\(% 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 angle between a line with gradient \(m\) and the positive \(x\)-axis is \(\arctan m\). Hence, we must have \begin {align*} \arctan m_1 - \arctan m_2 & = \pm \frac {\pi }{4} \\ \tan \left (\arctan m_1 - \arctan m_2\right ) & = \tan \left (\pm \frac {\pi }{4}\right ) \\ \frac {m_1 - m_2}{1 + m_1 m_2} & = \pm 1, \end {align*}
as desired.
We have \(y = \frac {x^2}{4a}\), and hence \(\DiffFrac {y}{x} = \frac {x}{2a}\). Hence, the tangent to the point \(\left (p, \frac {p^2}{4a}\right )\) is given by \begin {align*} y - \frac {p^2}{4a} & = \frac {p}{2a} \left (x - p\right ) \\ 4ay - p^2 & = 2p (x - p) \\ 4ay & = 2px - p^2, \end {align*}
with gradient \(\frac {2p}{4a} = \frac {p}{2a}\), and the tangent to the point \(\left (q, \frac {q^2}{4a}\right )\) is given by \(4ay = 2qx + q^2\), with gradient \(\frac {q}{2a}\).
Hence, when they intersect, it must be the case that \begin {align*} 2px - p^2 & = 2qx - q^2 \\ 2 (p - q) x & = p^2 - q^2 \\ 2 (p - q) x & = (p + q) (p - q) \\ x & = \frac {p + q}{2} \end {align*}
since \(p \neq q\).
The \(y\)-coordinate is given by \begin {align*} y & = \frac {2px - p^2}{4a} \\ & = \frac {p^2 + pq - p^2}{4a} \\ & = \frac {pq}{4a}. \end {align*}
If the two curves meet at \(\frac {\pi }{4}\), the gradients must satisfy that \begin {align*} \frac {\frac {p}{2a} - \frac {q}{2a}}{1 + \frac {p}{2a} \cdot \frac {q}{2a}} & = \pm 1 \\ \frac {2a (p - q)}{4a^2 + pq} & = \pm 1 \\ 2a (p - q) & = \pm \left (4a^2 + pq\right ) \\ 4a^2 (p - q)^2 & = (4a^2 + pq)^2 \\ 4a^2 p^2 - 8 a^2 pq + 4 a^2 q^2 & = 16 a^4 + 8 a^2 pq + p^2 q^2 \\ p^2 q^2 + 16 a^2 pq + 16 a^4 - 4 a^2 p^2 - 4 a^2 q^2 & = 0. \end {align*}
For the intersection of the two tangents, we consider \((y + 3a)^2 - (8a^2 + x^2)\).
\begin {align*} (y + 3a)^2 - (8a^2 + x^2) & = y^2 + 6ay + 9a^2 - 8a^2 - x^2 \\ & = y^2 + 6ay - x^2 + a^2 \\ & = \frac {p^2 q^2}{16a^2} + 6a \cdot \frac {pq}{4a} - \left (\frac {p + q}{2}\right )^2 + a^2 \\ & = \frac {p^2 q^2}{16a^2} + \frac {3pq}{2} - \frac {(p + q)^2}{4} + a^2. \end {align*}
We have the following being equivalent: \begin {align*} (y + 3a)^2 & = 8a^2 + x^2 \\ \frac {p^2 q^2}{16 a^2} + \frac {3pq}{2} - \frac {(p + q)^2}{4} + a^2 & = 0 \\ p^2 q^2 + 3pq \cdot 8a^2 - (p + q)^2 \cdot 4a^2 + a^2 \cdot 16 a^2 & = 0 \\ p^2 q^2 + 24 pq a^2 - 4 a^2 p^2 - 4 a^2 q^2 - 8 pq a^2 + 16 a^4 & = 0 \\ p^2 q^2 + 16 a^2 pq + 16 a^4 - 4 a^2 p^2 - 4 a^2 q^2 & = 0, \end {align*}
which was true due to the tangents intersecting at \(\frac {\pi }{4}\).
Hence, we must have the intersection of two tangents lie on \((y + 3a)^2 = 8a^2 + x^2\), which finishes our proof.
Let \(\theta \) be this acute angle, and from the previous part, we can see that \begin {align*} 4a^2 (p - q)^2 & = \tan ^2 \theta (4a^2 + pq)^2 \\ 4a^2 p^2 - 8 a^2 pq + 4 a^2 q^2 & = \tan ^2 \theta 16a^4 + \tan ^2 \theta 8 a^2 pq + \tan ^2 \theta p^2 q^2 \\ \tan ^2\theta p^2 q^2 + 8(\tan ^2 \theta + 1) a^2 pq + \tan ^2 \theta 16 a^4 & = 4a^2 p^2 + 4 a^2 q^2 \end {align*}
Given \((y + 7a)^2 = 48 a^2 +3x^2\) for the intersection of the two tangents, we have \begin {align*} (y + 7a)^2 - \left (48 a^2 +3x^2\right ) & = 0 \\ \left (\frac {pq}{4a} + 7a\right )^2 - \left (48 a^2 + 3 \left (\frac {p + q}{2}\right )^2\right ) & = 0 \\ \frac {p^2 q^2}{16 a^2} + \frac {7pq}{2} + 49 a^2 - 48a^2 - \frac {3 (p + q)^2}{4} & = 0 \\ p^2 q^2 + 8 a^2 \cdot 7pq + 16 a^4 - 3 (p + q)^2 \cdot 4 a^2 & = 0 \\ p^2 q^2 + 56 pq a^2 + 16 a^4 - 12 p^2 a^2 - 12 q^2 a^2 - 24 pq a^2 & = 0 \\ p^2 q^2 + 32 pq a^2 + 16 a^4 - 12 p^2 a^2 - 12 q^2 a^2 & = 0 \\ p^2 q^2 + 32 pq a^2 + 16 a^4 - 3 \left (\tan ^2\theta p^2 q^2 + 8(\tan ^2 \theta + 1) a^2 pq + 16 \tan ^2 \theta a^4\right ) & = 0 \\ (1 - 3 \tan ^2 \theta ) p^2 q^2 + 8 (1 - 3 \tan ^2 \theta ) pq a^2 + 16 (1 - 3 \tan ^2 \theta ) a^4 & = 0 \\ (1 - 3 \tan ^2 \theta ) \left [p^2 q^2 + 8 pq a^2 + 16 a^4\right ] & = 0 \\ (1 - 3 \tan ^2 \theta ) (pq + 4a^2)^2 & = 0. \end {align*}
Hence, either \(pq + 4a^2 = 0\), or \(1 - 3 \tan ^2 \theta = 0\). The former cannot always the case. Therefore, \(1 - 3 \tan ^2 \theta = 0\), which gives \(\tan \theta = \pm \frac {\sqrt {3}}{3}\).
Since \(\theta \) is acute, we have \(\tan \theta = \frac {\sqrt {3}}{3}\), and hence \(\theta = \frac {\pi }{6}\) is the acute angle between the two tangents.