Differential Amplifier with active load
Say you apply some differential signal \$V_{in}\$.

Due to the differential gain of the amplifier, \$V_X\$ may now be,

$$ V_X = V_N + \Delta $$

Similarly,

$$ V_Y = V_N - \Delta $$

Since \$R_1\$ = \$ R_2 \$, the voltage diver formed between \$V_X\$ and \$V_Y\$ yields \$V_n\$. 

Due to the symmetry of the differential amplifier, \$V_N\$ appears as an AC ground.