Let $\alpha$ be a regular curve in $\mathbb{R}^2$ and let all of its tangent lines pass through the origin. Also, let $\beta$ be a regular curve in $\mathbb{R}^2$ and let all of its normal lines pass through the origin.
How can I show that $\alpha$ is contained in a straight line through
the origin and that $\beta$ is contained in a circle around the origin?
I know that the tangent line $\alpha(t)$ is the line that points in the direction of the tangent vector $T(t)$ and that the normal line of $\beta(t)$ is the line that points in the direction of $N(t)$, but how can I put this all together to complete the proof?
Best Answer
Let $c$ be a regular parametrization of the curve $\beta$. At each point of the curve a unique normal line passing through it is well defined as we are in $\mathbb{R}^2$ and there is a tangency direction because of the regularity of the parametrization. As the normal line at $c(t)$ passes through the origin it has the direction $c(t)-0 = c(t)$. And then, we deduce $\langle \dot{c}(t) , c(t) \rangle = 0$ because of them being orthogonal.
Applying this relation we get:
$\frac{d}{dt} \| c(t)\|^2 = \frac{d}{dt} \langle c(t) , c(t) \rangle = 2\langle \dot{c}(t) , c(t) \rangle = 0$
As the reasoning works for every $t$ we have that $\| c(t)\|^2$ is constant as a function of $t$ and so is $\| c(t)\|$. Putting $R$ for this last constant we have that $\beta$ is contained in the circumference of center $0$ and radius $R$.