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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?

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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 $\dot{r}(t) = const$.

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a) It should be $1+\dot\tau(t)$ in the second equation. b) I think you're assuming that the curve is parametrized by arc length? Else $n\cdot \dot r=0$ need not hold. c) What does "a vector of the unit normal" mean? –  joriki Aug 8 '11 at 14:54
    
@Gortaur,Yeah,you are right.I am considering the 3-dim case.Your proof is very nice. –  user14242 Aug 8 '11 at 15:01
    
@joriki: thank you for the comments. a) misprint, b)+c) - lack of English practice in this filed. Thought that regular parametrized already means natural parametrization. Unit normal of course was not true since we even showed that it's zero. –  Ilya Aug 8 '11 at 15:48

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.

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