How to solve a system of non linear equations of the form $2ax-2x(ax^2+by^2) = 0$

Question

Solve this system for $x,y$: $$2ax-2x(ax^2+by^2) = 0\tag{1}$$ $$2by-2y(ax^2+by^2) = 0\tag{2}$$ where $a$ and $b$ are constants such that $a>b>0$.

I re-arranged (1) to $x = \sqrt{\frac{a-by^2}{a}}$ and tried substituting it into (2) but I did not know where to go from there. I'm looking for real solutions.

Answer 1

$2ax-2x(ax^2+by^2) = 2x(a-ax^2-by^2)=0 \implies x=0 $ or $ a(1-x^2)=by^2$ $2by-2y(ax^2+by^2) = 2y(a-ax^2-by^2)=0 \implies y=0 $ or $ b(1-y^2)=ax^2$

Then you can try this 4 cases.

Answer 2

If $x = 0$, the first equation is ok, and:

$$ 2by-2y(ax^2+by^2) = 0 \\ 2by-2y(by^2) = 0 \\ by-by^3 = 0 \\ y-y^3 = 0 \\ y(1-y^2) = 0 $$

Thus, either $y=0$ or $y=1$ or $y=-1$. Here we have three solutions.

If $x \ne 0$: $$ 2ax-2x(ax^2+by^2) = 0 \\ ax = x(ax^2+by^2) \\ a = ax^2+by^2 $$

Replacing in the second equation, we have: $$ 2by-2y(ax^2+by^2) = 0 \\ by - y(ax^2+by^2) = 0\\ by - ya = 0 \\ y(b-a) = 0 $$

As $a > b$, the only solution is $y=0$. So: $$ a = ax^2+by^2 \\ a = ax^2 \\ 1 = x^2 \\ $$

Thus, either $x=1$ or $x=-1$.

Therefore, the solutions $(x, y)$ are: $\{ (0, 0), (0, 1), (0, -1), (1, 0), (-1, 0) \}$.