Re: Graphics << Implicit vs ContourPlot

*To*: mathgroup at smc.vnet.net*Subject*: [mg122772] Re: Graphics << Implicit vs ContourPlot*From*: Stefan Salanski <wutchamacallit27 at gmail.com>*Date*: Thu, 10 Nov 2011 06:52:21 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j9doii$b4h$1@smc.vnet.net>

On Nov 9, 6:35 am, John Accardi <johnacca... at comcast.net> wrote: > Goal: Show students a plot of both an ellipse with x axis as ellipse's > transverse axis > and the same ellipse in an x'y' coordinate plane that is rotated some angle > with respect the the original xy coordinate plane. (All in one plot) > > I used: > > << Graphics`ImplicitPlot`; ImplicitPlot[{7 x^2 - 6 Sqrt[3] x y + > 13 y^2 - 16 == 0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, > y == -x}, {x, -3, 3}, AspectRatio -> 1.25] > > which works well but I had to hard fix the axis of rotation at 45 degrees > and plot it (y == x and y == -x). I also get an obsolete warning and > the suggestion to use the new ContourPlot for this in the future: > > General::obspkg: "\!\(\"Graphics`ImplicitPlot`\"\) is now obsolete. > The legacy version being loaded may conflict with current Mathematica > functionality. See the Compatibility Guide for updating information." > > So I try to accomplish the same graph with ContourPlot: > > ContourPlot[{7 x^2 - 6 Sqrt[3] x y + 13 y^2 - 16 == > 0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, y == -x, y == 0, > x == 0}, {x, -3, 3}, {y, -3, 3}, AspectRatio -> 1.25] > > which gets me close but I have lost traditional plotting of the xy axes > (no tick marks). Instead I get ContourPlots boxed style coordinate system. > > Question: How can I get my old style axes back in the context of > ContourPlot? > > Thank you for any insights. Under documentation for ContourPlots -> Options -> Axes, the first example right there is to turn off the framed tick marks and turn on the traditional axes. ContourPlot[{7 x^2 - 6 Sqrt[3] x y + 13 y^2 - 16 == 0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, y == -x}, {x, -3, 3}, {y, -3, 3}, AspectRatio -> 1, Axes -> True, Frame -> False] Also, in response to your other post about the aspect ratio needing to be 1.25 to get 90 degree intersections, this is not the case for me. aspect ratio 1 works as I would expect it to. maybe your monitor/ projector has a distorted aspect ratio? Try using AspectRatio -> Automatic, which should give a "natural" aspect ratio (as mentioned in ContourPlots->Options->AspectRatio) -Stefan Salanski