Let's roll. :)
Deriving the parametric equations of a straight line roulette isn't terribly complicated. As I mentioned in this previous answer, rolling is best decomposed as a rotation and a translation. For this case, I'll take the straight line to be the horizontal axis.
Let's start again with a convenient parabola parametrization:
$$\begin{pmatrix}2at\\at^2\end{pmatrix}$$
where $a$ is the focal length (the distance from vertex to focus). The focus of this parabola is at the point $(0,a)$.
We also require the arclength function for this parametrization of the parabola: $s(t)=a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))$.
The trick to rolling a parabola is to consider the transformations necessary for a point on the parabola to touch an appropriate point on the straight line it is rolling on. The parametrization I have chosen is particularly convenient, in that the vertex of the parabola already touches the horizontal axis at the origin.
Going forward, we can translate the parabola to be rolled so that the intended contact point coincides with the origin. We then perform a rotation such that the parabola is now tangent to the horizontal axis, and then horizontally translate by an amount equal to the parabola's arclength. (A similar derivation is done for the cycloid.)
Mathematically, we perform this sequence of transformations on the point $(0,a)$; here is the translation:
$$\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}$$
The rotation then needed is given by the tangential angle rotation matrix. I derived the expression for the parabola in my previous answer, so I won't repeat it here. The only difference from the previous answer is that to go forward, we require a clockwise rotation, and thus we must transpose the tangential angle rotation matrix. This now gives us
$$\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$
Finally, we translate horizontally with the arclength expression given earlier:
$$\begin{pmatrix}a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))\\0\end{pmatrix}+\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$
The parametric equations for the roulette surprisingly simplify to
$$\begin{align*}x&=a\operatorname{arsinh}(t)\\y&=a\sqrt{1+t^2}\end{align*}$$
Eliminating the parameter $t$ yields the usual equation for the catenary, $y=a\cosh\frac{x}{a}$.
Here's a picture I previously did:
A similar derivation can be done to show that the directrix of the same rolling parabola envelopes a reflection about the horizontal axis of the catenary being traced by the focus.
Given a point $(a,b)$ and a horizontal line $y=k$, in $\mathbb{R^2}$.
Let the locus of the points which are equally far away from the point and the line be denoted by $(x,y)$.
Then distance between $(x,y)$ and the line is just $|y-k|$
The distance between the point $(x,y)$ and $(a,b)$ is $\sqrt{(x-a)^2 + (y-b)^2}$
Equating the two:
$|y-k|=\sqrt{(x-a)^2 + (y-b)^2}$
$(y-k)^2=(x-a)^2+(y-b)^2$
Expanding and rearranging:
$y=\frac{x^2-2ax+(a^2+b^2-k^2)}{2b-2k}$
Given any quadratic function, you can find the unique value for $a,b,k$ (thus the diretrix and focus).
Best Answer
A purely geometrical proof can be given. Consider indeed a circle (centre $C$ and radius $R$) and a line $b=HK$. Taken any point $F$ on the circle, we can draw the parabola (blue in the figure) with focus $F$ and directrix $b$.
Line $CF$ intersects that parabola at two points $A$, $B$, labeled so that $C$ is between $A$ and $F$. If $H$ is the projection of $A$ on the directrix $b$, we have by definition $FA=AH$. But then we also have $CA=AD$, where $D$ is the projection of $A$ on line $r$, parallel to $b$ and at a distance $R$ from it on the same side as $C$, because $CA=FA-R$ and $AD=AH-R$. Hence $A$ also belongs to the parabola (red in the figure) with focus $C$ and directrix $r$. Moreover, both parabolas have the same tangent line at $A$ (the bisector of angle $\angle FAH$), hence they are tangent at $A$.
In a similar way we can show that point $B$ also belongs to the parabola (green in the figure) with focus $C$ and directrix $g$ parallel to $b$ and at a distance $R$ from it, on the opposite side of $C$. As before, green and blue parabolas are tangent at $B$.
Hence, all parabolas with focus $F$ are tangent to the red and blue parabolas, which are then (by definition) their envelope.
The above proof works if the distance between $C$ and $b$ is greater than $R$. The other cases can be treated in an analogous way.