# Showing that $\alpha$ is a straight line

Let $\alpha : I \to \mathbb{R^2}$ be a regular curve. Suppose that all the tangent lines intersect in a fixed point. Show that $\kappa = 0$.

My attempt:

Let $q$ be the fixed point.

So there exists a funtion $\lambda : I \to \mathbb{R}$ such that

$$q = \alpha(t) + \lambda(t)T(t) \quad(1)$$

where $T(t) = \alpha'(t)/\|\alpha'(t)\|$

Taking the derivative in $(1)$, we have:

$$\alpha'(t) + \lambda'(t)T(t) + \lambda(t)T'(t) = 0$$ $$\Rightarrow \|\alpha'(t)\|T(t) + \lambda'(t)T(t) + \lambda(t)T'(t) = 0$$ $$\Rightarrow \big(\|\alpha'(t)\|+ \lambda'(t)\big)T(t) + \lambda(t)T'(t) = 0$$

From Frenet formula we get $T'(t) = \kappa(t)N(t)$, then:

$$\big(\|\alpha'(t)\|+ \lambda'(t)\big)T(t) + (\lambda(t)\kappa(t))N(t) = 0$$

Since $\{T,N\}$ is a basis of $\mathbb{R^2}$, we must have $\lambda(t)\kappa(t)=0$

I couldn't see that $\kappa(t)= 0$, $\forall t \in I$. Why can't $\lambda$ be zero for some $t\in I$?

• If $\lambda(t)=0$ then $q=\alpha(t)$. Since $\alpha$ is regular, this can only happen for distinct $t$'s, so by continuity of $\kappa$ it has to equal zero everywhere. – String Aug 28 '16 at 14:06
• If you want to apply the Frenet formulas, you should start by assuming the curve is arclength parametrized (so $\|\alpha'(t)\|=1$). Otherwise, your formula for $T'(t)$ is incorrect. – Ted Shifrin Aug 28 '16 at 17:14
• @String: Better to say "isolated," rather than "distinct." I can choose all $t$ in an interval to be distinct, but not isolated. – Ted Shifrin Aug 28 '16 at 17:15
• @TedShifrin: Point taken! – String Aug 28 '16 at 17:48

Let me suggest another approach. Let $O$ be the point of common intersection of all tangents of the curve $\alpha(t)$. Now, define the vector field $V(x)$ on the real plane $\mathbb{R}^2$ as follows: at each point $x$ on the plane define the vector $V(x) = \overrightarrow{Ox} = x-O$. In Cartesian coordinates centered at $O$, this vector field is simply $V(x) = x$, i.e. $$V(x) = x^1 \frac{\partial}{\partial x^1} + x^2 \frac{\partial}{\partial x^2},$$ where $x = (x^1,x^2)$. Then the curves tangent to this vector field are obtained by solving the system of ODE $\,\,\frac{dx}{ds} = x$. Thus, the solutions are $x(s) = e^s\, x$ which are straight lines passing through the point $O$. Now, observe that by assumption $\alpha'(t)$ is a vector collinear with the vector $\alpha(t) = V(\alpha(t))$. In other words, $\alpha'(t)$ and $V(\alpha(t))$ are collinear which means that $\alpha(t)$ is a curve tangent to the vector field $V(x)$. However, all curves tangent to $V(x)$ are straight lines passing through $O$. Therefore, $\alpha(t)$ is also a (portion of a) straight line passing through $O$, which means its curvature is $0$.
A similar approach, basically the same philosophy, is to go as follows. Let $\beta_0$ be the common point of intersection of all tangents to $\alpha(t)$ (if you wish, for simplicity, you may think $\beta_0 = 0$ is the origin of your coordinate system). This means that $\alpha'(t)$ and $\alpha(t) - \beta_0$ are collinear. Hence, there exists a scalar function $\lambda(t)$ such that $\alpha'(t) = \lambda(t) \big(\alpha(t) - \beta_0\big)$. Write the equations as $$\alpha'(t) - \lambda(t) \big(\alpha(t) - \beta_0\big) = 0$$ and multiply both sides by $e^{-\int \lambda(t)}$ getting $$\Big(e^{-\int \lambda(t)}\Big) \, \alpha'(t) - \Big( e^{-\int \lambda(t)}\Big)\lambda(t) \, \big(\alpha(t) - \beta_0\big) = 0.$$ But this is the derivative $$\frac{d}{dt}\left( \Big(e^{-\int \lambda(t)}\Big) \, \big( \alpha(t)-\beta_0\big)\right) = - \Big( e^{-\int \lambda(t)}\Big)\lambda(t) \, \big(\alpha(t) - \beta_0\big) + \Big(e^{-\int \lambda(t)}\Big) \, \alpha'(t) = 0$$ which after integration yields the existence of a constant vector $\sigma_0$ such that $$\Big(e^{-\int \lambda(t)}\Big) \, \big( \alpha(t)-\beta_0\big) = \sigma_0.$$ Therefore $$\alpha(t) = \beta_0 + \Big(e^{\int \lambda(t)}\Big) \, \sigma_0.$$ If you reparametrize the latter expression by $s = e^{\int \lambda(t)}$ you get $\alpha(s) = \beta_0 + s \sigma_0$ which is a (portion of a) straight line.