\(% 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}}\)
By taking derivatives with respect to \(t\), we have \[ \DiffFrac {x}{t} = 2at, \] and \[ \DiffFrac {y}{t} = 2a, \] hence \[ \DiffFrac {y}{x} = \frac {2a}{2at} = \frac {1}{t}. \]
The gradient of the normal will hence be \(-t\), and hence the normal through \(P(ap^2, 2ap)\) will be \[ y - 2ap = -p (x - ap^2). \]
The point \(N(an^2, 2an)\) is also on this line, and hence \[ 2a (n - p) = -ap (n - p) (n + p). \]
Since \(n \neq p\), we must have \[ 2 = -p (n + p). \]
Given \(p \neq 0\), we have \[ n + p = -\frac {2}{p}, \] and hence \[ n = -p - \frac {2}{p} = -\left (p + \frac {2}{p}\right ). \]
The distance between \(P(ap^2, 2ap)\) and \(N(an^2, 2an)\) is given by \begin {align*} \abs *{PN}^2 & = \left (2ap - 2an\right )^2 + \left (ap^2 - an^2\right )^2 \\ & = a^2 \left [4 (p - n)^2 + (p - n)^2 (p + n)^2\right ] \\ & = a^2 (p - n)^2 \left [4 + 4 \left (-\frac {2}{p}\right )^2\right ] \\ & = a^2 \left [p + \left (p + \frac {2}{p}\right )\right ]^2 \cdot 4 \left (\frac {p^2 + 1}{p^2}\right ) \\ & = 4a^2 \cdot 4 \cdot \frac {(p^2 + 1)^2}{p^2} \cdot \frac {p^2 + 1}{p^2} \\ & = 16 a^2 \frac {(p^2 + 1)^3}{p^4}. \end {align*}
Let \(f(p) = \frac {(p^2 + 1)^3}{p^4}\). By differentiation, \begin {align*} f'(p) & = \frac {3 \cdot 2p \cdot (p^2 + 1)^2 \cdot p^4 - (p^2 + 1)^3 \cdot 4 \cdot p^3}{p^8} \\ & = \frac {2 (p^2 + 1)^2 p^3}{p^8} \left [3p^2 - 2(p^2 + 1)\right ] \\ & = \frac {2 (p^2 + 1)^2}{p^5} \left (p^2 - 2\right ). \end {align*}
This means that \(f'(p) = 0\) precisely when \(p^2 - 2 = 0\), i.e. \(p = \pm \sqrt {2}\).
When \(0 < p < \sqrt {2}\), \(f'(p) < 0\), and when \(\sqrt {2} < p\), \(f'(p) > 0\).
When \(p < -\sqrt {2}\), \(f'(p) < 0\), and when \(-\sqrt {2} < p < 0\), \(f'(p) > 0\).
This means that when \(p^2 - 2 = 0\) (i.e. \(p = \pm \sqrt {2}\)), \(f(p)\) has a minimum.
Since \(\abs *{PN}^2 = \frac {16}{a^2} f(p)\) is a positive multiple of \(f(p)\), we must have that \(\abs *{PN}^2\) is minimised when \(p^2 = 2\).
Since \(Q(aq^2, 2aq)\) is on the circle with diameter \(PN\), we must have that \(QP\) and \(QN\) are perpendicular.
The gradient of \(QP\) is given by \[ m_{QP} = \frac {2aq - 2ap}{aq^2 - ap^2} = \frac {2 (q - p)}{(q + p) (q - p)} = \frac {2}{q + p}, \] and the gradient of \(QN\) is given by \[ m_{QN} = \frac {2aq - 2an}{aq^2 - an^2} = \frac {2 (q - n)}{(q + n) (q - n)} = \frac {2}{q + n}. \]
Since \(QP\) and \(QN\) are perpendicular, we must have \begin {align*} m_{QP} \cdot m_{QN} = -1 & \iff \frac {2}{q + p} \cdot \frac {2}{q + n} = -1 \\ & \iff -4 = (q + p)(q + n) \\ & \iff q^2 + (p + n)q + pn = -4 \\ & \iff q^2 - \frac {2}{p} \cdot q - p^2 - 2 = -4 \\ & \iff p^2 - q^2 + \frac {2q}{p} = 2, \end {align*}
as desired.
When \(\abs *{PN}\) is a minimum, we have \(p = \pm \sqrt {2}\), and hence \[ 2 - q^2 \pm \sqrt {2}q = 2, \] which gives \[ q (q \mp \sqrt {2}) = 0. \]
Hence, \(q = 0\), or \(q = \pm \sqrt {2}\) (which means \(p = q\), which cannot be the case). When \(q = 0\), \(Q(0, 0)\) is at the origin, as desired.