\(% 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 hyperbola has parametric equation \[ \left \{ \begin {aligned} x & = a \sec \theta , \\ y & = b \tan \theta . \end {aligned} \right . \]
Hence, by differentiation, we have \begin {align*} \DiffFrac {y}{x} & = \frac {\DiffFrac {y}{\theta }}{\DiffFrac {x}{\theta }} \\ & = \frac {b \sec ^2 \theta }{a \sec \theta \tan \theta } \\ & = \frac {b \cos \theta }{a \sin \theta \cos \theta } \\ & = \frac {b}{a \sin \theta }. \end {align*}
The tangent to the hyperbola at \(P\) will be \[ y - b \tan \theta = \frac {b}{a \sin \theta } (x - a \sec \theta ), \] which simplifies to \[ ay \sin \theta - ab \tan \theta \sin \theta = bx - ab \sec \theta , \] and hence \[ bx - ay \sin \theta = ab (\sec \theta - \tan \theta \sin \theta ). \]
Notice that \[ \sec \theta - \tan \theta \sin \theta = \frac {1 - \sin ^2 \theta }{\cos \theta } = \frac {\cos ^2 \theta }{\cos \theta } = \cos \theta , \] and so the equation of the tangent is \[ bx - ay \sin \theta = ab \cos \theta , \] exactly as desired.
Let \(\frac {x}{a} = \frac {y}{b} = s\) for \(S\), we have \(x = as\) and \(y = bs\), and hence \[ abs - abs \sin \theta = ab \cos \theta , \] which gives \[ s = \frac {\cos \theta }{1 - \sin \theta }, \] and hence \[ S \left (a\frac {\cos \theta }{1 - \sin \theta }, b\frac {\cos \theta }{1 - \sin \theta }\right ). \]
Let \(\frac {x}{a} = - \frac {y}{b} = t\) for \(T\), we have \(x = at\) and \(y = -bt\), and hence \[ abt + abt \sin \theta = ab \cos \theta , \] which gives \[ t = \frac {\cos \theta }{1 + \sin \theta }, \] and hence \[ T \left (a\frac {\cos \theta }{1 + \sin \theta }, -b\frac {\cos \theta }{1 + \sin \theta }\right ). \]
We have \begin {align*} \frac {a\frac {\cos \theta }{1 - \sin \theta } + a\frac {\cos \theta }{1 + \sin \theta }}{2} & = \frac {a \cos \theta }{2} \left (\frac {1}{1 - \sin \theta } + \frac {1}{1 + \sin \theta }\right ) \\ & = \frac {a \cos \theta }{2} \left (\frac {2}{\cos ^2 \theta }\right ) \\ & = \frac {a}{\cos \theta } \\ & = a \sec \theta , \end {align*}
and \begin {align*} \frac {a\frac {\cos \theta }{1 - \sin \theta } - b\frac {\cos \theta }{1 + \sin \theta }}{2} & = \frac {b \cos \theta }{2} \left (\frac {1}{1 - \sin \theta } - \frac {1}{1 + \sin \theta }\right ) \\ & = \frac {b \cos \theta }{2} \left (\frac {2 \sin \theta }{\cos ^2 \theta }\right ) \\ & = \frac {b \sin \theta }{\cos \theta } \\ & = b \tan \theta . \end {align*}
This means the midpoint of \(ST\) is \((a \sec \theta , b \tan \theta )\), which is exactly \(P\).
Since the tangents are perpendicular, that means \[ \LEvalAt {\DiffFrac {y}{x}}{\theta } \cdot \LEvalAt {\DiffFrac {y}{x}}{\phi } = -1, \] and hence \[ \frac {b}{a \sin \theta } \cdot \frac {b}{a \sin \phi } = -1, \] which means \[ b^2 = - a^2 \sin \theta \sin \phi . \]
The two tangents are \[ bx - ay \sin \theta = ab \cos \theta \] and \[ bx - ay \sin \phi = ab \cos \phi . \]
Since \(bx = bx\), we have \[ ay \sin \theta + ab \cos \theta = ay \sin \phi + ab \cos \phi , \] and hence \[ y (\sin \theta - \sin \phi ) = b (\cos \phi - \cos \theta ), \] which gives \[ y = b \cdot \frac {\cos \phi - \cos \theta }{\sin \theta - \sin \phi }. \]
Hence, \begin {align*} x & = \frac {ab \cos \theta + ay \sin \theta }{b} \\ & = \frac {a}{b} \left (b \cos \theta + b \sin \theta \frac {\cos \phi - \cos \theta }{\sin \theta - \sin \phi }\right ) \\ & = a \left (\cos \theta + \sin \theta \frac {\cos \phi - \cos \theta }{\sin \theta - \sin \phi }\right ) \\ & = a \frac {\cos \theta (\sin \theta - \sin \phi ) + \sin \theta (\cos \phi - \cos \theta )}{\sin \theta - \sin \phi } \\ & = a \cdot \frac {\sin \theta \cos \phi - \cos \theta \sin \phi }{\sin \theta - \sin \phi } \\ & = a \cdot \frac {\sin (\theta - \phi )}{\sin \theta - \sin \phi }. \end {align*}
This means \[ \left \{ \begin {aligned} x^2 & = a^2 \cdot \frac {\sin ^2 (\theta - \phi )}{\left (\sin \theta - \sin \phi \right )^2}, \\ y^2 & = b^2 \cdot \frac {\left (\cos \phi - \cos \theta \right )^2}{\left (\sin \theta - \sin \phi \right )^2} = - a^2 \sin \theta \sin \phi \cdot \frac {\left (\cos \phi - \cos \theta \right )^2}{\left (\sin \theta - \sin \phi \right )^2}. \end {aligned} \right . \]
Notice that \[ a^2 - b^2 = a^2 + a^2 \sin \theta \sin \phi = a^2 (1 + \sin \theta \sin \phi ). \]
Hence, \begin {align*} x^2 + y^2 & = a^2 \left [\frac {\sin ^2 (\theta - \phi )}{\left (\sin \theta - \sin \phi \right )^2} - \sin \theta \sin \phi \cdot \frac {\left (\cos \phi - \cos \theta \right )^2}{\left (\sin \theta - \sin \phi \right )^2}\right ] \\ & =\frac {a^2}{(\sin \theta - \sin \phi )^2} \left [\sin ^2 (\theta - \phi ) - \sin \theta \sin \phi \left (\cos \phi - \cos \theta \right )^2\right ]. \end {align*}
What is desired is to show \[ (1 + \sin \theta \sin \phi ) (\sin \theta - \sin \phi )^2 = \sin ^2 (\theta - \phi ) - \sin \theta \sin \phi \left (\cos \phi - \cos \theta \right )^2. \]
We have \begin {align*} \RHS & = (\sin \theta \cos \phi - \cos \theta \sin \phi )^2 - \sin \theta \sin \phi (\cos ^2 \phi + \cos ^2 \theta - 2 \cos \phi \cos \theta ) \\ & = \sin ^2 \theta \cos ^2 \phi + \cos ^2 \theta \sin ^2 \phi - 2 \sin \theta \cos \theta \sin \phi \cos \phi \\ & \phantom {=} - \sin \theta \sin \phi \cos ^2 \phi - \sin \theta \sin \phi \cos ^2 \theta + 2 \sin \theta \cos \theta \sin \phi \cos \phi \\ & = \sin \theta \cos ^2\phi (\sin \theta - \sin \phi ) + \cos ^2 \theta \sin \phi (\sin \phi - \sin \theta ) \\ & = (\sin \theta \cos ^2 \phi - \cos ^2 \theta \sin \phi ) (\sin \theta - \sin \phi ). \end {align*}
Therefore, what is left to prove is that \[ (1 + \sin \theta \sin \phi )(\sin \theta - \sin \phi ) = \sin \theta \cos ^2 \phi - \cos ^2 \theta \sin \phi \]
Notice that \begin {align*} \LHS & = \sin \theta - \sin \phi + \sin ^2 \theta \sin \phi - \sin \theta \sin ^2 \phi \\ & = \sin \theta (1 - \sin ^2 \phi ) - \sin \phi (1 - \sin ^2 \theta ) \\ & = \sin \theta \cos ^2 \phi - \sin \phi \cos ^2 \theta \\ & = \RHS . \end {align*}
This shows that \[ \frac {1}{(\sin \theta - \sin \phi )^2} \left [\sin ^2 (\theta - \phi ) - \sin \theta \sin \phi \left (\cos \phi - \cos \theta \right )^2\right ] = 1 + \sin \theta \sin \phi , \] and hence \[ x^2 + y^2 = a^2 - b^2, \] as desired.