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Re: PolarPlot


PolarPlot[Flatten[Table[
   {2 a Sqrt[Cos@t], -2 a Sqrt[Cos@t]},
   {a, 7}]], {t, 0, 2 \[Pi]},
 AspectRatio -> 1,
 Frame -> True, Axes -> False]

Cos[ArcTan[y/x]]

1/Sqrt[y^2/x^2 + 1]

ContourPlot[Evaluate[Table[
   (x^2 + y^2) == 4 a^2 /Sqrt[y^2/x^2 + 1],
   {a, 7}]], {x, -15, 15}, {y, -10, 10},
 AspectRatio -> 1,
 ContourLabels -> (Text[Sqrt[Numerator[#3[[2]]]/4], {#1, #2}] &)]

Put the arrow over one of the lines in the ContourPlot.


Bob Hanlon

---- Bruce Colletti <vze269bv at verizon.net> wrote: 
> Re 6.0.2 under WinXP.
> 
> I'm using the command below to plot the polar equation r^2 = 4a^2 Cos@theta.  Is there a better way?  Thankx.
> 
> Bruce
> 
> PolarPlot[{2 a Sqrt[Cos@t], -2 a Sqrt[Cos@t]}, {t, 0, 2 \[Pi]}, 
>  PlotRange -> 7]
> 



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