How to find common tangents between 2 ellipses ?

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Adrien Luthi 2020-11-3

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/635464-how-to-find-common-tangents-between-2-ellipses

评论： Niki Amini-Naieni 2023-7-3

采纳的回答： Matt J

在 MATLAB Online 中打开

Hi everyone,

I am trying to find the common tangents between 2 ellipses by using solve function. I have f(x1) and g(x2) which are my 2 elipses (ellipse 1 the smallest and ellipse 2) equations and the 2 unknows (x1 and x2) and their derivative at each point which gives me the slope of their tangents at each point.

Then, my equations to solve are:

and here my Matlab code:

syms x1 x2
f_x1 = -sqrt(b1^2*(1-((x1-xC1)/a1)^2)) + yC1;
g_x2 = sqrt(b2^2*(1-((x2-xC2)/a2)^2)) + yC2;
f_x1_prime = b1^2*(x1-xC1)/(a1^2*sqrt(b1^2*(1-((x1-xC1)/a1)^2)));
g_x2_prime = -b2^2*(x2-xC2)/(a2^2*sqrt(b2^2*(1-((x2-xC2)/a2)^2)));
eq1 = (g_x2 - f_x1)/(x2 - x1) == f_x1_prime;
eq2 = f_x1_prime == g_x2_prime;
[S] = solve(eq1, eq2, x1, x2, 'ReturnConditions', true, 'Real', true);
assume(S.conditions)
restriction = [S.x1>xC1, S.x1<xC1+a1, S.x2<xC2, S.x2>xC2-a2];
solx = solve(restriction, S.parameters);
x1 = subs(S.x1, S.parameters, solx);
x2 = subs(S.x2, S.parameters, solx);

As result, I have something in S but the warning and no solution after having assume the condition and solve the parameter.

S = 
  struct with fields:
            x1: [1×1 sym]
            x2: [1×1 sym]
    parameters: [1×1 sym]
    conditions: [1×1 sym]

and the warning:

    Warning: Unable to find explicit solution. For options, see help. 
> In sym/solve (line 317)
In findTangentPoints (line 24)
In testTangent (line 11) 

Does anyone can help me to use this function solve or see any errors ? I am going to be crazy...

Thanks a lot everyone and have a nice evening!

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John D'Errico 2020-12-25

编辑：John D'Errico 2020-12-25

Note that as you have written f and g, you are restricting an ellipse to a specific branch of that curve. That happens as soon as you take a sqrt. In fact, a pair of non-intersecting ellipses will have FOUR common tangent lines.

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采纳的回答

Matt J 2020-12-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/635464-how-to-find-common-tangents-between-2-ellipses#answer_584222

编辑：Matt J 2021-2-1

在 MATLAB Online 中打开

alpha1=30; alpha2=-45; %EDIT

a1=0.015; b1=0.020; xC1=0.13; yC1=0.09;

a2=0.08; b2=0.10; xC2=0.27; yC2=0.11;

T1=eye(3); T1(1:2,end)=[-xC1;-yC1]; %translation matrix

T2=eye(3); T2(1:2,end)=[-xC2;-yC2];

E1=diag(1./[a1^2,b1^2 -1]);

E2=diag(1./[a2^2,b2^2,-1]);

R=@(t) [cosd(t), -sind(t) 0; sind(t) cosd(t) 0; 0 0 1]; %EDIT

C1=T1.'*R(alpha1).'*E1*R(alpha1)*T1; %homogeneous ellipse equation matrix

C2=T2.'*R(alpha2).'*E2*R(alpha2)*T2;

syms Lx Ly Lz %line coefficients

equ1=[Lx,Ly,Lz]/C1*[Lx;Ly;Lz]==0; %tangent to ellipse 1

equ2=[Lx,Ly,Lz]/C2*[Lx;Ly;Lz]==0; %tangent to ellipse 2

equ3=Lx^2+Ly^2+Lz^2==1; %resolve the scale ambiguity in the coefficients

sol=solve([equ1,equ2,equ3]);

sol=double([sol.Lx,sol.Ly,sol.Lz]); %convert symbolic solutions to numeric doubles

sol=sol(2:2:end,:);%discard redundant solutions

%%% Display the solutions

fimplicit(@(x,y) qform(x,y,C1));

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

hold on

fimplicit(@(x,y) qform(x,y,C2));

for i=1:4

fimplicit(@(x,y) lform(x,y,sol(i,:))); %dumb warnings - ignore

end

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

hold off

axis equal, grid on

axis([0.1,0.4,0,.24])

function val=qform(x,y,Q)

sz=size(x);

xy=[x(:),y(:),x(:).^0];

val=reshape( sum( (xy*Q).*xy ,2) ,sz);

end

function val=lform(x,y,L)

sz=size(x);

val=reshape( [x(:),y(:),x(:).^0]*L(:) ,sz);

end

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Matt J 2023-7-3

编辑：Matt J 2023-7-3

@Niki Amini-Naieni You would have to have studied some projective geometry for any brief explanation to make sense to you. Basically, given an ellipse, E, the equation coefficients of lines tangent to E form a dual ellipse D. Therefore, given two ellipses E1 and E2, the common tangents must therefore lie in the intersection of their duals D1 and D2.

This wiki page talks about deriving the dual ellipse when the given ellipse is non-rotated,

https://en.wikipedia.org/wiki/Dual_curve

The extension to rotated ellipses isn't that hard.

Niki Amini-Naieni 2023-7-3