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

*To*: mathgroup at smc.vnet.net*Subject*: [mg99071] Re: [mg99051] ContourPlot, equation and R.H. side of*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sun, 26 Apr 2009 01:37:49 -0400 (EDT)*Reply-to*: hanlonr at cox.net

Try these ContourPlot[3 x^2 + 6 y^2 == 6, {x, -2, 2}, {y, -2, 2}, Axes -> True] Attributes[ContourPlot] {HoldAll,Protected} eqn1 = 3 x^2 + 6 y^2 == 6; ContourPlot[Evaluate[eqn1], {x, -2, 2}, {y, -2, 2}, Axes -> True] eqn2 = 3 x^2 + 6 y^2; ContourPlot[eqn2 == 6, {x, -2, 2}, {y, -2, 2}, Axes -> True] ContourPlot[eqn2, {x, -2, 2}, {y, -2, 2}, Axes -> True, Contours -> {6}] Bob Hanlon ---- Bill <WDWNORWALK at aol.com> 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.