Given the equations: \begin{align} \frac{dx}{dt}&=1-(b-1)x+ax^{2}y\\ \frac{dy}{dt}&= bx-ax^{2}y \end{align} a) Fix $a$ and vary $b$. Show that a Hopf Bifurcation occurs at $b=a+1$.
So I got two fixed points $(0,0)$ and $(1,\frac{b}{a})$ I computed the Jacobian at $(1,\frac{b}{a})$ and from there was able to characterise the fixed points depending on the values of a and b. Finally I set the real part of the equation $\frac{b-a-1}{2}=0$ and solved to get $b=a+1$.
My question is if this is correct and why does it only work at the second fixed point and not the first? Is it because $(0,0)$ is not an isolated fixed point? How could I use this to discuss the dependence of the period of the limit cycle close to this $b=a+1$ Hopf Bifurcation?