What we aim to investigate is what is the

We begin by defining some of the points, which are also marked on the diagram.

Since we are investigating the parabola's

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

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

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