Asked by Shoucheng Huai
on 29 Dec 2016

theta=0:pi/400:2*pi;

x=sin(pi*theta)+cos(pi*theta);

y=cos(pi*theta)-2*sin(pi*theta);

x1=sin(pi*theta)+cos(pi*theta)+2;

plot(y,x,y,x1);

hold on

Answer by jahanzaib ahmad
on 28 Jul 2019

[xx,yy]=polyxpoly(y,x,y,x1);

plot(xx,yy,'*');

hold on

Sign in to comment.

Answer by Walter Roberson
on 29 Dec 2016

You can also proceed symbolically if you have the symbolic toolbox:

syms theta Y

y = cos(pi*theta)-2*sin(pi*theta);

THETA = solve(y == Y, theta);

x = sin(pi*THETA(2)) + cos(pi*THETA(2));

x1 = sin(pi*THETA(1)) + cos(pi*THETA(1)) + 2;

y_common = solve(x == x1, Y);

x_common = simplify( subs(x, Y, y_common), 'step', 20)

Shoucheng Huai
on 30 Dec 2016

Thanks for your answer. I need not this .

Sign in to comment.

Answer by Roger Stafford
on 29 Dec 2016

Edited by Roger Stafford
on 29 Dec 2016

You can convert this into the equations of two slanted ellipses. The first ellipse is:

(x-y)^2+(2*x+y)^2 = 5*x^2+2*x*y+2*y^2 = 9

(Can you see why?)

The second ellipse must therefore be:

5*(x-2)^2+2*(x-2)*y+2*y^2 = 9

(Can you see why that is so?)

By subtracting the left sides of these two equations you get a straight line. Solving for y in terms of x in the line and substituting that back into the equation of the first ellipse gives you a quadratic equation in x which has two solutions, namely the x values of the two intersections of the two ellipses. You can then easily find the two corresponding y’s.

Shoucheng Huai
on 30 Dec 2016

Walter Roberson
on 30 Dec 2016

Please post the code for the curves, not just a partly-obscured picture of the code.

Sign in to comment.

Answer by Roger Stafford
on 31 Dec 2016

Edited by Roger Stafford
on 1 Jan 2017

You can use a modified version of the Newton-Raphson method for finding the intersection of two parameterized curves, provided the parameter functions defining the curves can be differentiated. For each intersection point the method requires an estimated value for each of the two parameters that would yield that point.

Let x1 = f1(t1), y1 = g1(t1) define one curve and x2 = f2(t2), y2 = g2(t2) define the second curve. Let their respective derivative functions be called df1dt1, dg1dt1, df2dt2, and dg2dt2. Let t1e and t2e be the respective estimated values of t1 and t2 for some intersection point. Then do this:

t1 = t1e; t2 = t2e;

tol = 1e-13 % Define acceptable error tolerance

rpt = true; % Pass through while-loop at least once

while rpt % Repeat while-loop until rpt is false

x1 = f1(t1); y1 = g1(t1);

x2 = f2(t2); y2 = g2(t2);

rpt = sqrt((x2-x1)^2+(y2-y1)^2)>=tol; % Euclidean distance apart

dt = [df1dt1(t1),-df2dt2(t2);dg1dt1(t1),-dg2dt2(t2)]\[x2-x1;y2-y1];

t1 = t1+dt(1); t2 = t2+dt(2);

end

x1 = f1(t1); y1 = g1(t1); % <-- These last two lines added later

x2 = f2(t2); y2 = g2(t2);

I tried this with a slightly modified version of your original ellipses:

x1 = sin(t1)+cos(t1); y1 = cos(t1)-2*sin(t1);

x2 = sin(t2)+cos(t2)+1.5; y2 = cos(t2)-2*sin(t2)+1;

using the estimates t1 = 1 and t2 = 2.4, determined by an appropriate inspection of the plots of the two curves, and had the following results:

x1 x2 y1 y2 t1 t2

1.3817732906760 1.4380694650099 -1.1426396637477 -1.0883200766435 1.0000000000000 2.4000000000000

1.3930006621432 1.3942125905603 -1.0625407651143 -1.0624834020886 0.9588192419883 2.4310674208882

1.3930927453049 1.3930928781613 -1.0617974598551 -1.0617968855823 0.9584414863689 2.4318614253186

1.3930928207253 1.3930928207253 -1.0617968503328 -1.0617968503328 0.9584411766348 2.4318614660485

As you can see, in three steps from the original estimates an intersection point was found to an accuracy of at least 13 decimal places. The same method can be used for the second intersection point of these curves, given an appropriate estimate of the corresponding parameters.

Sign in to comment.

Opportunities for recent engineering grads.

Apply Today
## 0 Comments

Sign in to comment.