Problem source

Given a curve described by $y=\mathrm{f}(x)$, and such that $y\geqslant0$, a \textit{push-off }of the curve is a new curve obtained as follows:
for each point $(x,\mathrm{f}(x))$ with position vector $\mathbf{r}$ on the original curve, there is a point with position vector $\mathbf{s}$ on the new curve such that $\mathbf{s-r}=\mathrm{p}(x)\mathbf{n},$ where $\mathrm{p}$ is a given function and $\mathbf{n}$ is the downward-pointing unit normal to the original curve at $\mathbf{r}$. 
\begin{questionparts}
\item For the curve $y=x^{k},$ where $x>0$ and $k$ is a positive integer, obtain the function $\mathrm{p}$ for which the push-off is the positive $x$-axis, and find the value of $k$ such that, for all points on the original curve, $\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|$. 
\item Suppose that the original curve is $y=x^{2}$ and $\mathrm{p}$ is such that the gradient of the curves at the points with position vectors $\mathbf{r}$ and $\mathbf{s}$ are equal (for every point on the original curve). By writing $\mathrm{p}(x)=\mathrm{q}(x)\sqrt{1+4x^{2}},$ where $\mathrm{q}$ is to be determined, or otherwise, find the form of $\mathrm{p}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose we have $y = x^k$, then the tangent at $(t,t^k)$ has gradient $\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}$ and the normal has gradient $-\frac1k x^{1-k}$. For the push-off to be the positive $x$-axis, we need $p(x)$ to be the length of the line. The line will have the equation:

\begin{align*}
&& \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\
y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\
&& x& =t + kt^{2k-1}
\end{align*}

The distance from $(t,t^k)$ to $(t+kt^{2k-1},0)$ is $\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}$, ie $p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}$

If  $\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|$, then we need $\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}$, but clearly this is satisfied when $k = 1$.

\item The points are $(t, t^2)$ and the normal has graident $-\frac{1}{2t}$, the normal vector is $\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}$. If we write $p(x) = q(x)\sqrt{4x^2+1}$ then the the new points are at $\binom{t+q(t)2t}{t^2-\q(t)}$ an the gradient will be $\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}$. We need it to be the case that

\begin{align*}
&& 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\
\Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\
\Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2}  \\
\Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C
 \\
\Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\
\Rightarrow && p(x) &= A
\end{align*}

So the push-off's are constants.

\end{questionparts}