I'm trying to create an ellipse in parametric form. I have just two Foci along the major axis. So, how shall I get r1, r2 etc.
point1 = 10 8
point2 = 25 20
The syntax is
t = linspace(0, 2*pi, 200);
xt = r1 * cos(t) + xc;
yt = r2 * sin(t) + yc;
cot = cos(theta); sit = sin(theta);
x = xt * cot - yt * sit;
y = xt * sit - yt * cot;
plot(x, y, '-');

1 comentario

Ramesh Bala
Ramesh Bala el 26 de Jul. de 2018
Editada: Matt J el 26 de Jul. de 2018
% or in other general forms :
% data
t = linspace(0,2*pi,80); % the domain of the parameter t.
x = @(t)r1.*cos(t);
y = @(t)r2.*sin(t);
figure(1)
ELLIPSE = plot(x(t),y(t));
hold off

Iniciar sesión para comentar.

 Respuesta aceptada

Matt J
Matt J el 26 de Jul. de 2018
Editada: Matt J el 26 de Jul. de 2018

1 voto

You cannot determine the ellipse with only the foci. You also need to know r1. Once you know that however, you can determine r2 readily,
xc_yc=(point1+point2)/2;
c=norm(point1-xc_yc);
r2 = r1^2-c^2;

6 comentarios

Ramesh Bala
Ramesh Bala el 27 de Jul. de 2018
Danke ! how to get that r1. I think r1 = 2a ? major axis length so how to get a from foci (xy ) coords??
Matt J
Matt J el 27 de Jul. de 2018
Editada: Matt J el 27 de Jul. de 2018
The major axis length needs to be given. You cannot derive it from the foci. Two different ellipses with different major axis lengths can have the same foci.
Ramesh Bala
Ramesh Bala el 30 de Jul. de 2018
Alright ! Lemme have a check
Thanks for the answer.
Ramesh Bala
Ramesh Bala el 6 de Sept. de 2018
I couldn't get the proper work out for this. Could you share any code regarding this case
Matt J
Matt J el 6 de Sept. de 2018
I already told you that it was impossible. The foci alone aren't enough.
Ramesh Bala
Ramesh Bala el 7 de Sept. de 2018
yeah I know that ,I thought to take the surrounding rectangle boundary as tangent and then to draw an ellipse ?

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Productos

Versión

R2018a

Preguntada:

el 26 de Jul. de 2018

Comentada:

el 7 de Sept. de 2018

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by