\(% 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}}\)
For \(y^2 = 4ax\), we have \(x = \frac {y^2}{4a}\), and therefore \[ \DiffFrac {x}{y} = \frac {2y}{4a}. \]
Therefore, the normal through \(Q\), \(l_Q\) satisfies that \[ l_Q: x - aq^2 = - \frac {4a}{2 \cdot 2aq} \cdot \left (y - 2aq\right ), \] i.e. \[ l_Q: q(x - aq^2) = -\left (y - 2aq\right ). \]
Since \(P \in l_Q\), we must have \begin {align*} q(ap^2 - aq^2) & = -\left (2ap - 2aq\right ) \\ aq(p+q)(p-q) & = -2a (p-q) \\ pq + q^2 & = -2 \\ q^2 + pq + 2 & = 0 \end {align*}
as desired.
We also have \[ r^2 + pr + 2 = 0. \] Since \(q \neq r\), \(q, r\) are the solutions to the equation \[ x^2 + px + 2 = 0, \] and therefore \(q + r = -p, qr = 2\).
Note that the equation for \(QR\) satisfies that \[ m_{QR} = \frac {2ar - 2aq}{ar^2 - aq^2} = \frac {2}{r + q}. \]
Therefore, \(l_{QR}\) satisfies that \begin {align*} l_{QR}: y - 2aq & = \frac {2}{r + q} (x - aq^2) \\ y & = \frac {2}{r + q} \left (x - aq^2 + \frac {r + q}{2}\cdot 2aq\right ) \\ y & = \frac {2}{r + q} \left (x - aq^2 + aq^2 + aqr\right ) \\ y & = \frac {2}{r + q} \left (x + aqr\right ) \\ y & = -\frac {2}{p} (x + 2a). \end {align*}
This passes through a fixed point \((-2a, 0)\).
\(OP\) has equation \(y = \frac {2ap}{ap^2}x\), which is \(y = \frac {2x}{p}\). Therefore, since \(T = OP \cap QR\), \(x_T\) must satisfy that \begin {align*} -\frac {2}{p} (x + 2a) & = \frac {2x}{p}, \\ -(x + 2a) & = x \\ x & = -a. \end {align*}
Therefore, \(y_T = -\frac {2a}{p}\), \(T\left (-a, -\frac {2a}{p}\right )\) lies on the line \(x = -a\) which is independent of \(p\).
The distance from the \(x\)-axis to \(T\) is \(\abs *{\frac {2a}{p}} = \frac {2a}{\abs *{p}}\).
Notice that since \(qr = 2\), \(q\) and \(r\) must take the same parity, and therefore \(\abs *{p} = \abs *{q} + \abs *{r}\). By the AM-GM inequality, we have \[ \abs *{q} + \abs *{r} \geq 2 \sqrt {\abs *{q} \cdot \abs *{r}} = 2\sqrt {2}, \] with the equal sign holding if and only if \(\abs *{q} = \abs *{r}\), \(q = r\), which is impossible.
Therefore, \(\abs *{p} > 2\sqrt {2}\) and therefore \(\frac {2a}{\abs *{p}} < \sqrt {2}\) as desired.