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How can I plot the hyperbola represented by the following complex equation:

$$\left|z-a\right|-\left|z-b\right|=2t\tag{1}$$

Should I convert (1) into the following form?

ContourPlot[x^2 - y^2 == 1, {x, -2, 2}, {y, -2, 2}]

contour plot of hyperbola

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  • $\begingroup$ You can start from ComplexExpand[Abs[x + I y - a] - Abs[x + I y - b], TargetFunctions -> {Re, Im}], which doe not give an hyperbola though. $\endgroup$ Commented Jul 21, 2013 at 8:59
  • $\begingroup$ Are you sure the complex expression above is a hyperbola? What do the $a, b, t$ stand for? Assuming that the $a$ and $b$ are the same as in $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, then that is not a hyperbola. For instance, what happens when $a=b$? $\endgroup$ Commented Jul 21, 2013 at 16:36
  • $\begingroup$ You can the function defined here: Coloring a shape according to a function, e.g. ParametricPlot[cc[u, 3/2], {u, 0, 2 Pi}, Axes -> False, PlotStyle -> Thick, ColorFunction -> Function[{x, y, u}, Hue[u/(2 Pi)]], ColorFunctionScaling -> False, Exclusions -> {Cos[u] == 2/3, Cos[u] == -2/3}] $\endgroup$ Commented Jul 21, 2013 at 18:21

1 Answer 1

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Since you're trying to plot an implicit equation, you would normally use ContourPlot. For example, to get two branches satisfying the equation for $t = 1$, just do this:

With[{a = 2, b = 1}, ContourPlot[ Abs[x + I y - a] - Abs[x + I y - b], {x, -1, 3}, {y, -2, 2}, Contours -> {1/2, -1/2}, ContourShading -> False, ContourStyle -> ColorData[1][1] ] ]

contours

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