Current mirrors as active loads in BJT differential amplifier, what does that mean?
Here is a unconventional explanation (you can find many other conventional explanations on the web)...

The basic idea is to oppose two current sources (transistor collectors); so we have to connect them in the same loop and to vary their currents in opposite directions:



As a result the voltage of the common point between the sources will "move" extremely vigorously (a fully geometrical phenomenon):

 

A good example of this configuration is the classic CMOS stage where the two transistors are connected with their sources to Vdd and Vss (ground) and their drains are joined:



But in the emitter-coupled (long-tailed) pair, the collectors of both transistors (T1 and T2 below) are turned up. So, we should turn the one collector (the T1's in this example) back down to stand against the other (T2's) collector. For this purpose, we have connected the simple current mirror (T3-T4) in the picture below:



See also this paper and its more sophisticated version.