= Solution
A <fixed point> satisfies
$$
x+y^2-a=0,
\qquad
y(4x-x^2-a)=0.
$$
The branch with $y=0$ is
$$
P_0(a)=(a,0).
$$
Every other fixed point can be parametrized by its $x$-coordinate:
$$
a=4x-x^2,
\qquad
y=\pm\sqrt{x(3-x)},
\qquad 0\leq x\leq3.
$$
Equivalently,
$$
x_\pm=2\pm\sqrt{4-a},
\qquad y=\pm\sqrt{x_\pm(3-x_\pm)}.
$$
The $x_-$ pair exists for $0\leq a\leq4$, and the $x_+$ pair for $3\leq a\leq4$.
The <Jacobian matrix> is
$$
J(x,y)=
\begin{pmatrix}
1&2y\\
y(4-2x)&4x-x^2-a
\end{pmatrix}.
$$
On $P_0(a)$ its <eigenvalues> are $1$ and $a(3-a)$, so this branch has a <nonhyperbolic equilibrium> at $a=0$ and $a=3$. On a nonzero-$y$ branch,
$$
\det J=-4y^2(2-x),
$$
which vanishes at $x=2$, hence at $a=4$ and $y=\pm\sqrt2$. The three bifurcation values and locations are therefore
$$
\boxed{
\begin{array}{c|c|c}
a&\text{bifurcation locations}&\text{type suggested by the fixed-point branches}\\ \hline
0&(0,0)&\text{pitchfork},\\
3&(3,0)&\text{pitchfork},\\
4&(2,\sqrt2),(2,-\sqrt2)&\text{saddle-node at each point}.
\end{array}}
$$
For the <center manifold> calculation, append $\dot a=0$ as in the <extended centre manifold for a parameter>.
At $(a,x,y)=(0,0,0)$ put $\mu=a$ and write the centre manifold as $x=h(y,\mu)$. Its <centre-manifold invariance equation> is
$$
h_y\,y(4h-h^2-\mu)=h+y^2-\mu.
$$
It gives $h=\mu-y^2+O(3)$, and hence
$$
\dot y=y(4h-h^2-\mu)=3\mu y-4y^3+O(4).
$$
This is the <pitchfork bifurcation normal form> with stable nonzero centre branches for $\mu>0$: the bifurcation at $a=0$ is \b[supercritical].
At $(a,x,y)=(3,3,0)$ put $u=x-3$ and $\mu=a-3$. Then
$$
\dot u=u+y^2-\mu,
\qquad
\dot y=y(-2u-u^2-\mu).
$$
Again $u=h(y,\mu)=\mu-y^2+O(3)$, so
$$
\dot y=-3\mu y+2y^3+O(4).
$$
The nonzero centre branches exist for $\mu>0$ and are unstable in the centre direction, whereas $y=0$ is centre-stable there. Thus the bifurcation at $a=3$ is a \b[<subcritical pitchfork bifurcation>] with reversed normal-form parameter.
Finally fix $y_0=\pm\sqrt2$ and put
$$
u=x-2,
\qquad v=y-y_0,
\qquad \mu=a-4.
$$
The translated system is
$$
\dot u=u+2y_0v+v^2-\mu,
\qquad
\dot v=-(y_0+v)(u^2+\mu).
$$
The extended centre manifold has
$$
u=-2y_0v+\mu+O(v^2,v\mu,\mu^2),
$$
and therefore
$$
\dot v=-y_0(\mu+8v^2)+O(v^3,\mu v).
$$
This is the <saddle-node bifurcation> normal form. For $\mu<0$ there are two nearby fixed points, which coalesce and disappear at $\mu=0$. The complete reductions are recorded in the <Extended centre-manifold reductions of the 2019 Cambridge reflection-symmetric system>.
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