Re: drawing tilted ellipses?

*To*: mathgroup at smc.vnet.net*Subject*: [mg20457] Re: drawing tilted ellipses?*From*: "Kai G. Gauer" <gauer at sk.sympatico.ca>*Date*: Wed, 27 Oct 1999 02:04:34 -0400*References*: <7v3bmr$5te@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Yes Use ImplicitPlot (or implicitplot3d). Then, all that you need to do is instead of typing Axx+Cyy==F, you'd type Axx+Bxy+Cyy+Dx+Ey==F, with A...F being your favorite and appropriate constants.Better yet, why not just define some g:{A,B,..,F)== k a bijective function that regulates your parameters when you take g inverse? The analagous case would also hold for 3+ space, with a bunch of notational changes (which, I for one must admit should be unnecesssary for mathematica standards). However, remember that by using three space coordinates, you get to add in much more terms such as xyz, xxy, zy, zx, etc (with each of those little terms of x and terms of y and terms of z being such that the sum of the degree of each of x,y,z taken seperately and then added together should each be less than 3) to eliminate those pesky unwanted terms such as Fx that shift you to the left/right of the axis, just set some of your values such as D, E ==0 or whatever else you may want to fix it to). I personally find that allowing the matrix transformation theory of coordinates is underused in Mathematica, and this is why I'd recommend sticking with my style. From then, you might even be able to get away with mapping single coordinate values of your matrix back and forth to a function that would then evaluate in Implicit plot (or ....Plot3d) For further info as to what I've been trying to do with 2&3 dim conic sections/ quadratic surfaces, check the archives for a few of my postings. (My concern is more about calculating the foci effeciently and other parameters related to it.) I have yet to find any transformation theory that covers what I am trying to indirectly possibly get to as a corolllary (other than the fact that there exists one such (3x3) matrix whose det gives the values of what type of conic a 2- space curve should be... puzzling... why wouldn't the "best matrix" for this job be a 2x2 form (or something relating the matrix dimensions to the space that the curve should exist in)...and why the det and not the trace or some other well behaved quadratic operator? I'd love to hear someone talk about the forthcomings of this side of the theory (and why there hasn't been much progress on this branch of coordinate geometry lately) a good author is Loney '50's if you can find him.) Joe Strout wrote: > snip > Any tips? > > Thanks, > -- Joe > > -- > ,------------------------------------------------------------------. > | Joseph J. Strout Biocomputing -- The Salk Institute | > | joe at strout.net http://www.strout.net | > `------------------------------------------------------------------' > Check out the Mac Web Directory! http://www.strout.net/macweb.cgi