Ra = Ua pp/ Ia pp = Ua p / Ia p = Ua rms / Ia rms

Power at the primary: Pa=Ua rms^2 / Ra = Ia rms^2 * Ra = Ua rms*Ia rms

Note that Ia p = Ia dc, Ia pp = 2*Ia dc, Ia rms = Ia dc / 1,4142

The "goal" is to avoid graphic analysis, and find the simple formulae, "good enough" for "everyday use".

Our triode output tube with its "bias" Ugk, can have max. peak AC input voltage in class A1 equal to Ugk, or Ugk rms = Ugk/1,4142.
With very high load Ra, AC voltage at the load Ra is:

Ua = mu * Ugk

But, our "real" load Ra form voltage divider with tube plate resistance rp, and voltage at the load Ra is actually lower:

Ua = (mu * Ugk) / (1 + rp/Ra)

And from Ra = Ua / Ia, we have Ua = Ra * Ia

If we put together these equations:

Ra*Ia = (mu*Ugk)/(1+rp/Ra), and Ra*Ia = (mu*Ugk)/((Ra+rp)/Ra), and

Ra*Ia = (Ra*mu*Ugk)/(Ra+rp), and Ia = (mu*Ugk)/(Ra+rp), and

Ra+rp = mu*Ugk/Ia, and finally:

Ra = ((mu * Ugk) / Ia) - rp FORMULAE FOR Ra

What does this mean in practice? If we have some DC operating point for our output triode, say 300B - Uak=400V, Ia=80mA, Ugk=-85V and we know (about) rp and mu from tube manuals (simplification, assumed that rp and mu are constant, but error is minimal and negligible). Say, rp = 700 Ohms and mu = 3,9. Then:

Ra = ((3,9 * 85) / 0,08) - 700 = 3443,75 Ohms ~3,5 kOhms

In agreement with load - line analysis and tube manual data!