Specification of Hurwitz's Theorem 
Hurwitz's Theorem in Number Theory states that for every
  irrational number $\xi$, the equation
  $$\left|\xi-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}$$ has infinitely
  many solutions $(p,q)$ such that $p$ and $q$ are relatively prime, and
  that this is not true of
  $$\left|\xi-\frac{p}{q}\right|<\frac{1}{(\sqrt{5}+\varepsilon)q^2}$$
  has finitely many relative prime solutions for any $\varepsilon>0$.

There is a meaningful sense in which this is a statement about $\phi=\frac{1+\sqrt{5}}{2}$, because the proof of the second part is a property of the number $\phi$, and if $\phi$ (plus a specific set of numbers that are equivalent to $\phi$ in a certain sense), then the bound changes to $2\sqrt{2}$. We might only care about some other irrational number, and so we might form the following problem:

Let $\mu(\xi)$ be the unique number (should it exist) that has the
  property that  $$\left|\xi-\frac{p}{q}\right|<\frac{1}{\mu(\xi)q^2}$$
  has infinitely many solutions $(p,q)$ such that $p$ and $q$ are
  relatively prime, and that
  $$\left|\xi-\frac{p}{q}\right|<\frac{1}{(\mu(\xi)+\varepsilon)q^2}$$
  has finitely many relative prime solutions for any $\varepsilon>0$.

I am interested in when $\mu(\xi)$ exists (I conjecture that this is equivalent to asking when does $\xi$ have bounded coefficients in its continued fraction expansion), and when it does exist, how to calculate it (presumably via solving a recurrence). Proofs, references, and commentary on the current state of the problem (if it is unsolved) are all welcome.
For certain instances of the problem, I have found the value of $\mu(\xi)$ via recurrences. For example, I have proven that $\mu(\sqrt{d})=2\sqrt{d}$ for square-free $d$. I know that this is related to Markov Numbers and the Markov Spectrum, however neither of those answer my question.
This is the same problem as here but I'm looking for the general case.
 A: Little unclear, it is very likely that, given the larger root of
$$ A x^2 + B xy + Cy^2,  $$
(root meaning either $x/y$ or $y/x$ so it becomes a function of one variable)
of discriminant
$$ \Delta = B^2 - 4AC > 0, $$
you want
$$ \sqrt \Delta $$
Notice that the discriminant of $x^2 - d y^2$ is $4d,$ and the square root of that is $2 \sqrt d.$
A: Because I already think this is what you are dealing with: given a binary form $A x^2 + B xy + C y^2$ with positive $\Delta = B^2 - 4 AC,$ not a square but permitted square factors. 
If you have a single solution to $A x^2 + B xy + C y^2 = W,$ there are infinitely many, and $|x|,|y|$ get arbitrarily large while $W$ stays fixed. 
Take any of the infinitely many solutions to
$$  \tau^2 - \Delta \sigma^2 = 4. $$ Create the matrix
$$
\left(
\begin{array}{rr}
\frac{\tau - B \sigma}{2} & -C \sigma \\
A \sigma & \frac{\tau + B \sigma}{2}
\end{array}
\right)
$$  
Given a solution $(x,y)$ to $A x^2 + B xy + C y^2 = W,$ we get infinitely many with 
$$ (x,y) \mapsto \left(\frac{\tau - B \sigma}{2} x -C \sigma y, \; \;  A \sigma x +  \frac{\tau + B \sigma}{2} y \right) $$
Oh, if we take the smallest $\sigma > 0$ in $  \tau^2 - \Delta \sigma^2 = 4, $ and call the resulting matrix $M,$ then all those matrices can be written as $M^n$ for $n$ a positive or negative integer.
