Consider the system $$a^2+b^2-c^2-d^2=0\\ac+bd=0$$ Prove that it can be solved for for $(a,c)$ in terms of $(b,d)$ near any nonzero solution.
I'm applying the Implicit Function Theorem to $F:\mathbb R_a\times \mathbb R_b\times \mathbb R_c\times\mathbb R_d\to \mathbb R^2,$ $F(a,b,c,d)=(a^2+b^2-c^2-d^2,ac+bd)$ at a point $(a_0,b_0,c_0,d_0)$ which is a nonzero solution. The Jacobian determinant is $2b^2+2d^2$, which is nonzero at the point. This means that there is a smooth $g: U\subset \mathbb R_b\times \mathbb R_d\to \mathbb R^2$ with $g(b_0,d_0)=(a_0,c_0)$ and ... (I don't know how to write the second condition; it should look like $F(t,g(t))=0$ for all $t\in U$ but here the order of coordinates is mixed... )
So the questions are whether what I did is correct, how to write the second condition, and how to see that this solves the problem?
You need to calculate the Jacobian determinant of $F(a,b,c,d)$ in $a$ and $c$, giving $2a^2 + 2c^2$. So you should check that this is never zero at any solution to $F(a,b,c,d) = 0$ other than $(0,0,0,0)$.
Then the implicit function theorem says that given any nonzero $(a_0,b_0,c_0,d_0)$ solving $F(a_0,b_0,c_0,d_0) = 0$, there is a $C^1$ function $g(b,d) = (g_1(b,d), g_2(b,d))$ defined near $(b_0,d_0)$ such that $F(g_1(b,d), b, g_2(b,d), d) = 0$ on a neighborhood of $(b_0,d_0)$. This $g(b,d)$ will satisfy $g(b_0,d_0) = (a_0,c_0)$.