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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?

Thanks

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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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    @agra94: that's an equation for the tangent line. You fix a parameter of the curve $t$, and get a point $r(t)$ together with the tangent vector $\dot r(t)$. Then $r(t) + \dot r(t)s$ is the tangent line at a point $r(t)$, parametrized by $s$. In the problem it is stated that for each $t$ there exists $s$ for which it holds that $r(t) + \dot r(t) s = a$. I just denote this $s$ as $\tau(t)$, since it depends on $t$.2018-02-05
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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.