----- > [!proposition] Proposition. ([[regular curve with tangent lines through fixed point is a line]]) > Suppose that [[regular curve|regular]] [[parameterized curve]] $\alpha:I \to \mathbb{R}^{n}$ has the property that all of its [[tangent line to a parameterized curve|tangent lines]] pass through a [[fixed point]]. Then the trace of $\alpha$ is a (segment of a) straight line. > [!proof]- Proof. ([[regular curve with tangent lines through fixed point is a line]]) > Let $p \in \mathbb{R}^{n}$ be the fixed point. By assumption, given $t \in I$ we can find $a_t \in I$ such that the [[tangent line to a parameterized curve|tangent line]] to $\alpha$ at $t$ equals $p$: $\alpha(t)+\alpha'(t) a_{t} = p.$This implies that $\alpha(t) - p = a_{t}\alpha'(t)$, i.e., $\alpha(t)-p$ is parallel to $\alpha'(t)$ and therefore $(\alpha(t)-p) \times \alpha'(t)=0 \text{ for all }t.$ [[derivative|Differentiating both sides]] via [[derivative of the cross product]] yields $(\alpha'(t) \times \alpha'(t)) + (\alpha(t)-p) \times \alpha''(t)=0,$ i.e., $(\alpha(t)-p) \times \alpha''(t)=0\text{ for all }t \in I.$ So both $\alpha'(t)$ and $\alpha''(t)$ are parallel to $\alpha(t)-p$; therefore, they are parallel to each other: $\alpha''(t) \times \alpha'(t)=0.$ Recall the characterization of [[curvature of parameterized curve|curvature]]: $\kappa(t)=\frac{\|\alpha''(t) \times \alpha'(t)\|}{\|\alpha'(t)\|^{3}},$ we see that $\kappa(t) \equiv 0$. Now by [[straight iff zero curvature]] we are done. **Note.** We needed regularity: consider the [[absolute value]] function $|x|$. ^a41f16 ----- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag