Ok, everything I wrote in Post # 12 is wrong. I'm going to try an entirely different approach: rotations. We start with a simple North-South hyperbola, which we can write as

$\displaystyle y^{2}-q^{2}x^{2}=a>0.$

Here $\displaystyle \pm q$ are the slopes of the asymptotes.

Here is a plot, where $\displaystyle q=1/4.$

What we would like to do is rotate this figure counter-clockwise through some angle such that the $\displaystyle -1/4$ asymptote lines up with the x axis, and the $\displaystyle +1/4$ asymptote becomes the slope of the line corresponding to your data. It turns out that the hyperbola to start with such that when you rotate it through a certain angle, you get one asymptote to be the x axis, and the other the line $\displaystyle y=m(x-p),$ is the following:

$\displaystyle y^{2}-\left(\tan\left(\frac{\tan^{-1}(m)}{2}\right)\right)^{\!\!2}x^{2}=a.$

Next, we rotate the coordinates counter-clockwise through the angle

$\displaystyle \frac{\tan^{-1}(m)}{2}.$

Define

$\displaystyle C=\cos\left(\frac{\tan^{-1}(m)}{2}\right),$ and

$\displaystyle S=\sin\left(\frac{\tan^{-1}(m)}{2}\right).$

Then our new coordinates become

$\displaystyle \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix} C &-S\\ S &C\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}.$

Plugging this into our hyperbola equation yields

$\displaystyle (Sx+Cy)^{2}-\left(\tan\left(\frac{\tan^{-1}(m)}{2}\right)\right)^{\!\!2}(Cx-Sy)^{2}=a.$

Finally, to translate the hyperbola to the right by $\displaystyle p$ units, we simply replace $\displaystyle x$ by $\displaystyle x-p$ to obtain

$\displaystyle (S(x-p)+Cy)^{2}-\left(\tan\left(\frac{\tan^{-1}(m)}{2}\right)\right)^{\!\!2}(C(x-p)-Sy)^{2}=a.$

Solving for y yields the upper sheet equation of

$\displaystyle y_{+}=\frac{\sqrt{4(4C^{2}-m^{2}S^{2})(4a+(C^{2}m^{2}-4S^{2})(x-p)^{2})+4C^{2}(4+m^{2})^{2}S^{2}(x-p)^{2}}+2C(4+m^{2})S(x-p)}{8C^{2}-2m^{2}S^{2}}.$

So this is the equation to which you're going to fit your data. It has three parameters: the slope $\displaystyle m,$ the x-intercept $\displaystyle p,$ and an arbitrary parameter $\displaystyle a$ that measures how close the hyperbola gets to the intersection of its two asymptotes.

I have plotted this equation, and it looks right to me.

That's all for now. Next step: least-squares fit!