Prove that a parametric curve does not intersect itself Consider the following parametric curve $(x(t), y(t))_{t \in \mathbb{R}_+}$:
$$x(t) = \frac{(1+e)(1 - \cos(t)) - t(e-1) \sin (t)}{e t^2 - t^2 \cos(t) - t \sin(t)} $$
$$y(t) = \frac{e(1+t^2)(1 - \cos(t))}{e t^2 - t^2 \cos(t) - t \sin(t)}, $$
where $e = \exp(1)$.
How to prove that this curve has no multiple points, except the two ones located on the axis $x = 0$ ?
 A: We have to prove that
$$\left. \begin{array}{l}
           0\le t_1 \le t_2 \\
           x(t_1) = x(t_2) \ne 0 \\
           y(t_1) = y(t_2) 
         \end{array}
 \right\}\quad \Longrightarrow \quad t_1 = t_2.$$
We first give the following result. The proof is given at the end.
Fact 1: For any $t > 0$, we have the following relation among $t, x=x(t), y=y(t)$:
$$\mathrm{e}^2 t^2 x^2+\mathrm{e}^2 x^2-2 \mathrm{e}^2 x y+
\mathrm{e}^2 y^2-2 \mathrm{e}^2 y+2 \mathrm{e} y-y^2 = 0. \tag{1}$$
Let us proceed. From Fact 1, we have
$$\left. \begin{array}{l}
           0 < t_1 \le t_2 \\
           x(t_1) = x(t_2) \ne 0 \\
           y(t_1) = y(t_2)
         \end{array}
 \right\}\quad \Longrightarrow \quad t_1 = t_2.$$
It remains to prove that
$$\left. \begin{array}{l}
           0\le t_2 \\
           x(0) = x(t_2) \ne 0 \\
           y(0) = y(t_2)
         \end{array}
 \right\}\quad \Longrightarrow \quad 0 = t_2.$$
It depends on the definition of $x(0)$ and $y(0)$, since when $t=0$, the denominator $\mathrm{e} t^2 - t^2\cos t - t\sin t$ is zero.
If we define $x(0)=0$ and $y(0)=0$, then there is nothing to prove.
If we define $x(0) = \lim_{t\to 0} x(t)$ and $y(0) = \lim_{t\to 0} y(t)$, then
$x(0) = \frac{3 - \mathrm{e}}{2\mathrm{e} - 4}$ and $y(0) = \frac{\mathrm{e}}{2\mathrm{e} - 4}$.
Suppose $t_2 > 0$, $x(t_2) = x(0)$ and $y(t_2) = y(0)$. Inserting them into (1), we get $t_2 = 0$. Contradiction.
We are done.
$\phantom{2}$
Proof of Fact 1: When $t > 0$, it is easy to prove that $\mathrm{e} t^2 - t^2\cos t - t\sin t > 0$,
and from the equations, we have
\begin{align}
(t^2 x - \mathrm{e} - 1)\cos t + (-\mathrm{e} + x + 1)t\sin t &= \mathrm{e}t^2 x - \mathrm{e} - 1, \\
(-\mathrm{e}t^2 + t^2y - \mathrm{e})\cos t + vt \sin t &= \mathrm{e}t^2 y - \mathrm{e} t^2 - \mathrm{e}
\end{align}
or
\begin{align}
a_1 \cos t + b_1 \sin t &= c_1, \tag{2}\\
a_2\cos t + b_2 \sin t &= c_2 \tag{3}
\end{align}
where $a_1 = t^2 x - \mathrm{e} - 1$, $b_1 = (-\mathrm{e} + x + 1)t$, 
$c_1 = \mathrm{e}t^2 x - \mathrm{e} - 1$,
$a_2 = -\mathrm{e}t^2 + t^2y - \mathrm{e}$, $b_2 = yt$,
and $c_2 = \mathrm{e}t^2 y - \mathrm{e} t^2 - \mathrm{e}$. From (2) and (3), we have
\begin{align}
(a_1 b_2 - a_2 b_1) \cos t &= b_2 c_1 - b_1 c_2, \\
(a_1b_2 - a_2 b_1) \sin t &= a_1c_2 - a_2 c_1.
\end{align}
By using the identity $\cos^2 t + \sin^2 t = 1$, we have
$$(a_1b_2 - a_2b_1)^2 = (b_2c_1 - b_1c_2)^2 + (a_1c_2 - a_2c_1)^2$$
which results in
$$\mathrm{e}^2 t^2 x^2+\mathrm{e}^2 x^2-2 \mathrm{e}^2 x y+
\mathrm{e}^2 y^2-2 \mathrm{e}^2 y+2 \mathrm{e} y-y^2 = 0.$$
We are done.
A: In fact, this is easy to prove once we note the following identity:
$$\forall t \geq 0,\quad \left(1-\frac{y(t)}{e} \right)^2 + 2 x(t) = x^2(t) t^2 + (1+x(t) - y(t))^2.$$
From there, we deduce that either $x(t) = 0$ (and then $y(t) = 0$ or $y(t) = 2e/(1+e)$; so $(0, 0)$ and $(0, 2e/(1+e))$ are the two multiple points) or, for a fixed value of $x$ and $y$, there is a unique $t$ which matches. This ends the proof.
A: 
At $(x=0,y=0)$ the $t$ derivative vanishes,
and at $(x=0,y=2e/(1+e))$ the $\dfrac{dy}{dx}= \dfrac{dy/dt}{dx/dt}$ derivative is constant.. as should be also included for double point in  the indeterminate situation by L'Hospital's Rule. Like this elsewhere both derivatives do not vanish.
