Too many solutions using solve function

Question

Shachar Givon 2019-1-13

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/439552-too-many-solutions-using-solve-function

评论： Shachar Givon 2019-1-13

采纳的回答： madhan ravi

在 MATLAB Online 中打开

I am trying to use the solve function to find the point where two hyperbolas intersect.

There should only be one solution (also possible to see that when lookin at a plot of the two hyperbolas)

However the slove function gives me 4 solution, one of them is the correct one.

What am I doing wrong that gets me 3 additional unrelated results?

syms x y  
EQ1 = sqrt((x-4.5)^2 + (y-19.5)^2 )-sqrt((x-14)^2 + (y-25)^2) == -0.9;
EQ2 = sqrt((x-4.5)^2 + (y-1.5)^2 )-sqrt((x-4.5)^2 + (y-10.5)^2) == 0.999;
%you can see in the figure there should only be one intersect.
figure
ezplot(EQ1,[-100 100])
hold on 
ezplot(EQ2,[-100 100])
R = solve(EQ1,EQ2,[x y]);

The results I get:

K>> R.x
ans =
   21.9990
   19.4080
   16.2100
   17.6200
K>> R.y
ans =
    3.9830
    7.7380
    7.4000
    4.4520

Only (16.2,7.4) is correct.

Thank you for the help

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Answer 1

madhan ravi 2019-1-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/439552-too-many-solutions-using-solve-function#answer_356202

编辑：madhan ravi 2019-1-13

在 MATLAB Online 中打开

Use vpasolve()

[x y] = vpasolve(EQ1,EQ2,[x y])

Gives:

x =
 
16.209551693222247107423390740782
 
y =
 
7.3999826408579772986633766002805

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Walter Roberson 2019-1-13

When you use vpasolve() with a non-polynomial, it chooses a starting point for the search and uses it to find one solution. You have some control over the starting point: you can request a random starting point, or you can provide a range of values to search over.

If you need more than one solution of a non-polynomial then you need to either use solve() or else use vpasolve with different starting points or different constraints on the search range.

Shachar Givon 2019-1-13