and a fifth point e on the conic the the
crossratio of the four lines|e,a|,|e,b|,|e,c|,|e,d| is independent on the choice of e. Thus we can call it the
crossratio of a,b,c,d with respect to C.
Thus we can choose an arbitrary point d on the conic.
d' on the circle must be chosen such that the
crossratio of a,b,c,d w.r.t C is the same as the crossratio of
a', b', c', d' w.r.t K.
So how do we get the point d'.
One possibility (also a tricky one) is as follows:
intersect the four lines|e,a|,|e,b|,|e,c|,|e,d| with a line.
your get four intersections a'', b'', c'', d''. Now define a Moebius transformation
that maps
a'' --> a'
b'' --> b'
c'' --> c'
the Moebius transformation maps d'' also to the circle K....
and (since the world is nice and miracles occur sometimes)
maps it to the position with exactly the right cross ratio. Call the mapped point
d', define the projective transformation
a' --> a
b' --> b
c' --> c
d' --> d
draw the arc a', b', c'
and map it through the projective transformation.
Et voila -- your done.
Uff it took me two minutes to do the construction. And 20 minutes to explain it.
