\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Consider the point on the curve whose gradient is equal to \(m\). Since on the curve, \(x = \frac {y^2}{4a}\), and hence \[ \DiffFrac {x}{y} = \frac {y}{2a} = \frac {1}{m}, \] which solves to \(y_0 = \frac {2a}{m}\), and hence \(x_0 = \frac {a}{m^2}\), the tangent to this point is \(y = mx + \frac {a}{m}\).
If \(\frac {a}{m} < c\) and \(mc > a\), this means that the line \(y = mx + c\) is above the tangent. Let \(\theta = \arctan m\), and we know the perpendicular distance between these lines will be \[ \left (c - \frac {a}{m}\right ) \cdot \cos \theta = \left (c - \frac {a}{m}\right ) \cdot \frac {1}{\sqrt {m^2 + 1}} = \frac {cm - a}{m \sqrt {m^2 + 1}}. \]
If \(\frac {a}{m} \geq c\) and \(mc \leq a\), this means that the line \(y = mx + c\) is the tangent (in the equal case) or below the tangent (in the less-than case), which both means the line \(y = mx + c\) intersects with the parabola.
Hence, when \(mc \leq a\), the shortest distance is always \(0\).
The distance \(d\) between \((p, 0)\) and \((at^2, 2at)\) can be expressed as \begin {align*} d^2 & = (at^2 - p)^2 + (2at)^2 \\ & = a^2 t^4 - 2apt^2 + p^2 + 4a^2t^2 \\ & = a^2 t^4 + 2a(2a - p)t^2 + p^2. \end {align*}
We would like to minimise \(d \geq 0\), which is the same as minimising \(d^2\).
The minimum of the quadratic function \[ f(x) = a^2 x^2 + 2a(2a-p)x + p^2 \] occurs when \[ x = -\frac {2a(2a - p)}{2 \cdot a^2} = \frac {p - 2a}{a} = \frac {p}{a} - 2. \]
However, \(d^2 = f(t^2)\) and \(t^2\) can only be non-negative.
If \(\frac {p}{a} - 2 \geq 0, \frac {p}{a} \geq 2\), then this value can be taken, and the minimum will be \[ d^2 = \frac {4 a^2 p^2 - \left [2a(2a - p)\right ]^2}{4 a^2} = p^2 - (2a - p)^2 = - 4a^2 + 4ap = 4a(p - a) \] and the minimal \(d\) will be \[ d = 2\sqrt {a(p - a)}. \]
In the other case where \(\frac {p}{a} < 2\), to let the \(t^2\) value to be as close as possible to the symmetric axis, we would like \(t^2 = 0\), at which point the minimal distance will be \[ d^2 = f(0) = p^2, \] and the minimal \(d\) will be \[ d = p. \]
The circle described is simply a circle centred at \((p, 0)\) with radius \(b\). Therefore, the shortest distance will be \(d - b\) if \(d > b\), and \(0\) otherwise.
To put this into cases,
If \(p \geq 2a\), \(d = 2\sqrt {a(p - a)}\).
Otherwise, \(p < 2a\), \(d = p\).