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I wanted add a Parabola to my Manipulate/Plot, that have the same length and the same hanging points as the Catenary, so that when I Manipulate(Modify) the position of the Points A and B of the Catenary, it will also modify the form of the Parabola. So I'm trying to interpose the Parabola of the Catenary.. Do you have an idea? Here is my code for Manipulating the Catenary:

Needs["PlotLegends`"]
catenary[x_] := a*Cosh[(x - c)/a] + y
m = Manipulate[Module[{root}, (root = FindRoot[catenary[#[[1]]] == #[[2]] & /@ pt, {{y, 0}, {c, 0}, {a, 1}}];
Show[Plot[catenary[x] /. root, {x, pt[[1, 1]], pt[[2, 1]]}, AspectRatio -> 3/2], 
 Graphics[{{Red, Point[pt]}, 
   Text[Style["A " <> ToString[pt[[1]]], "Subsection"], .2 + 
     pt[[1]]], 
   Text[Style["B " <> ToString[pt[[2]]], "Subsection"], .2 + 
     pt[[2]]], 
   Text[Style["C " <> ToString[pt[[3]]], "Subsection"], .2 + 
     pt[[3]]]}], PlotRange -> {{-10, 10}, {-10, 10}}, 
 AxesLabel -> {"X", "Y"}, PlotLabel -> "Kettenlinie (Catenary)", 
 LabelStyle -> Red, AspectRatio -> 1, 
 ImageSize -> 350])], {{pt, {{-5, 5}, {5, 5}, {0, 0}}}, Locator}, {{pt, {{-1, 1}, {1, 1}, {0, 0}}, ""}, InputField, Background -> Lighter[Yellow], FrameMargins -> 3, BaseStyle -> {"ControlStyle", 14}, ImageSize -> 360}]

Thanks a lot for your help..

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For starters: the arclength function of the catenary $y=a\cosh\frac{x}{a}$ is $a\sinh\frac{x}{a}$, while the arclength function for the parabola $x^2=4cy$ is $c(\mathrm{arsinh}(t)+t\sqrt{1+t^2})$, so you are going to have to solve a few nonlinear equations for this... my point being, this is more a math question than a Mathematica question. – J. M.♦ Jun 19 '12 at 12:45
sure, but how to plot and Manipulate both of them at the same time.. i mean interposing the parabola of the catenary in the Plot.. O_o – Zellel B Jun 19 '12 at 12:59
The math's the hard part; once it's done, the translation to Mathematica code is straightforward. Did you at least try to do the derivations for the formula of a parabola satisfying the constraints you gave? – J. M.♦ Jun 19 '12 at 13:14
@J.M. In fact, Mma can't solve the arclen equation as is, neither symbolically nor numerically for the three parabola parameters. – belisarius Jun 28 '12 at 22:33

closed as too localized by J. M.♦ Apr 27 at 5:20

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