Re: ContourPlot, equation and R.H. side of equation_Plotting problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg99092] Re: ContourPlot, equation and R.H. side of equation_Plotting problem*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Sun, 26 Apr 2009 01:41:46 -0400 (EDT)*References*: <gsuirn$dr2$1@smc.vnet.net>

Hi, ContourPlot[Evaluate[eqn1], {x, -2, 2}, {y, -2, 2}, Axes -> True] ? Regards Jens Bill wrote: > Hi: > > 1a.) When I assign 3 x^2 + 6 y^2 == 6 to eqn1, in Mathematica like this: > > eqn1=3 x^2 + 6 y^2 == 6; > > I can't get ContourPlot to plot eqn1 using this code: > > ContourPlot[eqn1, {x, -2, 2}, {y, -2, 2}, Axes -> True] > > > 1b.) If I assign the equation like this without the constant on the R.H. side in eqn2, > ContourPlot will plot the equation as expected, using the following syntax: > > eqn2=3 x^2 + 6 y^2; > ContourPlot[eqn2 == 6, {x, -2, 2}, {y, -2, 2}, Axes -> True] > > > Question: How can I get method 1a to work? Could you please give me code for this? > > > > Thanks, > > Bill > > > PS. I'm using Mathematica 6.0.1 w/ Win XP on a PC. >