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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg99051] ContourPlot, equation and R.H. side of equation_Plotting problem
  • From: Bill <WDWNORWALK at aol.com>
  • Date: Sat, 25 Apr 2009 04:50:35 -0400 (EDT)

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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