Distance between points on a curve

The curve in the first quadrant follows the equation $y = \frac{1}{x}$ , and using a derivative the slope of that curve at any point $P_0(p, \frac{1}{p})$ is $y = -\frac{1}{p^2}$ , so the tangent line to any point $P_0(p, \frac{1}{p})$ on the curve is $y = -\frac{1}{p^2}x + \frac{2}{p}$ .

By rotational symmetry, $P_{-1}$ is at $(\frac{1}{p}, -p)$ , and since this point is also on the tangent curve, we have $-p = -\frac{1}{p^2} \cdot \frac{1}{p} + \frac{2}{p}$ which solves to $p = \sqrt{-1 + \sqrt{2}}$ .

Therefore, $P_0$ , or $Q_1$ , is $(\sqrt{-1 + \sqrt{2}}, \frac{1}{\sqrt{-1 + \sqrt{2}}})$ , and by symmetry, $Q_2$ is $(\frac{1}{\sqrt{-1 + \sqrt{2}}}, \sqrt{-1 + \sqrt{2}})$ , which makes the distance $d(Q_1, Q_2) = \sqrt{-4 + 4\sqrt{2}}$ , so $a = -4$ , $b = 4$ , and $c = 2$ , and $\frac{ab}{c} = \boxed{-8}$ .

The curve $x^2y^2 =1$ $\implies y = \pm \dfrac 1x$ . The blue curves in the first and third quadrants are $y = \dfrac 1x$ , while the orange curves in the second and fourth quadrants are $y = - \dfrac 1x$ .

Let $Q_1$ in the first quadrant be $\left(a, \dfrac 1a\right)$ . Then its corresponding $P_{-1}$ is $\left(\dfrac 1a, -a\right)$ . The gradient of the line $Q_1P_{-1}$ , $m = \dfrac {\frac 1a+a}{a-\frac 1a} = \dfrac {a^2+1}{a^2-1}$ . Since $Q_1P_{-1}$ is also the tangent of $y=\frac 1x$ at $Q_1$ , then $m = \dfrac {dy}{dx} = - \dfrac 1{x^2}$ at $Q_1$ or $m = - \dfrac 1{a^2}$ .

Then we have:

$\begin{aligned} \frac {a^2+1}{a^2 - 1} & = - \dfrac 1{a^2} \\ a^4 + 2a^2 - 1 & =0 \\ \implies a^2 & = \sqrt 2 - 1 & \small \color{#3D99F6} \text{Since }a^2 > 0 \\ a & = \sqrt{\sqrt 2 - 1} \\ \implies \frac 1a & = \frac 1{\sqrt{\sqrt 2-1}} = \sqrt{\sqrt 2+1} \end{aligned}$

And $Q_1 \left(\sqrt{\sqrt 2-1}, \sqrt{\sqrt 2+1}\right)$ . Since $y = \dfrac 1x$ is symmetrical about $y=x$ , $Q_2$ has its $x$ -coordinate and $y$ -coordinate swapped with those of $Q_1$ or $Q_2 \left(\sqrt{\sqrt 2+1}, \sqrt{\sqrt 2-1}\right)$ .

Therefore,

$\begin{aligned} d(Q_1, Q_2) & = \sqrt{\left(\sqrt{\sqrt 2-1} - \sqrt{\sqrt 2+1}\right)^2 + \left(\sqrt{\sqrt 2+1} - \sqrt{\sqrt 2-1}\right)^2} \\ & = \sqrt{2\left(\sqrt 2-1 - 2\sqrt{(\sqrt 2+1)(\sqrt 2-1)} + \sqrt 2+1\right)} \\ & = \sqrt{-4+4\sqrt 2} \end{aligned}$

$\implies \dfrac {ab}c = \dfrac {-4\times 4}2 = \boxed{-8}$ .