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.