Impedance matching open collector output
So, if I use greater than a 6' BNC cable, I need to worry about impedance matching, correct? 


Yes. For really good results, you might use a \$\lambda/10\$ or \$\lambda/20\$ rule instead of \$\lambda/8\$.


  Assuming the BNC cable is 75 Ohm, the receiver is 75 Ohms, I need to make the source (open collector output) 75 Ohms as well? 


Matching the source is the best way to minimize ringing.

But it's also possible to design systems that terminate only the source or only the receiver. Then you are allowing the signal to reflect once and counting on the termination at the other end to eliminate the reflection when it reaches there.


  I get confused about the ability to drive a 75 Ohm load (i.e. current drive capability) with the need for impedance matching. 


Yes, it takes a reasonably high-power driver to drive a 75 or 50 ohm line.


  If this is the case, how do you impedance match an open collector output?


With open collector, you can do this with a 75-ohm pull-up. The transistor itself has fairly high output impedance. This is essentially how CML logic outputs work. 

Note this means the transistor itself is driving a load equivalent to 37.5 ohms, and needs to provide current accordingly.