Is it the frequency that determines the current in a LC Tank Parallel Circuit?
The Q of a tuned parallel L-C circuit stands for quality factor and is a predictor to how much current flows in the tuned circuit at resonance. Here's the formula: -

Q = \$ \frac{1}{R_S}\sqrt{\frac{L}{C}}\$

\$R_S\$ is in series with the inductor, L

This basically means that if \$R_S\$ is zero then Q is infinite and the circulating currents in L and C are infinite for a non-zero voltage applied. Of course this doesn't happen because resistance is never 0 other than in a superconductor but that's another story!

Significantly for the question, if R stays the same and L is decreased by ten and C increased by ten (to retain the same resonant frequency), Q reduces by 10. In other words the high peaky tuned circuit you once had has become rather shallow and not so peaky in comparison.

Reducing inductance by ten does not usually mean \$R_S\$ reduces by ten. For a typical core wound with say 1000 turns, inductance might be (say) 1H. To reduce this to 0.1H means the turns reduce to 316 and the resistance therefore only reduces by about one-third.

Remember inductance is proportional to turns squared.

So after reducing inductance by 10 (resistance decreases by about 3) and increasing capacitance by 10, the net effect is a tuned circuit that has a Q that is 3x smaller than before and not as peaky or resonant.

Regarding the link - this is for an RLC parallel circuit where the resistance is represented as a parallel component and this is less-often used because the dominant loss in an RLC circuit is in the inductor (as series resistance). A parallel resistance could depict dielectric loss in the capacitor but this will be a tiny fraction (in most cases) of the inductor losses.