Too many solutions using solve function

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

Answer 1

[madhan ravi]

Use vpasolve()

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

Gives:

x =
 
16.209551693222247107423390740782
 
y =
 
7.3999826408579772986633766002805

[Shachar Givon]

Great, thank you

[madhan ravi]

Anytime :)

[Shachar Givon]

In that case I have a follow up -

I have 2 new Hyperbolas.

this time they have two intersect. while the regular solve gives me again 4 results (2 correct and 2 additional) the vpasolve gives me only one.

why did it choose to give only that one and not the other correct one?

syms x y  

EQ1 = sqrt((x-4.5)^2 + (y-19.5)^2 )-sqrt((x-14)^2 + (y-16.5)^2) == -1.2;
EQ2 = sqrt((x-4.5)^2 + (y-10.5)^2 )-sqrt((x-14)^2 + (y-7)^2) == -2.3977;

figure
ezplot(EQ1,[-100 100])
hold on 
ezplot(EQ2,[-100 100])
hold on 

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

Gives:

K>> R.x

ans =

   -3.5590

K>> R.y

ans =

  -10.0650

[Walter Roberson]

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]

Ok, thanks