An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.
Now as $\theta\rightarrow\pi$,
$$
y = r\sin\theta
= \frac{\theta\sin\theta}{\pi-\theta}
= \frac{\theta\sin(\pi-\theta)}{\pi-\theta}
\rightarrow \theta
\rightarrow \pi
$$
so the horizontal asymptote is $y=\pi$.
For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
Horizontal and vertical tangents can be found by finding the roots of
$\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming
that they are never simultaneously zero, in which case we'd have to
resort to higher order derivatives). A simple way to find these roots
is to observe that
$$r = \left(\frac{\pi}{\theta}-1\right)^{-1}
\implies \frac{dr}{d\theta}
= -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot
\left(-\pi\theta^{-2}\right)
= \frac{\pi}{(\theta-\pi)^2}
$$
so that
$$
0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta
\iff
\tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
and
$$
0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta
\iff
-\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
which show us that they never vanish simultaneously.
From the shape of $\frac{\pi}{\theta(\pi-\theta)}$,
which is symmetric and positive on $(0,\pi)$
with global minimum at $\frac{\pi}{2}$
and vertical asymptotes at the endpoints,
we see that it will meet $\tan\theta$
at exactly one point $\theta_0\in(0,\frac{\pi}{2})$;
thus, there is one unique vertical tangent to our curve.
From the graph above, we might estimate
$$
\tan\theta_0\approx\frac{.4}{.25}=1.6
\implies
\theta_0
\approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)}
\approx .8609
$$
where we used the quadratic equation
to deduce $\theta$ from $\tan\theta$.
However, the actual solution seems to be
closer to $0.97803904765198235$:
t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)
For horizontal tangents, the same function
$\frac{\pi}{\theta(\pi-\theta)}$ must now meet
$-\cot\theta$, which is equivalent to the equation
$$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right) $$
which on $[0,\pi]$ has only solutions at the two endpoints,
i.e., on our curve, at the origin and
at the horizontal asymptote found above.
For parametrizing the cylindre you can parametrize a circle $(\cos(t),\sin(t),0)$ and then "move" along the axe passing through its center $(0,0,0)$ and orthogonal to the plane where the circle is contained (i.e. $(0,0,s)$). Thus the cylinder can be parametrized by $$c(s,t) = (\cos(t),\sin(t),s), \quad t,s \in \mathbb{R}$$ the curve you are considering is $$\gamma(t) = c(t,t),\quad t \in \mathbb{R}$$ and therefore must lie on the cylindre.
The expression $\cos^2(t)+\sin^2(t)=1$, shows you that a point $(\cos(t),\sin(t),t)$ on your curve is always at a distance $1$ from the point $(0,0,t)$ (the axis of the cylindre).
Best Answer
We have,
$$x^2 + y^2 = (4\sin(t))^2 + (4\cos(t))^2 = 16\sin^2(t) + 16\cos^2(t)$$
We know that
$$\sin^2(t) + \cos^2(t) = 1$$
so,
$$16(\sin^2(t) + \cos^2(t)) = 16$$
So, we have,
$$x^2 + y^2 = 16$$
Which can be written as,
$$\left(\frac x 4\right)^2 + \left(\frac y 4\right)^2 = 1$$
If that's what you meant in your question.