A curve where all tangent lines are concurrent must be straight line

I'm trying to solve this question in the classical Do Carmo's differential geometry book (page 23):

5. A regular parametrized curve $\alpha$ has the property that all its tangent lines pass through a fixed point. Prove that the trace of $\alpha$ is a (segment of a) a straight line.

My attempt

Following the statement of the question, we have $\alpha(t)+\lambda(s)\alpha'(s)=const$.

Taking the derivative of both sides we have $\alpha'(s)+\lambda'(s)\alpha'(s)+\lambda(s)\alpha''(s)=0$ which is equal to $(1+\lambda'(s))\alpha'(s)+\lambda(s)\alpha''(s)=0$.

Since $\alpha'(s)$ and $\alpha''(s)$ are linearly independent, we have $\lambda'(s)=-1$ and $\lambda(s)=0$ for every $s$ which I found strange, since the derivative of the zero function is zero.

I need a clarification at this point and a hand to finish my attempt of solution.

Solution 1:

I guess you assumed that $\alpha$ is parametrized by arc length (or constant length), so $|\alpha'(s)|=1$ and differentiating gives $\langle \alpha' , \alpha''\rangle = 0$. Thus, if $\alpha''(s)\neq \vec 0$, then $\alpha'(s), \alpha''(s)$ are linearly independent.

So like you said, you find $\lambda (s) = 0$ and $\lambda'(s) = -1$ whenever $\alpha''(s)\neq 0$.

The set $\{ s : \alpha''(s)\neq \vec 0\}$ is an open set. If it is nonempty, it contains some intervals $I$. But your assertion on $\lambda$ cannot be true on an interval. Thus

$$\{ s : \alpha''(s)\neq \vec 0\}$$

is empty. So $\alpha''\equiv \vec 0$ and $\alpha$ defines a straight line.