Finding all bifurcations in a 2D system I want to find all bifurcations for the system:
\begin{align}                                                                                                                        
x' =& -x+y\\                                                                                                                           
y' =& \frac{x^2}{1+x^2}-\lambda y                                                                                                     
\end{align}
So far, I am using the idea that the intersection of the nullclines ($y'=0$ or $x'=0$) gives equilibrium points. So:
\begin{align}                                                                                                                        
x'=& 0 \implies x=y \\                                                                                                               
y'=& 0 \implies y=\frac{x^2}{\lambda(1+x^2)}                                                                                         
\end{align}
Clearly $(0, 0)$ is an equilibrium point but there is also the possibility of two more equilibrium points when the parabola given
by the above equation intersects the line $y=x$. To find the intersection I solved:
\begin{align}                                                                                                                        
x =& \frac{x^2}{\lambda(1+x^2)}\\                                                                                                    
\lambda(1+x^2) =& x\\                                                                                                                
\lambda x^2-x+\lambda =& 0 \\                                                                                                        
x =& \frac{1\pm \sqrt{1-4\lambda^2}}{2\lambda}                                                                                       
\end{align}
I have these two intersection points. Now I need to vary $\lambda$ to find where the curve passing through the intersection points
becomes tangent to $y=x$ and hence we would expect one equilibrium point instead of these two at this particular value of $\lambda$. Then continuing the variation of $\lambda$ in the same direction we would expect no equilibrium points from these two.
Hence we originally had $2$ equilibrium  points and then they coalesced and finally annihilated each other. This is a saddle-node bifucation.
How do I show this for my variation of $\lambda$? Are there any other bifurcations?
EDIT: Consider the discriminant of the equation:
\begin{align}                                                                                                                        
  x = \frac{1\pm\sqrt{1-4\lambda^2}}{2\lambda}                                                                                       
\end{align}
\begin{align}                                                                                                                        
1-4\lambda^2 =& 0 \\                                                                                                                 
1 =& 4\lambda^2 \\                                                                                                                   
1 =& \pm 2\lambda \\                                                                                                                 
\pm\frac{1}{2} =& \lambda                                                                                                            
\end{align}
So, I plotted the system with $\lambda = \frac{1}{2}$: sage: P = plot(x^2/ .5*(1+x^2), 0, 2) + plot(x, 0, 2)

We no longer have two bifurcations and instead have one. I was expecting the curves to be tangent. When does it happen that the curves are tangent? Actually I just realized that SAGE was not dividing the $1+x^2$ term, so I added the extra set of parens and everything works as expected!

Notice that x is real only when the discriminant 1-4λ^2 > 0. I.e. The curves of x' and y' do not intersect at points when 1-4λ^2 > 0 does not hold.
