---- > [!definition] Definition. ([[convex plane curve]]) > A plane curve $\alpha:(a,b) \to \mathbb{R}^{2}$ is **convex** if for all $t$, the trace $\alpha\big((a,b)\big)$ of $\alpha(t)$ lies entirely on one side of the closed half-plane determined by the [[tangent line to a parameterized curve|tangent line]] at $t$. > \ > If $\alpha$ is a simple [[closed plane curve]], and $D$ is the [[topological interior|topological interior]] of its trace, then the above definition is equivalent to saying $D$ is a [[convex set|convex subset]] of $\mathbb{R}^{2}$. ^30658d > [!basicproperties] > Let $\alpha$ be a [[convex plane curve|convex,]] *simple* plane curve. Let $L$ be a straight [[line]] passing through two distinct points $p,q$ on $\alpha$. Then one of the following must hold: >1. $L$ intersects $\alpha$ at only two points: $L \cap \alpha\big((a,b)\big)= \{ p,q \}$ >2. An arc $\overline{pq}$ of $\alpha$ in between $p,q$ lies entirely on $L$, i.e., $L \cap \alpha\big( (a,b) \big)=\overline{pq}.$ > Moreover, in case (1), if we choose a [[parameterized curve|parameterization]] $\alpha:[0, \ell] \to \mathbb{R}^{2}$ such that $\alpha(0)=\alpha(\ell)=p$ and $\alpha(t_{0})=q$ for some $\ell > t_{0} >0$, then the arc $\alpha([0,t_{0}])$ and the arc $\alpha([t_{0},\ell])$ must locate on distinct sides of $L$. ^7f44ec > [!basicexample] > The [[ellipse]] that is the trace of $\alpha(t)=(a \cos t, b \sin t)$ is [[convex plane curve|convex]]. > We can show this directly from the definition. Fix $t_{0} \in \dom \alpha$. Compute $\begin{align} \alpha'(t)= &(-a\sin t, b \cos t) \\ \alpha''(t)= & (-a \cos t, -b \sin t). \end{align}$ Since the trace is a subset of $\mathbb{R}^{2}$, we can characterize the [[tangent line to a parameterized curve|tangent line]] to $\alpha$ at $t_{0}$ as the set of vectors orthogonal to the [[signed unit-normal vector to plane curve|signed normal]] $n_{\text{s}}(t_{0})$, translated by $\alpha(t_{0})$. > $L=\{(x,y \in \mathbb{R}^{2}: (x,y) \cdot n_{\text{s}}(t) =0\}+ \alpha(t_{0}),$ i.e., $L=\{ (x,y) \in \mathbb{R}^{2} : \big( (x,y ) - \alpha(t_{0}) \big) \cdot n_{\text{s}}(t)\}=0.$ In our case this can be simplified to $L=\{ (x,y) \in \mathbb{R}^{2}: bx \cos t_{0} + ay \sin t_{0} - ab = 0\}.$ Now plug a general point $\alpha(t)$ into $L$ and you will get the result. ![[CleanShot 2024-02-14 at [email protected]]] ^82f119 ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```