Internal resistance of LDo
If the input supply voltage to an "ideal" 5 volt regulator is 10 volts and the output load is 10 ohms (0.5 amp output current), then the regulator will "act" like a 10 ohm resistor in order to facilitate 5 volts at the output. It's just a simple potential divider.

If the load changes rapidly between 9 ohms and 11 ohms, then the regulator (due to feedback control), will keep the output voltage at 5 volts by rapidly altering its effective series resistance to counter the changing load impedance and current. So, when the load is at 9 ohms, the regulator will act like a 9 ohm resistor in order to keep the output at 5 volts.

If the input voltage reduces to (say) 8 volts and the load is 10 ohms then the effective series resistance of the regulator has to be 3 volts/0.5 amp = 6 ohm.

The effective series resistance of an ideal voltage regulator will change slowly or rapidly to counter the load and input voltage circumstances it faces.