Examine the following animation, where the blue parabola is rolling along a horizontal, and its focus (shown by the black dot) traces out a red curve.
What we aim to investigate is what is the equation of the red curve, i.e. what is the locus of the focus of a rolling parabola? We can begin by using a simplified diagram which I have drawn below.
We begin by defining some of the points, which are also marked on the diagram.
$F$ - $(0,a)$, which is the focus of the parabola initially. $F'$ is the focus of the parabola after it has rolled.
$P$ - $(t,t^2)$, which is a general point on the parabola initially. $P'$ is the same point on the parabola, just after it has rolled.
$V$ - $(0,0)$, which is the vertex of the parabola (i.e. its minimum turning point). $V'$ is the vertex of the parabola after it has rolled.
Let's also call $s=VP$ (curve length), and $d=FP$ (line), simply for ease of notation.
Since we are investigating the parabola's focus and we know the focus of the parabola initially $F$, all we require is the coordinates of $F'$. If we can find $F'=(x(t),y(t))$ - both $x$ and $y$ in terms of the parameter $t$, then we can eliminate $t$ for which we will be able to obtain $y(x)$. So, that's the plan; let $F'=(x,y)$.
We know $FP=F'P'$ clearly. Also, $VP\text{ (curve length)}=VP'\text{ (straight line)}=s$, since it's like the curve 'straightens out', as it rolls without slipping.
The gradient of the tangent at point $P$ is $2t$, and the gradient of $FP$ is $\frac{t^2-a}{t}$.
We define $\theta$ as the angle between $FP$ and the tangent at $P$.Note that the angle between a line and the horizontal $x$-axis is given by $\arctan\text{(gradient)}$.
$$\therefore\hspace{5pt}\theta=\arctan\left(2t\right)-\arctan\left(\frac{t^2-a}{t}\right)$$
$$\implies\tan\theta=\frac{2t-\frac{t^2-a}{t}}{1+2t^2-2a}=\frac{t^2+a}{2t^3+(1-2a)t}$$
By simple geometry, we should see from the diagram that for $F'=(x,y)$, we have
$$x(t)=s-d\cos\theta$$
$$y(t)=d\sin\theta$$
Now, we can evaluate $d,s,\sin\theta,\cos\theta$ all in terms of $t$. First, for $d$, using Pythagoras' between points $F$ and $P$,
$$d=\sqrt{\left(t^2-a\right)^2+t^2}=\sqrt{t^4+(1-2a)t^2+a^2}$$
For $s$, we use the arclength formula.
$$s=\int_0^t\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\hspace{2pt}\mathrm{d}x=\int_0^t\sqrt{1+4x^2}\hspace{2pt}\mathrm{d}x$$
This is now an integral which we can evaluate using a trigonometric substitution: $x=\frac{1}{2}\tan u$.
$$\int_0^t\sqrt{1+4x^2}\hspace{2pt}\mathrm{d}x = \int_0^t \frac{\sec^2(u)\sqrt{\tan^2(u)+1}}{2}\hspace{2pt}\mathrm{d}u = \frac{1}{2}\int_0^t\sec^3 u\hspace{2pt}\mathrm{d}u$$
For this transformed integral, let $I=\int\sec^3 u\hspace{2pt}\mathrm{d}u$. Let's proceed with by parts,
$$I=\int\sec u\left(\sec^2 u\right)\hspace{2pt}\mathrm{d}u$$
$$\hspace{54pt}=\sec u\tan u - \int\sec u\tan^2 u\hspace{2pt}\mathrm{d}u$$
$$\hspace{78pt}=\sec u\tan u - \int\sec u\left(\sec^2 u-1\right)\hspace{2pt}\mathrm{d}u$$
$$\hspace{86pt}=\sec u\tan u - \int\sec^3 u\hspace{2pt}\mathrm{d}u + \int\sec u\hspace{2pt}\mathrm{d}u$$
$$\hspace{113pt}=\sec u\tan u - \int\sec^3 u\hspace{2pt}\mathrm{d}u + \ln\left(\sec u + \tan u\right)$$
$$\therefore\hspace{5pt}I=\sec u\tan u + \ln\left(\sec u + \tan u\right) - I \implies \underline{I=\frac{1}{2}\sec u\tan u + \frac{1}{2}\ln\left(\sec u + \tan u\right)}$$
Now, going back to our original expression for $s$, and recalling $u=\arctan 2x$,$\tan(\arctan(2x))=2x$, $\sec(\arctan(2x))=\sqrt{4x^2+1}$ we can finally evaluate it as
$$s(t)=\frac{1}{2}t\sqrt{4t^2+1}+\frac{1}{4}\ln\left(\sqrt{4t^2+1}+2t\right)$$
We are now almost there at our goal of finding $x(t)$ and $y(t)$ - we only require $\sin\theta$ and $\cos\theta$ now. We know $\tan\theta$ already, so this shouldn't be too hard, if a little messy.
$$\cos\theta=\frac{2t^3+(1-2a)t}{\sqrt{4t^6+(5-8a)t^4+(4a^2-2a+1)t^2+a^2}}$$
$$\sin\theta=\frac{t^2+a}{\sqrt{4t^6+(5-8a)t^4+(4a^2-2a+1)t^2+a^2}}$$
$$\therefore\hspace{5pt}x(t)=s-d\cos\theta=\frac{1}{4}\left(\frac{2(4a-1)t}{\sqrt{4t^2+1}}+\ln\left(2t+\sqrt{4t^2+1}\right)\right)$$
$$y(t)=d\sin\theta=\frac{\left(\sqrt{t^4+(1-2a)t^2+a^2}\right)\left(t^2+a\right)}{\sqrt{4t^6+(5-8a)t^4+(4a^2-2a+1)t^2+a^2}}=\frac{t^2+a}{\sqrt{4t^2+1}}$$
As shown by both the animation and the diagram, our parabola is symmetric about the $y$-axis, and so we will stick with this parabola of $y=x^2$, and so $a=\frac{1}{4}$. We could use a more generic form of parabola, but this yields very messy algebra, so we take the $y=x^2$ case for a clean paramaterisation.
$$\therefore\hspace{5pt}x(t)=\frac{1}{4}\ln\left(2t+\sqrt{4t^2+1}\right),\hspace{5pt}y(t)=\frac{1}{4}\sqrt{4t^2+1}$$
We have our parametric equations, and now let's go ahead and convert into Cartesian form, expressing $y$ in terms of $x$. From the $x(t)$ equation,
$$2t+\sqrt{4t^2+1}=e^{4x}\implies4t\hspace{1pt}e^{4x}=e^{8x}-1\implies t=\frac{1}{4}\left(e^{4x}-e^{-4x}\right)$$
$$\therefore\hspace{5pt}y=\frac{1}{4}\sqrt{4\cdot\frac{1}{16}\left(e^{4x}-e^{-4x}\right)^2+1}$$
$$\hspace{8pt}=\frac{1}{4}\sqrt{\frac{1}{4}e^{8x}+\frac{1}{4}e^{-8x}+\frac{1}{2}}$$
$$\hspace{46pt}=\frac{1}{4}\left(\frac{1}{2}e^{4x}+\frac{1}{2}e^{-4x}\right)=\frac{1}{4}\cosh4x$$
$$\therefore\hspace{5pt}\boxed{y=\frac{1}{4}\cosh4x}\hspace{10pt}\square$$
There we are; the locus of the focus of a rolling parabola is a cosh curve.
