The smallest distance between any point on a curve and the parabola $y=x^2$ is 1. What is the equation of the curve? Find the equation of the function $f(x)$, where:


*

*the optimal distance between any one point $P$ on the curve $y=f(x)$ and the parabola $y=x^2$ is always equal to $1$

*$f(x)>x^2$ for all $x$ (looking for the solution curve above the parabola, not below)


Here is a visual approximation of the construction on Desmos: https://www.desmos.com/calculator/pyomsrazo7
This problem arose when attempting to find the equation of motion of a ball with radius $1$ as it rolls along the concave side of the parabola $y=x^2$. Point $P$ is the center of the ball whose path of motion is the curve $y=f(x)$.

Let a point on the curve be $(a,f(a)$ and the tangent point on the parabola be $(b,b^2)$. Two conditions must be true for all $a \in \mathbb R$
First, the distance is $1$
$$ (b-a)^2 + (b^2-f(a))^2 = 1 $$
Second, the slope between the two points must be normal to the tangent slope, which is $2b$
$$ \frac{b^2-f(a)}{b-a} = -\frac{1}{2b} $$
Substituting $b^2-f(a) = -\frac{b-a}{2b}$, we get
$$ (b-a)^2 = \frac{4b^2}{4b^2+1} $$
Since $(a,f(a))$ is always on the interior of the parabola, we have $b-a > 0$ if $b > 0$, and $b-a < 0$ if $b < 0$. Therefore $b-a$ should have the same sign as $b$ and we can simplify
$$ b-a = \frac{2b}{\sqrt{4b^2+1}} $$
and 
$$ b^2 - f(a) = -\frac{1}{\sqrt{4b^2+1}} $$
From here, you have the solution in parametric form
\begin{align} x &= b - \frac{2b}{\sqrt{4b^2+1}} \\
y &= b^2 + \frac{1}{\sqrt{4b^2+1}} \end{align}
Here is a visualization. Two "bends" occur when $b = \pm \frac12$

Also note that this does not represent one continuous motion in your original proposed problem, as the ball is too large and would get stuck at the peak, where the tangent points are $b = \pm \frac{\sqrt{3}}{2}$
EDIT: A natural question you might ask is, what radius does the ball need to be to not get stuck at the peak? To find out, redo the problem with a radius of $r>0$ to get the parametric solution
\begin{align} x &= b - \frac{2br}{\sqrt{4b^2+1}} \\
y &= b^2 + \frac{r}{\sqrt{4b^2+1}} \end{align}
A simple fix is to restrict $r$ so we have only one tangent point at $x=0$. Solving $x=0$ gives
$$ (4b^2 + 1 - 4r^2)b^2 = 0 $$
The only solution needs to be $b=0$, which means
$$ 1-4r^2 \ge 0 $$
Therefore $r \le \frac12$
Here's a solution curve with $r=0.4$

Actually, $r=\frac12$ does not result in a smooth curve, so you may not want to include it. But $\forall r < \frac12$ is valid
EDIT 2: You can also obtain an "outer" solution by switching the signs
\begin{align} x &= b + \frac{2br}{\sqrt{4b^2+1}} \\
y &= b^2 - \frac{r}{\sqrt{4b^2+1}} \end{align}
This solution has no restriction on $r$ (the curve is always smooth no matter how large the radius)
