Find the equation of the chord of the hyperbola $x^2-y^2=9$ which is bisected at $(5,-3)$
This is solved in my reference as: $$ T=S_1\implies x(5)-y(-3)-9=5^2-(-3)^2-9\\ 5x+3y-16=0 $$ What is the logic behind such a substitution, $T=S_1$ ?
And how can I solve the problem without making this shortcut ?
It is $y=-\frac53x+\frac{16}3$.
Details: Let the equation be $y=ax+b$. It passes through $(5,-3)$ and crosses the hyperbola at two points: $$\begin{cases}-3=5a+b \\ x^2-(ax+b)^2=9\end{cases} \Rightarrow \begin{cases}b=-3-5a\\ (1-a^2)x^2+2a(3+5a)x-(3+5a)^2-9=0\end{cases} \Rightarrow \\ x_{1,2}=\frac{-a(3+5a)\pm \sqrt{a^2(3+5a)^2+(1-a^2)((3+5a)^2+9)}}{1-a^2}$$ From the midpoint: $$\frac{x_1+x_2}{2}=5 \Rightarrow \frac{-a(3+5a)}{1-a^2}=5 \Rightarrow a=-\frac53,b=\frac{16}{3}.$$ Reference: Desmos graph.
$$ P(\sec a,\tan a)\quad;\quad Q(\sec b,\tan b)\\ (5,-3)=(\frac{\sec a+\sec b}{2},\frac{\tan a+\tan b}{2})=(\frac{1}{2}\frac{\cos a+\cos b}{\cos a\cos b},\frac{1}{2}\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b})\\ =(\frac{2\cos\frac{a+b}{2}\cos\frac{a-b}{2}}{2\cos a\cos b},\frac{2\sin\frac{a+b}{2}\cos\frac{a+b}{2}}{2\cos a\cos b})\\ m=\frac{\tan b-\tan a}{\sec b-\sec a}=\frac{\sin(a-b)}{\cos a-\cos b}=\frac{2\sin\frac{a-b}{2}\cos\frac{a-b}{2}}{2\sin\frac{a-b}{2}\sin\frac{a+b}{2}}=\frac{\cos\frac{a-b}{2}}{\sin\frac{a+b}{2}}\\ y+3=\frac{\cos\frac{a-b}{2}}{\sin\frac{a+b}{2}}(x-5)=\frac{-5}{3}(x-5)\\ \implies 3y+9=-5x+25\implies5x+3y=16 $$