Let $P$ be a point on a curve $\alpha$ in $\mathbb{E}^2$, $T$ the intersection of the tangent line to $\alpha$ in $P$ and the $x$-axis and $F$ the orthogonal projection of $P$ onto the $x$-axis. Determine all curves such that the vector $\overline{TF}=F-T$ is constant (and does not depend on $P$).
What I already tried is:
Suppose that the curve is given as $\alpha:I\subset\mathbb{R}\rightarrow\mathbb{E}^2:t\mapsto(\alpha_1(t),\alpha_2(t))$. For a random point $P$ on the curve $\alpha$, there exist a $t_0\in I$ such that $P=(\alpha_1(t_0),\alpha_2(t_0))$. The tangent to $\alpha$ in $P$ is then given as $$l\leftrightarrow P+\lambda \alpha'(t_0),$$ where $\lambda\in\mathbb{R}$ and $\alpha'(t_0)=(\alpha_1'(t_0),\alpha_2'(t_0))$. It is easy to see that $F=(\alpha_1(t0),0)$ and that $T$ will be of the form $(x,0)$, since both lie on the $x$-axis. To find $T$ we rewrite $l$ as $$l=\left\{(\alpha_1(t_0)+\lambda\alpha_1'(t_0), \alpha_2(t_0)+\lambda\alpha_2'(t_0))|\lambda\in\mathbb{R}\right\}$$ and we immediately see that $\alpha_2(t_0)+\lambda\alpha_2'(t_0)=0$. This means $\lambda=-t_0/\ln{\alpha_2(t_0)}$ and so $$T=(\alpha_1(t_0)-t_0\alpha_1'(t_0)/\ln\alpha_2(t_0),0).$$ But we want $\overline{TF}=(t_0\alpha_1'(t_0)/\ln\alpha_2(t_0),0)$ to be a constant vector, independent of the choice of P, i.e. independent of $t_0$. So we need $$\dfrac{t_0\alpha_1'(t_0)}{\ln\alpha_2(t_0)}=\mu,$$ where $\mu\in\mathbb{R}$ is some constant. Hence we find that such curves have the restriction $$\alpha_1'(t)=\mu\dfrac{t}{\ln\alpha_2(t)}.$$
But suppose now that $\alpha(t)=(0,t)$, i.e. the $y$-axis. This means $P=(0,t_0)$, $l=\alpha$ and $T=F$. Hence $\overline{TF}=(0,0), \forall t\in I$ and $\alpha_1'(t)=0$. This means $$0=\mu\dfrac{t}{\ln\alpha_2(t)}$$ and so $\mu=0\dots$
What am I missing here?
We consider a general inclination. To find a curve of constant sub-tangent length projected on x-axis if integration is permitted, we have
$$ \frac{y}{y'}= c \to \frac{dy}{dx}=\frac{y}{c}\;$$
Integrating, $\log y $ is linear in $x$.
$$ \log y = \frac{x}{c} + \log a $$
$$ y = a e^{\frac{x}{c}}$$
where $a$ is an arbitrary constant. $c$ can be positive or negative, so all increasing/decreasing exponential functions have a constant sub-tangent.