How to plot an Ellipse

I want to plot an Ellipse. I have the verticles for the major axis: d1(0,0.8736) d2(85.8024,1.2157) (The coordinates are taken from another part of code so the ellipse must be on the first quadrant of the x-y axis) I also want to be able to change the eccentricity of the ellipse.

[muhammad arfan]

dears!!!

i have asigned to write a matlab code for 8 point to fit it in ellipse by using least square method..

i am new in using matlab and try my best but my points are not fit on ellipse. i use annealing method so that i have satisfied my teacher by my work. please chk my work and help me.

thanks

arfan khan

clc;
clear all;
close all;

r1 = rand(1);
r2 = [1+rand(1)];  % r2>r1
x0 = 0;
y0 = 0;
N = 8;
n= 100;
x1 = 1;
x2 = 2;
y1 = 1;
y2 = 2;
for i = 1:n
    x = x1 +(x2-x1).*rand(N,1);
    y = y1 +(y2-y1).*rand(N,1);

    f = ((((x./r1).^2) +(y./r2).^2)-1).^2;
    [m,l] = min(f);
    z =.001* exp(10*(1-i/n));
    v = z/2;
%     disp('v');
%     disp(v)
    x1 = x(l)*v;
    x2 = x(l)*v;
%     disp('x1')
%     disp(x1)
%     disp('x2')
%     disp(x2)
%      ay = v./y(l);
%      by = v./y(l);
%     disp(v);
%     
% %     hold on;
       disp('f');
       disp(f);
end
% plot(f,'or')
   plot(x,y, '*b');
x=((x(i)-x0)*cos(z)) - ((y(i)-y0)*sin(z))
    y=(x(i)-x0)*sin(z)-(y(i)-y0)*cos(z)
    xa(i)=rand(1)
    x(i)= a+(b-a)*rand(1);
    y(i)= rand(1);
    for 
        m(i) = ((((x).^2)/a^2) + (((y).^2)/b^2)-1).^2
        end
         hold on;

Answer 1

[Roger Stafford]

Let (x1,y1) and (x2,y2) be the coordinates of the two vertices of the ellipse's major axis, and let e be its eccentricity.

 a = 1/2*sqrt((x2-x1)^2+(y2-y1)^2);
 b = a*sqrt(1-e^2);
 t = linspace(0,2*pi);
 X = a*cos(t);
 Y = b*sin(t);
 w = atan2(y2-y1,x2-x1);
 x = (x1+x2)/2 + X*cos(w) - Y*sin(w);
 y = (y1+y2)/2 + X*sin(w) + Y*cos(w);
 plot(x,y,'y-')
 axis equal

[Dimitris Arapidis]

that worked nicely, thank you

[Image Analyst]

Several other ellipse and circle entries are in the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_ellipse.3F - check it out.

[Benjamin]

what are the "extra" -y*sin(w) and y*cos(w) terms for in lines 7 and 8?

[Roger Stafford]

@Benjamin. They are to produce the needed rotation by angle w of the ellipse to the slant of the given major axis with respect to the x-y axes. (The angle t in X and Y is needed to produce the required eccentricity.)

[Ramesh Bala]

but in the solution provided above x1 x2 ,y1 y2 are fine. How to get the eccentricity value to get b?

[Ramesh Bala]

or one assume it as e→1 ,it tends to a line,

 e→0 ,it tends to a circle.

[Walter Roberson]

If you already have the minor axis length then you can use that directly in b. If you do not have the minor axis length then you need to calculate it, and you need the information to calculate it; the main other way of characterizing it is to specify the eccentricity. If you do not know the minor axis length and you do not know the eccentricity, then what do you know?

[Ramesh Bala]

ok fine let me put it in parametric type having two foci as x1 y1, x2 y2

determining r1,r2 to get the elliptical form? I think from r1 can get r2.

So how to determine r1.

xt = r1 * cos(t) + xc;

yt = r2 * sin(t) + yc;

[Walter Roberson]

The foci are not enough information to determine the ellipse.

http://www.purplemath.com/modules/ellipse3.htm

[Image Analyst]

Here's a full demo:

% Define parameters.
fontSize = 15;
x1 = 1;
x2 = 20;
y1 = 2;
y2 = 8;
eccentricity = 0.85;
numPoints = 300; % Less for a coarser ellipse, more for a finer resolution.

% Make equations:
a = (1/2) * sqrt((x2 - x1) ^ 2 + (y2 - y1) ^ 2);
b = a * sqrt(1-eccentricity^2);
t = linspace(0, 2 * pi, numPoints); % Absolute angle parameter
X = a * cos(t);
Y = b * sin(t);
% Compute angles relative to (x1, y1).
angles = atan2(y2 - y1, x2 - x1);
x = (x1 + x2) / 2 + X * cos(angles) - Y * sin(angles);
y = (y1 + y2) / 2 + X * sin(angles) + Y * cos(angles);

% Plot the ellipse as a blue curve.
subplot(2, 1, 1);
plot(x,y,'b-', 'LineWidth', 2);	% Plot ellipse
grid on;
axis equal
% Plot the two vertices with a red spot:
hold on;
plot(x1, y1, 'r.', 'MarkerSize', 25);
plot(x2, y2, 'r.', 'MarkerSize', 25);
caption = sprintf('Ellipse with vertices at (%.1f, %.1f) and (%.1f, %.1f)', x1, y1, x2, y2);
title(caption, 'FontSize', fontSize);
xlabel('x', 'FontSize', fontSize);
ylabel('y', 'FontSize', fontSize);

% Plot the x and y.  x in blue and y in red.
subplot(2, 1, 2);
plot(t, x, 'b-', 'LineWidth', 2);
grid on;
hold on;
plot(t, y, 'r-', 'LineWidth', 2);
legend('x', 'y', 'Location', 'north');
title('x and y vs. t', 'FontSize', fontSize);
xlabel('t', 'FontSize', fontSize);
ylabel('x or y', 'FontSize', fontSize);
% Set up figure
g = gcf;
g.WindowState = 'maximized';
g.NumberTitle = 'off';
g.Name = 'Ellipse Demo by Roger Stafford and Image Analyst'

[Maite Osaba]

@Image Analyst I found this implementation really useful! Thanks! Do you think there is a way to plot this so the ellipse is filled with color?

Answer 2

[Azzi Abdelmalek]

a=5; % horizontal radius
b=10; % vertical radius
x0=0; % x0,y0 ellipse centre coordinates
y0=0;
t=-pi:0.01:pi;
x=x0+a*cos(t);
y=y0+b*sin(t);
plot(x,y)

[Ivailo Ivanov]

It was very simple and comprehensible.

[Cynthia Dickerson]

Thanks! This code worked for me perfectly. :)

[Sandy M]

Hi, I do my ellipse graph

A=10; 
B=7.5;
X=-10:.1:10;
Y=(7.5/10)*(1-x^2)^(1/2)
z=-(7.5/10)*(1-x^2)^(1/2)
Plot(x,y,x,z)

Its ok but i need it in cm units cause if i change properties of figure and paper to cm i get deference’s about 3 or 5 mm How can I justify the unit

Answer 3

[Kommi]

For ellipse

>> clc

clear all

%length of major axis

a = input('please enter the length of major axis: ');

b = input('please enter the length of minor axis: ');

x1 = input('please input the x coordinate of the ellipse: ');

y1 = input('please enter the y coordinate of the ellipse: ');

t = -pi:0.01:pi;

x = x1+(a*cos(t));

y = y1+(b*sin(t));

plot(x,y);

Answer 4

[Kate]

how would you plot a ellipse with only knowing some co-ordinates on the curve and not knowing the y radius and x radius?

[Roger Stafford]

That involves more work, Kate. I'll show you a first, easy step and let you struggle through the rest.

It requires a minimum of five points to uniquely determine a conic section - that is, an ellipse, a parabola, or a hyperbola. One way to approach the problem is to find a set of values A, B, C, D, E, and F such that

 A*x^2 + B*x*y + C*y^2 + D*x + E*y + F

is as close to zero as possible for the points (assuming the six constants have been normalized by having the sum of their squares be unity.) With five points it should be essentially zero for each of the points. For more points, it depends on how close they all are to a valid conic section. If they actually lie on some conic, then with the appropriate constants A, B, etc., the above would also be zero at each point.

These constants can be found using matlab's 'svd' singular value decomposition function. Let x be a column vector of all the points' x-coordinates and y a column vector of their corresponding y-coordinates. Do this:

 [U,S,V] = svd([x.^2,x.*y,y.^2,x,y,ones(size(x))]);

Then the desired constants will be in the vector V(:,6), namely, A = V(1,6), B = V(2,6), C = V(3,6), D = V(4,6), E = V(5,6), and F = V(6,6). These are associated with the smallest singular value in S. If that value is zero, then there is a precise match with all points.

Probably the easiest way to proceed from here on in plotting the conic section, would be to solve for y as a function of x, (or possibly x as a function of y,) in the quadratic equation

 A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0

The type of conic section is determined by the discriminant, B^2-4*A*C. If it is negative, the curve is an ellipse, if it is zero, the curve is a parabola, and if it is positive, it is a hyperbola. Recalling high school algebra, there will in general be two y solutions or none for each x, (and two x solutions or none for each y,) using the famous quadratic formula.

Making a successful plot would require a determination of the range of x (or of y) for which there are solutions and then generating a set of x values (or y values,) maybe with linspace, and making two separate plots of the two possible solutions for y (or for x).

Another approach to handling such a quadratic equation involves making the appropriate translation and rotations so as to place the equation in a standard form from which plots can readily be made. This, however, probably involves more effort than the above, though it is more informative.

[Image Analyst]

Perhaps of some interest, if you need to find the ellipse points: http://www.ecse.rpi.edu/homepages/qji/Papers/ellipse_det_icpr02.pdf

[Devi Satya Cheerla]

in the equation of ellipse X2/a2 + Y2/b2 = 1. knowing the points on ellipse, can find a and b. then enter the code below to mathematically compute y and to plot x,y.

code:

x=(0:.01:a); # x value is from 0 to 'a' and discrete with 0.01 scale#
i=1:(a*100+1); # i is to calculate y at every discrete value. it should be for 1 i.e first x value to the last x value.. as it does not have a zero, add 1#
clear y # to clear any previous y value#
for i=1:(a*100+1)
    y(i)=(b^2*(1-(x(i)^2)/a^2))^.5; #from the ellipse equation y=sqrt(b2(1-(x2/a2))#
end
plot(x,y)
hold on
plot(x,-y)
hold on
plot(-x,y)
hold on
plot(-x,-y)

[Sandy M]

hi why u product the nmber with 100?

and, if i want the graph with cm units, what i do? cause i change garaph and paper properties but i still defreces about 4 mm when i prented it

Answer 5

[Omar Maaroof]

you can use

Ellipse2d

[Walter Roberson]

MATLAB does not offer Ellipse2d plotting directly. Instead, the Symbolic Toolbox's engine, MuPAD, offers plot::Ellipse2d https://www.mathworks.com/help/symbolic/mupad_ref/plot-ellipse2d.html which can only be used from within a MuPAD notebook . R2018b was intended to be the last release that included the MuPAD notebook, but it was carried on to R2019a as well.

Answer 6

[Matt J]

Using this FEX download,

https://www.mathworks.com/matlabcentral/fileexchange/87584-object-oriented-tools-for-fitting-conics-and-quadrics

 [x1,y1,x2,y2, e]=deal(1,2,20,8 ,0.85); %hypothetical input

 a = 1/2*sqrt((x2-x1)^2+(y2-y1)^2);
 b = a*sqrt(1-e^2);
 center=[x1+x2,y1+y2]/2;
 theta=atan2d(y2-y1,x2-x1); %rotation angle

 obj=ellipticalFit.groundtruth([], center,[a,b], theta);


 plot(obj);   hold on;
 plot([x1,x2],[y1,y2],'xk'); hold off; 
 axis padded