For an orthographic map projection (of radius $r$) centered at latitude $\varphi_0$, I believe the ellipse defining the line of latitude $\varphi$ is centered at $y$ coordinate $r\cos(\varphi_0)\sin(\varphi)$, with major axis radius $r\cos(\varphi)$ and minor axis radius $r\sin(\varphi_0)\cos(\varphi)$. What I'm having trouble with is identifying the coordinates of the "vanishing points" where this ellipse intersects the horizon—where the ellipse is tangent to the circle defining the earth.
Equivalently (if less intuitively) I'm looking for the zero, one, or two points that satisfy both
$$ x^2 + y^2 = r $$
and
$$ \frac{x^2}{(r\cos(\varphi))^2} + \frac{(y - r\cos(\varphi_0)\sin(\varphi))^2}{(r\sin(\varphi_0)\cos(\varphi))^2} = 1 $$
It seems like there should be some (simple?) relation between $\varphi_0$, $\varphi$, and the longitude at which the line of latitude disappears, from a minimum of 90° ($\frac{\pi}{2}$) at $\varphi = 0$ to a maximum of 180° ($\pi$) at $\varphi = \pi/2 - \varphi_0$. (It's not hard to work out approximate values by brute force, and the curve looks vaguely arcsin-ish, but the analytical solution escapes me.)
I might be completely missing the point here but I will give it a quick try anyway.
I will take the part of your question stating that you want "the longitude at which the line of latitude disappears" and ignore the rest since I am not sure if it is all correct.
Referencing the Wikipedia page you linked, the equations for $x$ and $y$ are
$$ \begin{align} x & = R \cos{\varphi} \sin{(\lambda - \lambda_0)} \\ y & = R (\cos{\varphi_0} \sin{\varphi} - \sin{\varphi_0} \cos{\varphi} \cos{(\lambda - \lambda_0)}) \end{align} $$
and the equation for "the angular distance $c$ from the center of the orthographic projection" is
$$ \cos{c} = \sin{\varphi_0} \sin{\varphi} + \cos{\varphi_0} \cos{\varphi} \cos{(\lambda - \lambda_0)} $$
meaning that the projection vanishes at the point where $c$ is either $- \frac \pi 2$ or $\frac \pi 2$.
Therefore, we can find the "vanishing point" longitude in terms of the latitude as
$$ \begin{align} 0 & = \sin{\varphi_0} \sin{\varphi} + \cos{\varphi_0} \cos{\varphi} \cos{(\lambda - \lambda_0)} \\ \cos{\varphi_0} \cos{\varphi} \cos{(\lambda - \lambda_0)} & = - \sin{\varphi_0} \sin{\varphi} \\ \cos{(\lambda - \lambda_0)} & = - \frac {\sin{\varphi_0} \sin{\varphi}} {\cos{\varphi_0} \cos{\varphi}} \\ \cos{(\lambda - \lambda_0)} & = - \tan{\varphi_0} \tan{\varphi} \\ \lambda - \lambda_0 & = \arccos{(- \tan{\varphi_0} \tan{\varphi})} \\ \lambda & = \lambda_0 + \arccos{(- \tan{\varphi_0} \tan{\varphi})} \end{align} $$
which isn't very nice but is simple enough.
If you need the corresponding $x$ and $y$ these come out fairly reasonable as
$$ \begin{align} y & = R \left(\cos{\varphi_0} \sin{\varphi} - \sin{\varphi_0} \cos{\varphi} \left(- \frac {\sin{\varphi_0} \sin{\varphi}} {\cos{\varphi_0} \cos{\varphi}}\right)\right) \\ & = R \sin{\varphi} \left(\cos{\varphi_0} - \sin{\varphi_0} \left(- \frac {\sin{\varphi_0}} {\cos{\varphi_0}}\right)\right) \\ & = R \sin{\varphi} \left(\cos{\varphi_0} + \left(\frac {\sin ^ 2{\varphi_0}} {\cos{\varphi_0}}\right)\right) \\ & = R \sin{\varphi} \left(\frac {\cos ^ 2 {\varphi_0} + \sin ^ 2{\varphi_0}} {\cos{\varphi_0}}\right) \\ & = R \frac {\sin{\varphi}} {\cos{\varphi_0}} \end{align} $$
and
$$ \begin{align} x & = \pm R \cos{\varphi} \sqrt {1 - \left(- \frac {\sin{\varphi_0} \sin{\varphi}} {\cos{\varphi_0} \cos{\varphi}}\right) ^ 2} \\ & = \pm R \cos{\varphi} \sqrt {1 - \frac {\sin ^ 2{\varphi_0} \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi}}} \\ & = \pm R \cos{\varphi} \sqrt {\frac {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi} - \sin ^ 2{\varphi_0} \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi}}} \\ & = \pm R \sqrt {\frac {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi} - \sin ^ 2{\varphi_0} \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0}}} \\ & = \pm R \sqrt {\frac {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi} - (1 - \cos ^ 2{\varphi_0}) \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0}}} \\ & = \pm R \sqrt {\frac {\cos ^ 2{\varphi_0} \cos ^ 2{\varphi} + \cos ^ 2{\varphi_0} \sin ^ 2{\varphi} - \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0}}} \\ & = \pm R \sqrt {\frac {\cos ^ 2{\varphi_0} (\cos ^ 2{\varphi} + \sin ^ 2{\varphi}) - \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0}}} \\ & = \pm R \sqrt {\frac {\cos ^ 2{\varphi_0} - \sin ^ 2{\varphi}} {\cos ^ 2{\varphi_0}}} \\ & = \pm R \sqrt {1 - \left(\frac {\sin{\varphi}} {\cos{\varphi_0}}\right) ^ 2} \end{align} $$