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):


  
*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.

Let's assume that the tangents pass trough the point $c$ in $\mathbb{R}^2$. Then we the vectors $\alpha(s) - c$ and $\alpha'(s)$ are proportional, that is
$(\alpha(s)-c)  + \lambda(s) \alpha'(s)=0$
for some scalar function $\lambda(s)$. Replacing $\alpha(s)$ with $\alpha(s) - c$, we may assume that $c=0$, that is
$$\alpha(s)
  + \lambda(s) \alpha'(s)=0$$
The set $\{s \ | \ \lambda(s) \ne 0\}$ is open and dense in the domain (since the curve is regular).  On this domain  we can write
$$\alpha'(s) = \mu(s) \alpha(s)$$
and so on each connected component
$$\alpha(s) = e^{M(s)} \cdot \mathbb{a}$$
where $M(s)$ is an antiderivative of $\mu(s)$. It is now easy to conclude (since $\gamma$ is regular) that the vector constants $\mathbb{a}$ are the same for all the connected components. Therefore, $\gamma$  describes a (part of ) the line of direction $\mathbb{a}$ through the origin.
