\(% 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}}\)
The line through \(P\) and \(Q\) has gradient \[ \frac {aq^2 - ap^2}{2aq - 2ap} = \frac {q^2 - p^2}{2 (q - p)} = \frac {p + q}{2}, \] and so it has equation \begin {align*} y - ap^2 & = \frac {1}{2} (p + q) (x - 2ap) \\ y & = \frac {1}{2} (p + q) x + ap^2 - ap^2 - apq \\ y & = \frac {1}{2} (p + q) x - apq. \end {align*}
The line is tangent to the circle with centre \((0, 3a)\) and radius \(2a\), if and only if its distance from \((0, 3a)\) is \(2a\).
The line has equation \[ 2y - (p + q) x + 2apq = 0 \] and hence the distance is \begin {align*} d & = \frac {\abs *{2 \cdot 3a - (p + q) \cdot 0 + 2apq}}{\sqrt {2^2 + (p + q)^2}} \\ & = \frac {\abs *{6a + 2apq}}{\sqrt {4 + p^2 q^2 + 6pq + 5}} \\ & = \frac {\abs *{2a (3 + pq)}}{\sqrt {(pq + 3)^2}} \\ & = \frac {2a \abs *{3 + pq}}{\abs *{3 + pq}} \\ & = 2a, \end {align*}
and so the distance from \(l\) to \((0, 3a)\) is \(2a\) as desired.
We rearrange the condition to an equation in \(q\) \begin {align*} p^2 + 2pq + q^2 & = p^2 q^2 + 6pq + 5 \\ \left (p^2 - 1\right ) q^2 + 4pq + \left (5 - p^2\right ) & = 0, \end {align*}
and since \(p^2 \neq 1\), this must be a quadratic.
We examine the discriminant, \(\Delta \): \begin {align*} \Delta & = (4p)^2 - 4 \left (p^2 - 1\right ) \left (5 - p^2\right ) \\ & = 16p^2 - 4\left (-p^4 - 5 + 6p^2\right ) \\ & = 4p^4 - 8p^2 + 20 \\ & = 4 \left (p^4 - 4p^2 + 5\right ) \\ & = 4 \left [\left (p^2 - 2\right )^2 + 1\right ] \\ & \geq 4 \cdot 1 \\ & = 4 \\ & > 0, \end {align*}
and so \(\Delta > 0\), meaning there will be two distinct real values of \(q\) satisfying the condition.
By Vieta’s Theorem, we have \(q_1 + q_2 = - \frac {4p}{p^2 - 1}\), and \(q_1 q_2 = \frac {5 - p^2}{p^2 - 1}\).
Notice that \[ \left (q_1 + q_2\right )^2 = \frac {16p^2}{\left (p^2 - 1\right )^2}, \] and \begin {align*} q_1^2 q_2^2 + 6 q_1 q_2 + 5 & = \frac {\left (5 - p^2\right )^2}{\left (p^2 - 1\right )^2} + \frac {6 \cdot \left (5 - p^2\right )}{\left (p^2 - 1\right )} + 5 \\ & = \frac {\left (5 - p^2\right )^2 + 6 \left (5 - p^2\right ) \left (p^2 - 1\right ) + 5 \left (p^2 - 1\right )^2}{\left (p^2 - 1\right )^2} \\ & = \frac {25 - 10p^2 + p^4 - 6p^4 + 36 p^2 - 30 + 5p^4 - 10p^2 + 5}{\left (p^2 - 1\right )^2} \\ & = \frac {16 p^2}{\left (p^2 - 1\right )^2}, \end {align*}
and so \(\left (q_1 + q_2\right )^2 = q_1^2 q_2^2 + 6 q_1 q_2 + 5\).
Let \(P\left (2ap, ap^2\right )\) for some \(p \neq 1\), and let the corresponding solutions to the condition be \(q_1, q_2\). Define the points \(Q_1 \left (2aq_1, aq_1^2\right )\) and \(Q_2 \left (2aq_2, aq_2^2\right )\).
The previous part of the question shows that \(Q_1\) and \(Q_2\) exists and are distinct.
The first part ensures that \(PQ_1\) and \(PQ_2\) are tangents to the circle.
But since \(q_1\) and \(q_2\) satisfies the conditions as well, we must have \(Q_1 Q_2\) being a tangent to the circle as well.
Hence, triangle \(P Q_1 Q_2\) has all vertices on \(x^2 = 4ay\), and that all three sides are tangent to the desired circle.