Time vs magnification in a binary lens For a single lens: we got 

\begin{equation}
 u(t)=\sqrt{\left(\frac{t-t_0}{t_E}\right)^2 + u_0^2},\end{equation}

so the magnification is 

\begin{equation}
A = \frac{u^2+2}{u\sqrt{u^2+4}} \end{equation}

Therefore the magnification for a single lens depends on three parameters: $u_0$ (minimum separation lens-source), $t_0$ (time where the magnification is maximum) and $t_E$ is the characteristic event timescale, which is the time it takes for the source to move with respect to the lens by one Einstein ring radius.

For a binary lens, however, the formulae of $u$ and $A$ above no longer hold. $A$ can be calculated for a point $u=(u_1,u_2)$ using numerical methods (Press et al. 2002) but, how is calculated $u$ for a certain time $t$? According to Tsapras (2018) $u$ depends on $u_0$, $t_E$, $t_0$ but also the angle $\alpha$ at which
the source trajectory crosses the binary axis, the mass ratio between the two components, $q$, and the separation between them, $s$, projected on the lens plane. ¿How these last three parameters come into play in the calculation of $u$ for a binary lens?
This answer applies to the simplest case: a static binary and a point source.

In the binary case, $u(t)$ describes how close each point in the source trajectory comes to (what you have defined as) the origin of the binary axis.

The solution is non-trivial. It involves finding the roots of a 5th order complex polynomial. You can look for example at how Etienne Bachelet's pyLIMA package codes it in Python. If you want to do it yourself, you first need to identify the location of the caustics and critical curves for the given arrangement. For that you need your two masses (or just the mass ratio) and their separation in units of the Einstein radius. The coefficients of the polynomial (A, B, C) are described in Schneider &amp; Weiss 1986 (eqn 9b).

Then you need to define the trajectory of your source relative to the binary axis. You also have to choose your origin (some people use the midpoint between the masses, others use the largest of the two masses or the barycenter).

For each point in the trajectory (instantaneous source position) you have to evaluate the corresponding magnification, that is, find the roots of the polynomial for that position. You will then get five solutions, which correspond to the images of the source. However, not all 5 solutions correspond to real images. To find out which are the real ones, you have to substitute them in the inverted equation and see if they map back to the given source position. You will always have three (outside the caustics) or five (inside the caustics) real images.

You can see an example below.


You can see how different source trajectories generate different light curves at this link.