Thank you all for giving me a more clear sight on how this works as I was confused with the capacitance stated in the datasheet, now it is more clear to me. I also looked up some video's on youtube on this topic and basically I learned that you can drive a coax with whatever impedance you want, as long as the far end is terminated exactly in the characteristic impedance there will be no/verry little reflection distorting the wave form (at any point on the line). If this is not the case and the far end is not exactly matched, there will be a reflected wave arriving back at the source, and then you have a problem when this impedance is not matched because it will reflect again and so on.. to me this is basically the reason why they use the source impedance matching to "absorb" any reflections coming back from the far end.
As FvM mentioned, I understand that in the on-state the transistor will be on and will have a very low on resistance, maybe a few ohms, and therefore the line is not matched which is important for a improper terminated far end as this will cause a reflected wave travelling back towards the source and this will neither be terminated at the source because of the low on impedance of the switch. Possibly a push-pull output stage with a 75 Ohm series resistor might provide a better driver circuit? It should be better matched in the on & off state and will provide equal rise & fall times?