I found the following problem:
Suppose all normal lines of a curve pass through a fixed point. Show that the trace is contained in a circle.
I spent a lot of time thinking about this but couldn't find an answer. I looked for the answer in the book and it says:
Differentiate $\alpha(s)=\lambda(s) n(s)= constant$
I am a bit confused: What is $\alpha$? Up to now, the book is using $\alpha$ to name a curve but here, it's doesn't seems to be the case.
Best Answer
Let $\alpha(s)$ be a point on the curve, then there is a scaling factor $\lambda(s)$ such that the quantity:
$$ \alpha(s) + \lambda(s) n(s)=c$$
Evaluates to the constant point at each point on the curve. Differentiating both sides,
$$ \alpha'(s) + \lambda'(s) n(s) +\lambda(s) n'(s)=0$$
We know $\alpha'(s),n'(s)$ is in the direction of unit tangent vector. Dot both side with $n(s)$:
$$ \lambda'(s)=0$$
Meaning $\lambda=C$