2D bifurcation problem I come across this problem which is about bifurcation. I am trying to take all the cases. I am expecting Hopf bifurcation to occur here but the last case I could not find the fixed point. Could you please help me?
Consider the vector field on plane
\begin{align}
\dot{x} &= x - xy +1\\ \label{sys1}
\dot{y} &= \alpha y + \beta x^2,\nonumber
\end{align}
where $\alpha$ and $\beta$ are parameters. Study the bifurcation of the dynamical system in as much details as possible. You have to study the situation in the neighbourhood of the Bogdonov-Takens bifurcation points.
[Case 1:] When $\alpha=0$ and $\beta =0$ we have not equilibrium point.
[Case 2:] Assume $\beta=0$.
\begin{align}
\dot{z} &= z - zy +1\\ \label{sys2}
\dot{y} &= \alpha y\nonumber
\end{align}
When $\alpha=0$ we go back to case 1, however when $\alpha \not = 0$ we have  a equilibrium point $(-1,0)$. Shifting the fixed point to the origin by $z= x-1$ we have the system
\begin{equation}
\begin{aligned}
\dot{x} &= x+y - x y \\
\dot{y} &= \alpha y
\end{aligned}
\end{equation}
which has the fixed point $(0,0)$. Finding the Jacobian for (\ref{sys3})
$$A_{(0,0)} = 
\begin{pmatrix}
1&&1\\
0&& \alpha
\end{pmatrix},
$$
we can see that the eigenvalues are $\lambda_1=1$, and $\lambda_2=\alpha$ with the eigenvectors
$$v_1 = \begin{pmatrix}
1\\
0
\end{pmatrix}, \text{ and } 
v_2 = \begin{pmatrix}
\frac{1}{\alpha -1} \\1
\end{pmatrix}$$
When $\alpha \geq 0$ the equilibrium point (0,0) unstable (source) with two dimension unstable manifold, and when $\alpha <0$ the equilibrium point $(0,0)$ is a hyperbolic unstable (saddle) equilibrium point; and the stable and unstable eigenspaces are;
$$\mathcal{E}^s = \{ (x,y) \in \mathbb{R}^2 | y=  \frac{1}{\alpha - 1} x \}, \text{ and } \mathcal{E}^u = \{ (x,y) \in \mathbb{R}^2 | y= 0 \}$$
with one dimension stable manifold and one dimension unstable manifold.
Case 3: Assume $\alpha=0$. When $\beta =0$ we go to case 1. When $\beta \not =0$ we have the system has no fixed points.
Case 4: When $\alpha \not = 0 $, and $\beta \not = 0$

Hint.
Analyzing the intersections of
$$
\cases{
x-x y+1 =0\\
\lambda y +x^2=0
}
$$
we can foresight the equilibrium points qualification.
For $\lambda \approx -1 $

For $\lambda\approx -5$

Here the tangency point is solved easily as follows
$$
\lambda x+x^3+\lambda = (x-x_1)^2(x-x_2)
$$
so equating to zero the $x$'s powers coefficients
$$
\cases{
\lambda+x_1^2x_2 = 0\\
\lambda -x_1^2-2x_1x_2 = 0\\
2x_1+x_2=0
}
$$
we get $x_1 = -\frac 32, x_2=3, \lambda = -\frac{27}{4}$
For $\lambda\approx -35$

For $\lambda \approx 1$

NOTE
For $\lambda > 0$ we have one equilibrium point.
For $-\frac{27}{4}<\lambda< 0$ we have one equilibrium point.
For $\lambda < -\frac{27}{4}$ we have three equilibrium points.
For a given equilibrium point $(x_0,y_0)$ the Jacobian is
$$
J=\left(
\begin{array}{cc}
 1-y_0 & -x_0 \\
 2 x_0 & \lambda  \\
\end{array}
\right)
$$
with eigenvalues
$$
\frac 12\left(\lambda+1-y_0\pm\sqrt{(y_0-1+\lambda)^2-8x_0^2}\right)
$$
