4
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fig1 = ParametricPlot[{(1 + r -12 r^2 + 4 r^3)/(2 r - 2 ), 
   Sqrt[(4 r^2)/(r - 1)^2 (r - r^2 (r - 3)^2)]}, {r, 2, 4}, 
  Axes -> True, Frame -> False, 
  PlotStyle -> {Directive[Yellow, Thick]}]

fig2 = ParametricPlot[{(1 + r - 12  r^2 + 4 r^3)/(
   2 r - 2 ), -Sqrt[(4 r^2)/(r - 1)^2 (r - r^2 (r - 3)^2)]}, {r, 2, 
   4}, Axes -> True, Frame -> False, 
  PlotStyle -> {Directive[Yellow, Thick]}, 
  PlotRange -> {{-12, 12}, {-8, 8}}, AspectRatio -> Automatic]

Show[fig1,fig2]

I dont draw of the inner region of this function (only one-parameter)? please help.

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  • $\begingroup$ I think you just need Show[fig1, fig2, PlotRange -> All] $\endgroup$ – george2079 Mar 22 '16 at 17:36
  • $\begingroup$ Hi, thank you your answer but this isnt satisfy. ı need like a regionplot. $\endgroup$ – Fatmath Mar 22 '16 at 17:47
  • $\begingroup$ This function doesnt transform to the cartesian format!! $\endgroup$ – Fatmath Mar 22 '16 at 17:53
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Dr. belisarius Mar 22 '16 at 17:53
  • $\begingroup$ i found the answer of my question mathematica.stackexchange.com/questions/10811/… thenk you. $\endgroup$ – Fatmath Mar 24 '16 at 18:41
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We "invent" the second parameter:

f[r_, n_] := {(1 + r - 12 r^2 + 4 r^3)/(2 r - 2), 
              n Sqrt[(4 r^2)/(r - 1)^2 (r - r^2 (r - 3)^2)]}

ParametricPlot[u f[r, 1] + (1 - u) f[r, -1], {r, 2, 4}, {u, 0, 1}, 
               Axes -> True, Frame -> False, PlotRange -> All, PlotPoints -> 50]

Mathematica graphics

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4
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using RegionPlot ( this is awful slow.. )

region = ParametricRegion[{
    { (1 + r - 12 r^2 + 4 r^3)/(2 r - 2),
     z Sqrt[(4 r^2)/(r - 1)^2 (r - r^2 (r - 3)^2)]}, 
    2 < r < 4 && -1 < z < 1} , {r, z}];
RegionPlot[region, PlotRange -> {{-7, 7}, {-7, 7}}]

enter image description here

oddly if we supply the exact bounds on r:

 2 (1 + Sin[\[Pi]/18]) < r < 2 (1 + Cos[(2 \[Pi])/9])

its even slower..

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  • $\begingroup$ Does anybody else get something like this from above code on Windows 10 (64 Bit) using Mathematica 10.4 or am I unique? $\endgroup$ – gwr Mar 22 '16 at 18:32
  • $\begingroup$ Maybe that ugly result also explains why it is not slow at all on my machine? ;-) $\endgroup$ – gwr Mar 22 '16 at 18:33
  • $\begingroup$ I checked it again, v10.1 takes a minute and a half. $\endgroup$ – george2079 Mar 22 '16 at 18:40
  • $\begingroup$ Do you have an explanation for the result I am getting using your solution? (If you are on 10.1 then we might have an explanation for me getting a result in a blink -- which is wrong?) $\endgroup$ – gwr Mar 22 '16 at 18:41
  • 2
    $\begingroup$ @gwr, I get the same result as you, 10.4 $\endgroup$ – RunnyKine Mar 22 '16 at 19:10

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