Here's my system, \begin{gather*} N_{t+2}=N_t\exp{[r(1-\frac{N_t}{K})]}\frac{1-e^{-aP_t}}{aP_t} \\ P_{t+1}=N_t[1-\frac{1-e^{-aP_t}}{aP_t}] \end{gather*}
In the research paper, it states that we can approximate $\frac{1-e^{-aP_t}}{aP_t}\approx(aP_t)^{-1}$. That is, by a function resembling negative binomial form.
Solving for steady state solution yields, $N^*=N^*\exp{[r(1-N^*)]}\frac{1-e^{-aP^*}}{aP^*} $ and $ P^*=N^*[1-\frac{1-e^{-aP^*}}{aP^*}]$
Which we follow by replacing the term with negative binomial (ignoring trivial case where $N=P=0$) we have, \begin{equation*} \begin{array}{cc} 1=\exp{[r(1-N^*)]}\frac{1}{aP^*}\\ P^*=N^*[1-\frac{1}{aP^*}] \end{array} \end{equation*}
And this is where I'm stuck, the paper states that when $(N^*_1,P^*_1)=(1,0)$ that we would be able to deduce stability through rudimentary ways.
Using auxiliary equations, and scaling, I got the equations, $f(x,y)=xe^{r(1-x)}\frac{1}{ay}$ and $g(x,y)=x[1-\frac{1}{ay}]$
And the Jacobian as follows, \begin{equation*} \mathbf{J_2}= \begin{pmatrix} \frac{e^{r(1-x)}(1-rx)}{ay} & -\frac{xe^{r(1-x)}}{ay^2} \\[0.5ex] 1-\frac{1}{ay} & \frac{x}{ay^2} \end{pmatrix} \end{equation*}
Now, the paper goes on to say that if this is the case, then at the equilibrium point we should be able to deduce the necessary conditions we need to solve the eigenvalue problem. But I can't have $ay=0$!!! This problem is a predator-prey system, so using the carrying capacity as one of the points makes sense, but I'm confused. BY the way, $a$ is our varying parameter in which we obtain bifurcation diagram with, I can provide plot if needed.
The approximation $$\frac{1-e^{-aP_t}}{aP_t} \approx \frac{1}{aP_t} $$ is only good when $aP_t$ is large. If $P_t \to 0$ then $$\frac{1-e^{-aP_t}}{aP_t} \to 1.$$