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For the parabola y = a x^2+ b x + c I want to see the locus of the vertex as b varies (with a and c to be parameters). What are the ways to plot the locus of the parabola dynamically in Mathematica?

I am very new to Mathematica, so I could figure out only basic Manipulate:

Manipulate[
  Plot[a x^2 + b x + c, {x, -10, 10}, PlotRange -> 40, AspectRatio -> 1], 
   {a, -20, 20}, {b, -20, 20}, {c, -10, 10}
 ]

I would like to see the locus parabola appearing in the same manner as in the post, if possible.

Thanks.

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  • $\begingroup$ Can you give more details about exactly what you want? As it is, this is likely to be closed as being unclear what you're asking. $\endgroup$ – march Mar 25 '16 at 17:43
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    $\begingroup$ I want something like this. A animation where the locus of the vertex of the parabola is plotted as i manipulate a,b,c. $\endgroup$ – Lokesh Jaddu Mar 26 '16 at 1:59
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Manipulate[
 plot1 = Plot[a x^2 + b x + c, {x, -10, 10}, PlotRange -> 40, 
   AspectRatio -> 1, ImageSize -> Small];
 plot3 = If[a == 0, Graphics[], 
   ParametricPlot[{-u / (2 a), a (-u / (2 a))^2 + u (-u / (2 a)) + c}, {u, -b, b}, 
    PlotStyle -> {Red, Thick}]];

 Show[plot1, plot3],

 {{a, -2}, -20, 20}, {b, -20, 20}, {c, -10, 10}, 
 TrackedSymbols :> True]

enter image description here

Edit Another method to visualize locus:

Manipulate[
  f[x_, i_] := a x^2 + (b + i) x + c;
  tmp = Table[f[x, i], {i, -20, 20, 5}];

  plot1 = Plot[tmp, {x, -10, 10}, PlotRange -> 40, AspectRatio -> 1];
  plot2 = Plot[Tooltip[-a x^2 + c, "Locus"], {x, -10, 10}, PlotStyle -> {Black, Thick}];

  Show[plot1, plot2],

{a, -20, 20}, {b, -20, 20}, {c, -10, 10}, TrackedSymbols :> True]

enter image description here

Edit2 Note on connection between first and second parts. Notice in 'plot3' we have used parametric form.

Solve[{x == -b / (2 a), y == a x^2 + b x + c}, {x, y}] // FullSimplify

$y=c-\frac{b^2}{4 a}$ and $x^2=\frac{b^2}{4 a^2}$.

So, we may come back from parametric to cartesian equation $y= - a x^2 + c$

(this equation is used in plot2).

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  • $\begingroup$ Thank you. And how to find the equation of that locus? $\endgroup$ – Lokesh Jaddu Mar 26 '16 at 15:52
  • $\begingroup$ @LokeshJaddu, in the second visualisation it is -a x^2 + c. In practice it is more complicated, but in Manipulate your a and c change sign from - to +, so we may use this very simple representation. $\endgroup$ – garej Mar 26 '16 at 16:03
  • $\begingroup$ I understood that its -a x^2 + c . But isn't there a way to let mathematica figure that equation out for me? $\endgroup$ – Lokesh Jaddu Mar 26 '16 at 16:08
  • $\begingroup$ You can use GroebnerBasis[] for eliminating parameters. $\endgroup$ – J. M.'s ennui Mar 27 '16 at 0:43

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