[?] Solve powerless polynomial equations?

Questions
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1

Hello everyone.
Not much to say, paste the Wolfram Mathematica code

Solve[FullSimplify@
  TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. 
       CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. 
           Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, {Sin[\[Theta]],
                 Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, 
             y}])[[1]], {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, 
       b1, c1, d1, e1} /. {\[Theta] -> 2 ArcTan[u]}], {a, b, x0, y0, 
  u}]

Unfortunately, Wolfram took a long time and didn’t solve it (even I used the meta-form)
I have to use the maple to solve the problem, the maple is very good.
Maple code

solve([-(4*a^2*u^2+b^2*(u^2-1)^2)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = a1, -(4*(a-b))*(a+b)*u*(u^2-1)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = b1, c1+(2*b^2*(u^2-1)*x0+4*a^2*u*y0)/((u^2+1)*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = 0, -(4*b^2*u^2+a^2*(u^2-1)^2)/((u^2+1)^2*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = d1, (4*b^2*u*x0-2*a^2*(u^2-1)*y0)/((u^2+1)*(-b^2*x0^2+a^2*(b+y0)*(b-y0))) = e1], [a, b, x0, y0, u])

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2

Are you sure that Maple gives you a correct answer?

I tried to figure out what you are doing and with

eq = Simplify /@ 
      Thread[TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. 
            CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. 
                Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, {Sin[\[Theta]], Cos[\[Theta]]}}, 
             {x + x0, y + y0}] == {x1, y1}, {x, y}])[[1]],
             {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, 
            e1} /. {\[Theta] -> 2 ArcTan[u]}]]

I arrive at five equations for your five unknowns.

But note that u is of 4th order and with

Solve[eq[[1, 1]]/eq[[4, 1]] == eq[[1, 2]]/eq[[4, 2]], u] // FullSimplify

( or

Solve[  List @@ eq[[1]]/List @@ eq[[4]] /. List -> Equal , u] // FullSimplify

)

you get four more or less complicated expressions for u.

Each of these has to be inserted in the other equations to solve for a, b, x0, y0, which are quadratic and therefore give in general two solutions. So things are for sure not simple and I guess there is a set of different solutions and not a single one.

Did you check it numerically?

3

In fact I think Maple did NOT arrive at a solution of your problem.

As before with

eq = Simplify /@ 
   Thread[TrigExpand[({{2, 0}, {1, 1}, {1, 0}, {0, 2}, {0, 1}} /. 
         CoefficientRules[((x^2/a^2 + y^2/b^2 - 1) /. 
             Solve[Dot[{{Cos[\[Theta]], -Sin[\[Theta]]}, \
{Sin[\[Theta]], Cos[\[Theta]]}}, {x + x0, y + y0}] == {x1, y1}, {x, 
               y}])[[1]],
          {x1, y1}]) == (x0^2/a^2 + y0^2/b^2 - 1) {a1, b1, c1, d1, 
         e1} /. {\[Theta] -> 2 ArcTan[u]}]];
eq // ColumnForm

you have 5 equations for your five unknowns. By forming some quotients you get another 4 equations

eq[[1, 1]]/eq[[2, 1]] == eq[[1, 2]]/eq[[2, 2]]
eq[[1, 1]]/eq[[3, 1]] == eq[[1, 2]]/eq[[3, 2]]
eq[[1, 1]]/eq[[4, 1]] == eq[[1, 2]]/eq[[4, 2]]
eq[[1, 1]]/eq[[5, 1]] == eq[[1, 2]]/eq[[5, 2]]

and by dividing eq1 by eq4 you get a complicated solution (in fact 4 solutions) for u

Solve[List @@ eq[[1]]/List @@ eq[[4]] /. List -> Equal,u] // FullSimplify

which in fact is not very helpful at this stage.
Proceeding further you can get rid of some inconveniernt numerators

eq1 = Map[# a^2 b^2 (1 + u^2)^2 &, eq, {2}];
eq1 // ColumnForm

Using again the quotient-trick you get

q12 = eq1[[1, 1]]/eq1[[2, 1]] == eq1[[1, 2]]/eq1[[2, 2]]
q13 = eq1[[1, 1]]/eq[[3, 1]] == eq1[[1, 2]]/eq1[[3, 2]]
q14 = eq1[[1, 1]]/eq1[[4, 1]] == eq1[[1, 2]]/eq1[[4, 2]]
q15 = eq1[[1, 1]]/eq[[5, 1]] == eq1[[1, 2]]/eq1[[5, 2]]

By inspection you see that q13 and q15 contain x0 and y0, so these may be extracted / solved for

lsgxy = Solve[{q13, q15}, {x0, y0}] // Flatten

and q12 and q14 contain only a, b and u.
So

lsga = Solve[q12, a]

gives two solutions for a.
But inserted in q14 this equation is free of b. ???

Maybe b is a free parameter? Or the equations are chosen wrongly?

Ignoring this one could decide to insert the meanwhile known parameters into e.g. the first of the equations

eq1a = eq[[1]] /. lsgxy /. lsga[[1]] // FullSimplify // Together

giving an expression which contains indeed only b and u as unknowns,

Variables[List @@ eq1a]

but u in a really high order. Impossible to solve for analytically. So I really doubt that Maple could come up with a solution of your problem.