Problem source

\begin{questionparts}
  \item The line $L$ has equation $y=mx+c$, where $m > 0$ and $c > 0$.
    Show that, in the case  $mc > a > 0$, the shortest distance between $L$ and the parabola $y^2=4ax$ is 
    \[ \frac{mc-a}{m\sqrt{m^2+1}}\,.\]
    What is the shortest distance in the case that $mc\le a$?
  \item Find the shortest distance between the point $(p,0)$, where    $p > 0$, and the parabola $y^2=4ax$, where $a > 0$, in the different cases that arise according to the value of $p/a$.  
[\textit{You may wish to use the parametric coordinates $(at^2, 2at)$ of points on the parabola.}]
    Hence find the shortest distance between the circle $(x-p)^2 + y^2    =b^2$, where $p > 0$ and $b > 0$, and the parabola $y^2=4ax$, where $a > 0$, in the different cases that arise according to the values of $p$, $a$    and $b$.
  \end{questionparts}

Solution source

\begin{questionparts}
\item Suppose we have the shortest distance between the two curves, and the path between the points is not a normal to both curves. Then we could shift the endpoints to reduce the distance. (Assuming we're not at a point of intersection). Therefore, the normal to the curves must be the same (or in other words) the gradients of the curves must be the same. ie we are at a point where $2y y' = 4a$ we must have $y' = m$, so $y = \frac{2a}{m}$ and $x = \frac{a}{m^2}$ and the distance from this point to the line $y=mx+c$ is $\frac{|m \frac{a}{m^2} - \frac{2a}{m}+c|}{\sqrt{m^2+1}} = \frac{|mc-a|}{m\sqrt{m^2+1}} = \frac{mc-a}{m\sqrt{m^2+1}}$. If $mc \leq a$ then we find $\frac{a-mc}{m\sqrt{m^2+1}}$

However, we must check that the two curves do not intersect (otherwise the closest distace is $0$). ie we need to check if $(mx+c)^2 = 4ax$ has any solutions, this quadratic has discriminant $(2mc-4a)^2 - 4 \cdot m^2 \cdot c^2 = 16a^2-16amc = 16a(a-mc)$ which is clearly greater than $0$ when $a \geq mc$. Therefore the shortest distance in this case is $0$.

\item The distance squared between the point $(p,0)$ and a point of the form $(at^2,2at)$ is $D^2 = (at^2-p)^2+4a^2t^2 = a^2t^4+(4a^2-2ap)t^2+p^2$ 
\begin{align*}
&& \frac{D^2}{a^2} &= t^4 + 2\left(2-\frac{p}{a}\right)t^2 + \frac{p^2}{a^2} \\
&&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 + \frac{p^2}{a^2} - \left (2-\frac{p}{a} \right)^2 \\
&&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 +\frac{4p}{a} -4 \\
\end{align*}
Therefore if $2 \leq \frac{p}{a}$ then we can find a $t$ such that we attain the minimum for $D^2/a^2$ of $\frac{4p}{a}-4$ and so $D = \sqrt{4pa-4a^2} = 2\sqrt{a(p-a)}$ . If not the smallest value will be when $t = 0$ and we will have $|p|$


Now consider all the lines joining points on the parabola to the centre of the circle. The shortest distance from the parabola to the circle will be normal to the circle and therefore will also be a line through the center. Therefore we need only consider the shortest distance from $(p,0)$ to the parabola $-b$.

Case 1: If $p \geq 2a$ we have $2\sqrt{a(p-a)} - b$ or $0$ if $b \geq 2\sqrt{a(p-a)}$

Case 2: If $p < 2a$ we have $p-b$ or $0$ if $b \geq p$
\end{questionparts}