MATLAB Answers

# Using the ellipse graph.

95 views (last 30 days)
John Lutz on 30 Sep 2017
Answered: Ali Nafar on 13 Jun 2019
In polar coordinates (r,t), the equation of an ellipse with one of its foci at the origin is r(t) = a(1 - e2)/(1 - (e)cos(t)) I'm confused how to set this up, as I have never occurred an ellipse graph before. where a is the size of the semi-major axis (along the x-axis) and e is the eccentricity. Plot ellipses using this formula, ensuring that the curves are smooth by selecting an appropriate number of points in the angular (t) coordinate Thank you.
function untitled3
a = 1/2(b);
e = 0.5;
t = linspace(0,2*pi);
r = a(1 - e.^2)./(1 - (e)*cos(t));
plot(r,t)
axis equal
end
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Answers (3)

Image Analyst on 30 Sep 2017
You need to define b using a and e, not assume b is already defined like you did.
##### 1 CommentShowHide None
John Lutz on 30 Sep 2017
Still does not work, really confused on this.
t = linspace(0,2*pi);
e = 2*pi;
a = 1/2.*(2*pi);
r = (a.*(1 - e.^2) ./ (1 - ((e)*cos(t))));
axis equal
plot(r,t);
also this doesn't work either, tried many times to get this to work but it doesn't.

Sign in to comment.

Henry Giddens on 30 Sep 2017
Your equation ends up with some negative values - (which I'm not sure can be correct?), but if you are using polar coordinates, then use the polarplot or polar commands:
polarplot(t,abs(r))
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

Ali Nafar on 13 Jun 2019
L=0.5;
e=0.5;
phi0=0;
phi=linspace(0,2*pi);
rho=L*(1-e^2)./(1-e*cos(phi-phi0));
polar(phi,rho)
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Community Treasure Hunt

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

Start Hunting!