Contact order of a space curve with one of it's tangent lines

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Definition $1$: Let $\alpha: I \mapsto \mathbb{R^3}$ and $\beta: \overline{I} \mapsto \mathbb{R^3}$ be two regular curves such that $\alpha(t_0) = \beta(t_0)$, where $t_0 \in I \cap \overline{I}$. $\alpha$ and $\beta$ are said to have contact of order $n$ at $t_0$ ($n$ being an integer $\geq 1$) if all derivatives of order $\leq n$ of the functions $\alpha$ and $\beta$ are equal at $t_0$ and derivatives of order $n + 1$ are different.

Definition $2$: $\alpha: I \mapsto \mathbb{R^3}$ is said to be a regular curve if $|| \alpha'(t) || \neq 0, \forall t \in I$

Let $\alpha: I \mapsto \mathbb{R^3}$ be a regular unit speed curve.

a) If $k(s) > 0, \forall s \in I$, then prove that the tangent line to $\alpha$ at $s$ has contact of order $1$ with $\alpha$.

b) Give an example of a regular curve that has contact of order $2$ with one of it's tangent lines.

c) Is it possible to get a regular curve that has contact of order $\geq n$ with one of it's tangent lines for all integer $n \geq 1$?

a) is easy, for b) I think the example (that was) given in one of the answers (a parabola) will do, it's c) I'm having trouble with. I think the only possible example for such a curve is a line, since a curve having contact of order, say, $2$ at $t_0$ with one of it's tangent lines means the curvature at that point is $0$. I'm having trouble proving that, though. Any help would be appreciated.