Show that the isotropic foliation of a spheroid in $\mathbb R^4$ is regular if and only if the spheroid is actually a sphere This is part (i) of exercise 5.4.4 from McDuff and Salamon's book.

Consider $\mathbb R^4$ with its standard symplectic structure $\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2$, and the coisotropic submanifold
$$Q_a = \{ (x_1, y_1, x_2, y_2) \in \mathbb R^4 \mid x_1^2 + y_1^2 + a(x_2^2 + y_2^2) = 1 \}$$
for some $a > 0$. Its isotropic foliation is generated by the vector field $v = v_1 + v_2$, where
$$
\newcommand \p [1] {\frac \partial {\partial #1}}
v_1 = x_1 \p {y_1} - y_1 \p {x_1}, \qquad v_2 = a \left( x_2 \p {y_2} - y_2 \p {x_2} \right)
$$
I want to show that this foliation is regular if and only if $a = 1$. Of course, when $a = 1$, the isotropic foliation is the Hopf fibration. Being a fibration, it is as regular as it gets.

Let $V_i \simeq \mathbb R^2$ be the symplectic subspace of $\mathbb R^4$ with coordinates $x_i, y_i$. Each vector field $v_i$ is well-defined in the corresponding subspace $V_i$. Hence $v$'s integral curves are of the form $\gamma = \gamma_1 + \gamma_2$, where each $\gamma_i$ is an integral curve of $v_i$.
A generic $\gamma_i$ is a circle centered at the origin in $V_i$. Rotate each $V_i$ independently so that $\gamma_i$ lies in the positive half $x_i > 0$ of the line $y_i = 0$. This operation is a symmetry of the problem: it preserves both $\omega$ and $Q_a$. Hence the rotated version of $\gamma$ is still a symplectic leaf of $Q_a$.
There are two degenerate cases, when each $\gamma_i$ becomes degenerate, i.e., $\gamma_i = 0$ at all times. However, both cannot be simultaneously degenerate. In particular, when either is degenerate, $\gamma$ is equal to the other, hence $\gamma$ is a closed regular curve.
The curve $\gamma_1$ returns to its original position precisely at $t_n = 2 \pi n$ for every $n \in \mathbb Z$. Then $\gamma$ is a closed curve if and only if $\gamma_2(t_n)$ is a periodic sequence, if and only if $a$ is rational. In particular, if $a$ is irrational, then $\gamma$ is not a closed submanifold of $Q_a$, hence the isotropic foliation is not regular.
When $a$ is rational but different from $1$, the isotropic leaves are closed curves, so showing that the isotropic foliation is not regular requires a more sophisticated argument. In fact, so sophisticated that I am not seeing it. What could this argument possibly be?
 A: Let's first recall what is meant here by a regular (isotropic foliation of a) coisotropic submanifold (c.f. McDuff & Salamon, Introduction to Symplectic Topology, 3rd edition, p. 219):

A coisotropic submanifold $Q$ is called regular if it satisfies the
  following condition.
(R) For every $p_0 \in Q$ there exists a submanifold $S \subset Q$ containing $p_0$ (called a local slice through $p_0$) that intersects every isotropic
  leaf of $Q$ at most once and is such that $T_pQ = T_pS \oplus T_pQ^{\omega}$ for every $p \in S$. Moreover, the quotient space $Q/\sim$ is Hausdorff.

To spell out what is meant by this definition, we shall simply take as a slice a submanifold $S \subset Q$ which is transverse to every isotropic leaf it intersects. Then we say that $Q$ is regular if it is possible to find for every $p_0$ a sufficiently small slice containing $p_0$ which intersect each leaf at most once (and if the quotient is Hausdorff).
Remark: Let's observe that if $Q$ has codimension $k$ inside a symplectic manifold of dimension $2n$, $Q$ has dimension $2n-k$ and the leaves of its isotropic foliation have dimension $k$. Hence a slice has dimension $2(n-k)$.
We consider the coisotropic submanifold 
$$Q_a = \{ (x_1, y_1, x_2, y_2) \in \mathbb{R}^4 \, : \, x_1^2 + y_1^2 + a(x_2^2 + y_2^2) = 1 \}$$
for $a > 0$. We observe that $Q_a$ and $Q_{1/a}$ are related by a dilation and by a rotation, so that they have essentially identical isotropic foliations; we shall thus assume that $a \le 1$.
The isotropic leaf going through $(x,y) \in Q_a$ can be parametrized by (for $t \in \mathbb{R}$) $$(\cos(t) x_1 -  \sin(t) y_1, \cos(t) y_1 +  \sin(t) x_1, \cos(at)x_2 - \sin(at)y_2, \cos(at)y_2 + \sin(at)x_2).$$
As you mentioned, the orbit going through $p_0 = (1, 0, 0,0)$ is $(\cos t, \sin t, 0, 0)$, which has period $2\pi$. Nearby leaves can be parametrised by $(\sqrt{1-\epsilon^2}\cos t, \sqrt{1-\epsilon^2}\sin t, (\epsilon/\sqrt{a}) \cos (at + \delta), (\epsilon/\sqrt{a}) \sin (at + \delta))$ with $\epsilon, \delta$ being the (small) parameters.
A (2-dimensional) slice through $p_0$ is thus given by the time $t=0$ of these nearby trajectories, namely by the set of points $(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (\delta), (\epsilon/\sqrt{a}) \sin (\delta))$ with $\epsilon, \delta$ being the (small) parameters for the slice.
After one period $2\pi$ of the "short" orbit, the nearby orbit which started at $(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (\delta), (\epsilon/\sqrt{a}) \sin (\delta))$ is now at $(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (a 2\pi + \delta), (\epsilon/\sqrt{a}) \sin (a2\pi+\delta))$. These are two different points if $a$ is not an integer; since $a \le 1$, the two points are equal only if $a = 1$. Hence $Q_a$ is not regular when $a \neq 1$.
Remark: Beside $(\cos t, \sin t, 0, 0)$, the other "simple orbit" is $(0,0,\cos(at), \sin(at))$. This latter orbit is "slower" than the former when $a < 1$ and "faster" when $a > 1$. Hence, had we not restricted our attention to $a \le 1$, we would have had to consider the fastest of the two "simple orbits" in order to make sure that its nearby orbits intersect a slice more than once.
