My intention is to demonstrate on any conic various metrics properties on points belonging to the conic and connected by lines.
I want the points to be dynamically moved not the usual way, indirectly with the help of any Manipulate controls like Slider, but by an artifice, hiding the Locator control object and let one feel that by dragging any point on the curve you manipulate directly that point (though of course it is the locator that you manipulate).
Here is my embryonic code so far, just a circle with two points (and I made here the 2 locators visible):

findY[ptX_] := Module[{},
Which[((First[ptX] <= 0 && ptX[[2]] <=  0) || (First[ptX] > 0 && 
   ptX[[2]] <=  0) ),                  
   y /. Solve[x^2 + y^2 == 4, {y}][[1]] /. {x -> First[ptX]},
             ((First[ptX] <= 0 && 
   ptX[[2]] >  0) || (First[ptX] > 0 && ptX[[2]] >  0) ) ,                  
     y /. Solve[x^2 + y^2 == 4, {y}][[2]] /. {x -> First[ptX]}]]

 "Locator for the blue point `1` Blue point coords{`2`,`3`}", 
 Dynamic[ptA], First[ptA], findY[ptA]]]},
 "Locator for the green point `1` Green point coords{`2`,`3`}", 
 Dynamic[ptB], First[ptB], findY[ptB]]]}}, 
Row[{LocatorPane[Dynamic[{ptA, ptB}], 
ContourPlot[x^2 + y^2 == 4, {x, -3, 3}, {y, -3, 3}, 
 PlotRange -> Automatic, AspectRatio -> 1, Axes -> True, 
 Frame -> None, ImageSize -> 300, 
 Epilog -> {{PointSize[Large], Blue,  
    Point[Dynamic[{First[ptA], findY[ptA]}]]},
                     {PointSize[Large], Green, 
    Point[Dynamic[{First[ptB], findY[ptB]}]]}}],
Appearance -> Style["\[CirclePlus]", 20, Red]]}], 
(* Replace with Appearance->None to hide the Locators *)
{{ptA, {-2, 0}},None}, {{ptB, {2, 0}}, None}] (* ptA Blue ptB Green *)     

You can see that by clicking on any of the two locators each point will move along the whole circle, even cross each other: if you move your mouse slowly the locator will stay very close to the point but if you do it too quickly the point will move too but the locator may be far off giving an erratic effect to the movement of the point.
And an issue too, having difficulty of retrieving it then with your mouse if the locator is not visible.
My question: On a PC when you want to drag a point which is to be moved along a curve, you start by clicking on the point and then dragging where you want it to be (and not something else). Can the same effect be obtained be obtained with MMA? If internally this means dragging a MMA Locator it should be as far as possible transparent.

I took pains to check that my question is not a duplicate,sorry if it is. Errare humanum est.

In short: How can I constrain a locator to move along a curve, similar to the goal in Constrain movement of a locator inside Manipulate, except that here the curve is defined by an equation instead of parametrically?

  • $\begingroup$ You say, "I took pains to check that my question is not a duplicate". Hard to believe when searching this site with "constrain locator" finds several pertinent hits. I think this is a duplicate of this question $\endgroup$
    – m_goldberg
    May 3, 2014 at 14:17
  • $\begingroup$ @SigismondKmiecik I'm sorry for my words. I was too harsh. Let me ask a question: you want to constrain locator within some radius around the point, the point which coordinates are calculated based on locator position. Am I right? p.s. I'm in EU too :) $\endgroup$
    – Kuba
    May 3, 2014 at 19:03
  • $\begingroup$ @SigismondKmiecik I took the liberty of rephrasing your question. If I missed the point, please feel free to roll back the edit. $\endgroup$
    – Michael E2
    May 5, 2014 at 13:32
  • $\begingroup$ @MichaelE2 Yes, you can update my question like that if you are sure that it can only be answered by restricting a locator (that I would like not to see). My question is more general and open I don't assume that restricting a locator is the only way, even if I started coding like that because I know of no other. Kuba gave me a nice answer based on restricting transparently a locator. $\endgroup$
    – Sigis K
    May 5, 2014 at 22:57
  • $\begingroup$ @SigismondKmiecik You may want to delete old not relevant comments. $\endgroup$
    – Kuba
    May 6, 2014 at 19:32

1 Answer 1


Here's a way to constrain the locators to the curve. We use ContourPlot to generate a list of points on the curve; the nearest one, returned by the NearestFunction nf, is used as a starting point for FindRoot to solve for the nearest point on the curve.

Clear[findP, loc];
findP[p0_, conicEqn_, {x0_, y0_}] :=
  {x, y} /. FindRoot[
    {Cross[{x, y} - p0].D[Subtract @@ conicEqn, {{x, y}}] == 0, 
    {{x, x0, -3, 3}, {y, y0, -3, 3}},
    AccuracyGoal -> 3, PrecisionGoal -> 4];

loc[Dynamic[pts_], conic_] := DynamicModule[{nf},
   nf = Nearest[Cases[
      ContourPlot[conic, {x, -3, 3}, {y, -3, 3}, PlotPoints -> 100, 
        MaxRecursion -> 4, Axes -> None, Frame -> None],
      {_Real, _Real}, Infinity]];
   pts = findP[#, conic, First@nf[#]] & /@ pts;
      Dynamic[pts, (pts = findP[#, conic, First@nf[#]] & /@ #) &], 
        {x, -3, 3}, {y, -3, 3},
        AspectRatio -> 1, Axes -> True, Frame -> None, 
        ImageSize -> 200],
      Appearance -> Style["\[CirclePlus]", 20, Red]]}

Manipulate[{ptA, ptB},
 {{conic, x^2 + y^2 == 4}, InputField},
   loc[Dynamic[{ptA, ptB}], conic],
   TrackedSymbols :> {conic}],
 {{ptA, {-2, 0}}, None}, {{ptB, {2, 0}}, None}

Mathematica graphics

Mathematica graphics

  • $\begingroup$ Michael I understand your idea: the locator is the point that is dragged.When I manipulate your solution it seems that the locator is always precisely on the conics.But after having modified your answer by Appearance -> Style[".", 40, Red]] the point can be slightly off the conic and this baffles me. Any idea why ?(By the way there is a comma too much in your last line). See youtube.com/watch?v=h9VCsX54irM which gave me the idea to reproduce something similar with MMA (but far for me to compare Geogebra with MMA) . $\endgroup$
    – Sigis K
    May 6, 2014 at 19:20
  • $\begingroup$ It seems off center because the period lies below the center point of character. Look at Framed[".", FrameMargins -> 0]. Use something like Graphics[{Red, Disk[]}, ImageSize -> 8] if you want a dot on the curve. (Thanks for pointing out the typo.) $\endgroup$
    – Michael E2
    May 6, 2014 at 19:38
  • $\begingroup$ @SigismondKmiecik The error might originate in fline. I can get it to show a line with my own definition. (Comments are not the best place to work this out. Another question, maybe, or someone might help in chat.) $\endgroup$
    – Michael E2
    May 10, 2014 at 1:55

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