# How to find the intersection of two curves

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Shoucheng Huai 2016 年 12 月 29 日

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

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### 回答 (4 件)

jahanzaib ahmad 2019 年 7 月 28 日
[xx,yy]=polyxpoly(y,x,y,x1);
plot(xx,yy,'*');
hold on #### 0 件のコメント

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Walter Roberson 2016 年 12 月 29 日
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)

#### 1 件のコメント

Shoucheng Huai 2016 年 12 月 30 日

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Roger Stafford 2016 年 12 月 29 日

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.

#### 2 件のコメント

Shoucheng Huai 2016 年 12 月 30 日
Thanks for your answer. But I donot want to convert ellipse.Because I need the solve for complex two curves. You can see that.It is very difficult to convert this into the equations of two slanted ellipses. Walter Roberson 2016 年 12 月 30 日
Please post the code for the curves, not just a partly-obscured picture of the code.

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Roger Stafford 2016 年 12 月 31 日

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.

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