Differential Amplifier Common Mode Thevenin Equivalent

[anvoice]

TL;DR

A sanity check on a text I am using to learn EE. Cannot replicate the result for the common mode Thevenin resistance in a figure in the text.

I am currently reading about small signal analysis of a MOSFET differential amplifier. The text I am using has the below two figures for a common-mode equivalent circuit for the amplifier. The first makes sense to me except where it calls it a "Norton equivalent circuit", whereas I thought a Norton between either the vx or vy terminals would need to have a single resistance in parallel with a current source, and I don't see how this qualifies.
More importantly, the second figure has the Thevenin equivalent resistance as just RL. Seeing that exciting the subcircuits on the leftmost figure with a voltage and measuring the current seems to be a reasonable way of finding Rth in this case, I did that and ended up with a Rth of
2RL / (2 + (RL) x (vc)). Am I missing something major or is this a typo in the text? Granted it doesn't invalidate the main result (vx - vy = 0), it bothers me.

 

[anvoice]

I think I figured it out. Looking at just this subcircuit, I forgot to shut down the signal vc (common-mode input) to the MOSFETs. If vc is off, naturally the dependent current source becomes an open circuit and the resistance is indeed RL. It is also a warning sign in this case that if you forget to shut down vc, the Thevenin equivalent resistance is actually dependent on the excitation voltage, which shouldn't be the case. The hints were there, but just managed to piece them together.