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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.
>
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