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Re: ImplicitPlot errors


Hello:
I'm using Mathematica v5 and I'm not having problems

ImplicitPlot[{x^2 + x*y + y^2 == 7, y == 2 Sqrt[7/3]}, {x, -5, 5}]
   It plots an inclined ellipse and an horizontal tangent line.
I suspect it's because you've defined "y" before, so Mathematica uses 
this value
in the first equation and doesn't see the other variable it needs.
Type
   Clear[y];
and the execute the plotting. This should work. The other thing you 
could do is using variable "y" in a local enviroment with Block or 
Module..
Hope this helps
However, I do get your error message when I try to plot with respect 
to y {y,-5,5}. I guess because is because equation y==2 sqrt[7/3] 
doesn't have  a distinct (other than y) variable.


Atte. Andres Guzman


David Rees <w3bdevilREMOVE at THISw3bdevil.com> ha escrito:

> Hi,
>
> I've been trying to get ImplicitPlot to plot an implicit function (naturally
> ;) ), but it throws errors to cryptic for me, even when copying and pasting
> from the Mathematica function reference.
>
> In[40]:= ImplicitPlot[{x^2 + x*y + y^2 == 7, y == 2*Sqrt[7/3]}, {x, -5, 5}]
>
> ImplicitPlot::var :
>
> Equation x^2+x
> Function[x,x^2+2xy-3y^2-16]+Function[x,x^2+2xy-3<<1>>-16]^2==7 does not have
> a single variable other than x
>
> ImplicitPlot::var :
>
> Equation Function[x,x^2+2xy-3y^2-16]==2Sqrt(7/3) does not have a single
> variable other than x
>
> Out[40]:=\!\(ImplicitPlot[{x\^2 + x\ Function[x, x\^2 + 2\
>           xy - 3\ y\^2 - 16] + Function[
>               x, x\^2 + 2\ xy - 3\ y\^2 - 16]\^2 == 7, Function[x, x\^2 + 2\
>               xy - 3\ y\^2 - 16] == 2\ \@\(7\/3\), x == 2\ \@\(7\/3\)}, {x,
> \
> \(-5\), 5}, {Function[x, x\^2 + 2\ xy - 3\ y\^2 - 16], \(-5\), 5}]\)
>
>
>
> What am I doing wrong? Even ImplicitPlot[x+y==2,{x,-4,4}] fails.
>
> Thanks
>
>
>
>



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