Parametric Intersection point calculation of two curves

Hi all. I have a problem with the code below,

inter = solve('x-((y+a)/(x+b))^2','y-((y-a)/(x-b))^2','x','y'); inter.x; inter.y

where the given error is;

_Error using ==> mupadengine.mupadengine>mupadengine.feval at 144 MuPAD error: Error: Recursive definition [See ?MAXDEPTH];

during evaluation of 'ff'

Error in ==> solve at 77 sol = eng.feval('symobj::solvefull',eqns,vars);

Error in ==> newm at 18 inter = solve('x-((y+a)/(x+b))^2','y-((y-a)/(x-b))^2','x','y');_

On the other hand, if I change the parameters a and b with the values 2 and 1 respectively, it works great and reveals 7 intersection points, but I have to find the intersection point functions consisting of a and b. Such as a+2b, a^2-b^2 ....

All help is greatly appreciated.

[Roger Stafford]

As you have stated your problem, there are no simultaneous solutions to your two equations unless a = 0, because you are requiring both y = -a and y = a. If matlab gives you solutions with the substitution a = 2, then matlab would be lying to you. Are you sure you have described your problem accurately?

[Adelante]

First thank you for your comment. When I solve inter = solve('x-((y+2)/(x+1))^2','y-((y-2)/(x-1))^2','x','y'); one of the intersection point is x=1.6074447143648524831096114040652 and y=1.3058499038651999353460876987318. I am sure that the x,y values make both equations zero since I checked, however as I mentioned before, I do not need the real solutions but the equations consisting of a and b.

Answer 1

[Roger Stafford]

Now that you've corrected your equations, they do have a solution - in fact seven solutions. These can be derived from the roots of a seventh-order polynomial equation. It is impossible for matlab to give a symbolic solution to such an equation, but it can be solved numerically using matlab's 'roots' function. That is the reason you received an error message with symbolic values of a and b, but did receive the solutions when a and b were specified.

The numeric solution can be obtained either by:

   t = roots([1,-1,-b,-(a+2*b),4*a-b^2,(2*a-b)*b,(4*a+b^2)*b,-a*(4*a+b^2)]);
   x = t.^2;
   y = -a+(x+b).*t;

or by specifying a and b values before calling on 'solve', as you did.

As has been proved mathematically, you will never find a symbolic expression for the solution in terms of a and b involving only elementary operations including involution - in other words, those kinds of mathematical operations 'solve' could ordinarily furnish.

[Adelante]

Thank you for your kind answer. Yes you're right the was a misspelling in the first equation but i am grateful that you got my concern. Nonetheless I understand it right now, perfectly clear :)