Point where two hyperbolas touch each other

Question

This is a simple question that showed up in my research, but it has been vexing me a fair bit.

Consider the unit square $\{(x,y): 0\leq x,y\leq 1\}$ and define two hyperbolas restricted to the unit square region:

$H_1: \alpha = \frac{\beta}{x} + \frac{\gamma}{y}$ where $\alpha,\beta,\gamma>0$ and $\alpha>\beta+\gamma.$

$H_2: (1-x)(1-y)=\sigma^2$

Given $\alpha,\beta,\gamma,$ I want to determine the unique value of $\sigma_0\in [0,1]$ for which the two hyperbolas $H_1,H_2$ touch in a point.

Below, I have shown a picture for the case $\alpha = 5, \beta = 3, \gamma = 1,$ in which case the value of $\sigma_0$ turns out to be numerically $0.2192.$ (The blue region is the "interior" of $H_1$ and the red region is the "interior" of $H_2.$)

Let the point of touching be $(x_0,y_0).$ By equating the slopes of the two curves at $(x_0,y_0),$ we get that

$$\frac{1-y_0}{1-x_0} = \frac{y_0^2}{x_0^2}\frac{\beta}{\gamma}$$

Furthermore, $(x_0,y_0)$ must lie on both curves, so that

$$\alpha = \frac{\beta}{x_0} + \frac{\gamma}{y_0}$$

and

$$(1-x_0)(1-y_0) = \sigma_0^2.$$

Is it easy to solve this system of three equations in three unknowns $(x_0,y_0,\sigma_0)$ in terms of the known variables $(\alpha,\beta,\gamma)$? In fact, I am only interested in determining $\sigma_0$ explicitly in terms of $\alpha,\beta,\gamma.$ Would switching to a different co-ordinate system make this easier?

Thanks.

Answer 1

The single Mathematica command

Reduce[{(1 - y0)/(1 - x0) == y0^2 b/(x0^2 g), a == b/x0 + g/y0,
(1 - x0) (1 - y0) == s, a > 0, b > 0, g > 0, a > b + g, 0 <= s <= 1,
0 <= x0 <= 1, 0 <= y0 <= 1}, {x0, y0, s}, Reals]

gives the solution $$x_0 = \frac{\beta}{\alpha} + \sqrt{\frac{(\alpha-\beta)\beta\gamma}{\alpha^2(\alpha-\gamma)}}, \quad y_0 = \frac{\gamma x_0 }{\alpha x_0 - \beta}, \quad \sigma^2 = (1-x_0)(1-y_0),$$ although I have not checked this result for correctness. It does suggest, however, that it is possible to algebraically simplify the system by hand.