Show that the equation system
$$x^2 + y^2 + 3u^2 + 4v^2 = 1$$ $$x^2 + y^2 - u^2 - v^2 = 0$$
can be solved by two positive functions $u(x,y)$ and $v(x,y)$ and calculate their partial derivatives.
I think that this is related to the implicit function theorem, but I do not see how I could solve this. Could you help me?
By subtracting the second equation from the first yo get $4u^2+5v^2=1$. Now, keeping the second one, you have the system $$ \left\{ \begin{gathered} 4u^2 + 5v^2 = 1 \hfill \\ u^2 + v^2 = x^2 + y^2 \hfill \\ \end{gathered} \right. $$ which get $$ \left\{ \begin{gathered} 4u^2 + 5v^2 = 1 \hfill \\ - 4u^2 - 4v^2 = - 4x^2 - 4y^2 \hfill \\ \end{gathered} \right. $$ Therefore, by adding the equation you get $$ v^2 = 1 - 4x^2 - 4y^2 $$ and now, you have $$ u^2 = 5x^2 + 5y^2 - 1 $$ From these you can find $u(x,y)$ and $v(x,y)$