Hyperbola + Locus = Deadly Combination!

Relevant wiki: Locus

Any point on the hyperbola $xy=1$ can be written as $(t,\frac{1}{t})$ .

Let for a particular line the intersection points be $\left(t_1,\frac{1}{t_1}\right)$ and $\left(t_2,\frac{1}{t_2}\right)$ $\therefore \frac{\frac{1}{t_2}-\frac{1}{t_1}}{t_2-t_1}=4 \implies t_1t_2=-\frac{1}{4}$ Required point $\text{P}\equiv\left(\frac{2t_1+t_2}{3},\frac{\frac{2}{t_1}+\frac{1}{t_2}}{3}\right)$
Substituting $t_2=-\frac{1}{4t_1}$
$\text{P} \equiv \left(\frac{2t_1-\frac{1}{4t_1}}{3},\frac{\frac{2}{t_1}-4t_1}{3}\right)\equiv(x,y)$ $3x=2t_1-\frac{1}{4t_1} \quad ; \quad 3y=-4t_1+\frac{2}{t_1}$ Eliminating $t_1$ we get,
$2x+y=\frac{1}{2t_1}\quad...[1]$ Eliminating $\frac{1}{t_1}$ we get
$8x+y=4t_1\quad...[2]$ $[1]\times[2]\implies (8x+y)(2x+y)=2$ $\implies 16x^2+y^2+10xy=2$ Thus, answer is $\boxed{24}$