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Fimplicit not plotting properly

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Hello,
Im trying to implicity plot conic sections using this deffinition in the app desinger.
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
after inputing
a=2 , 2*h=0, b=0, 2*g=3, 2*f=1, c=7
fimplicit is returning a blank graph although this is a perfectly fine parabola.
when plotting it with a linspace as such:
xx=-10:10;
yy=-2.*xx.^2-7.*xx-7;
plot(xx,yy)
Matlab plots the function perfectly fine. So why does't the fimplicit plot it aswell.
thanks

Accepted Answer

Stephan
Stephan on 3 Dec 2020
Edited: Stephan on 3 Dec 2020
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
a=2;
h=0;
b=0;
g=3/2;
f=1/2;
c=7;
fimplicit(fun1,[-3 2 -12 -5])
  2 Comments
Stephan
Stephan on 3 Dec 2020
No, i also did not get it to work in an automatic way...

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More Answers (1)

Walter Roberson
Walter Roberson on 3 Dec 2020
a=2; h=0/2; b=0; g=3/2; f=1/2; c=7;
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
fimplicit(fun1, [-10 10 -10 10])
Looks plausible to me.
  2 Comments
Walter Roberson
Walter Roberson on 3 Dec 2020
Setting the xlim or ylim has no effect on the range that fimplicit uses to plot, and changing the limits to auto after the call does not trigger an already existing fimplicit to go back and analyze the function to find a range of bounds that will give an interesting plot. fimplicit only draws according to the range passed in or the default for the fimplicit call.
You are not passing in limits and the default limits happen to have no interesting content for this function.

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