Regular curve which tangent lines pass through a fixed point How to prove that if a regular parametrized curve has the property that all its tangent lines passs through a fixed point then its trace is a segment of a straight line?
Thanks
 A: In his answer, user14242 used a vector multiplication for two vectors - while I believe it's only defined for 3-dim case.
If you talk not only about the 3-dim problem then just writing explicitly the equation of the tangent line you obtain
$$
r(t)+\dot{r}(t)\tau(t) = a
$$
where $a$ is a fixed point and $\tau(t)$ denotes the value of parameter when the tangent line crosses $a$.
Let's assume that $t$ is a natural parametrization of the curve.
Taking the derivative w.r.t. $t$ you have
$$
\dot{r}(t)(1+\dot{\tau}(t))+n(t)\tau(t) = 0
$$
where $n(t) = \ddot{r}(t)$ is a normal vector. Also $n\cdot\dot{r} = 0$ and then
$$
\tau(t)\|n(t)\|^2 = 0.
$$
You have $\tau(t) = 0$ iff $r(t) = a$ and for all other points $n(t) = 0$ which gives us $\dot{r}(t) = const$.
A: Suppose the curve is $r(s)$ where $s$ is the regular parameter.WLOG,we may assume that all the tangent lines pass through the origin.Thus $r//T$ ($T=r(s)'$ is the unit tangent vector).i.e.$r\times T=0$.Differetiate both sides by $s$,we get $r\times N=0$ ($N$ is the unit normal vector).If $T$ is not a constant vector,then $N\neq 0$,and $N\bot T$,thus $r=0$,a contradiction.Hence $T$ is const,which implies that $r$ is a straight line.
