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using fimplicit function to plot

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I'm trying to make a plot using the fimplicit function but the figure is empty.
zf(1) = figure(1);
za(1) = axes;
c = -4:2:4;
fimplicit(@(x,y) y.^2 - x.^2- c);

Accepted Answer

Ameer Hamza
Ameer Hamza on 2 Oct 2020
For a single graph, c must be a scalar. If you want to plot the graph for all values in c, you can use for-loop or arrayfun() (kind of implicit for-loop)
zf(1) = figure(1);
za(1) = axes;
hold(za(1))
C = -4:2:4;
arrayfun(@(c) fimplicit(@(x,y) y.^2 - x.^2 - c), C);
  2 Comments
Maria Galle
Maria Galle on 2 Oct 2020
How would I plot the graph for all values of c using a for loop? I tried writing the code but get the plot for one value of c.
for c=-4:2:4
zf(1) = figure(1);
za(1) = axes;
fimplicit(@(x,y) y.^2 - x.^2- c);
end
Ameer Hamza
Ameer Hamza on 2 Oct 2020
Move figure() and axes() out of for loop
zf(1) = figure(1);
za(1) = axes;
hold(za(1));
for c=-4:2:4
fimplicit(@(x,y) y.^2 - x.^2- c);
end

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More Answers (1)

John D'Errico
John D'Errico on 2 Oct 2020
Edited: John D'Errico on 2 Oct 2020
As an alternative to the use of fimplicit, you can simply think of this as a contour plot. That is essentially all fimplicit does.
c = -4:2:4;
fxy = @(x,y) x.^2 - y.^2;
H = fcontour(fxy,[-5,5, -5,5],'LevelList',c);
Either way works. A contour plot has the virtue that you can create all level lines in one call and no need to loop.

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