Audio EQ Cookbook without frequency warping
here is a quick look at how the 5 degrees of freedom for the parametric EQ can be viewed.  it's my take on what Knud Christensen of tc electronic came up with about a decade ago at an AES convention and this patent.

so, forget about the Cookbook (and the issues of Q and bandwidth therein) and consider (in the s-plane) the parametric EQ as the sum of a bandpass filter (with a $Q$ value) in parallel with a wire:

$$ H(s) = (G_\text{boost} - 1)\frac{\frac{1}{Q}\frac{s}{\omega_0}}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} + 1 $$

$G_\text{boost} = 10^{\frac{dB}{20}}$ is the gain of the peak (or valley, if $dB&lt;0$).  the gain at DC and at Nyquist is 0 dB. that's a 2nd-order IIR and there are 3 independent parameters. 2 more to go.  so we next add an overall gain parameter:

$$ H(s) = G_0 \left((G_\text{boost} - 1)\frac{\frac{1}{Q}\frac{s}{\omega_0}}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} + 1 \right) $$

that's 4 knobs to twist.  one more knob to add (without raising the filter order) and we will be done with adding more independent parameters.

so what Knud does here is replace that "wire" (that trailing "$1$" in the transfer function) with a prototype shelving filter that must have the same poles, the same $Q$ and $\omega_0$ as the BPF, so that the denominator is the same.  the transfer function of that shelf is:

$$ H_\text{shelf}(s) = \frac{R\left(\frac{s}{\omega_0}\right)^2 + \frac{\sqrt{R}}{Q}\frac{s}{\omega_0} + 1}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} $$

where $R \triangleq 10^{\frac{tilt}{20}}$ and $tilt$ is the gain differential of the shelf in dB.  this is what offsets the gain at Nyquist to be different than the gain at DC.  After bilinear transformation, Nyquist gets boosted by $tilt$ dB and the gain at DC remains unchanged.  Like the $dB$ boost parameter, the $tilt$ parameter can be either positive or negative.  $G_0 \cdot R$ is the linear gain at Nyquist.

put that all together and you get:

$$ \begin{align}
H(s) &amp; = G_0 \left((G_\text{boost} - 1)\frac{\frac{1}{Q}\frac{s}{\omega_0}}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} + H_\text{shelf}(s) \right) \\
 &amp; = G_0 \left((G_\text{boost} - 1)\frac{\frac{1}{Q}\frac{s}{\omega_0}}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} + \frac{R\left(\frac{s}{\omega_0}\right)^2 + \frac{\sqrt{R}}{Q}\frac{s}{\omega_0} + 1}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1} \right) \\
&amp; = G_0 \frac{R\left(\frac{s}{\omega_0}\right)^2 + (G_\text{boost}+\sqrt{R}-1)\frac{1}{Q}\frac{s}{\omega_0} + 1}{\left(\frac{s}{\omega_0}\right)^2 + \frac{1}{Q}\frac{s}{\omega_0} + 1}  \\
\end{align}$$

no matter how you look at that, this has 5 degrees of freedom and those 5 biquad coefficients are fully defined from these 5 parameters.  doesn't matter if you map from $s$ to $z$ using the blinear transform or the trapezoidal rule (effectively the same thing) or any other method that does not change the order of the filter.  you may have to fudge the definition of $Q$ or bandwidth, you may have to compensate $\omega_0$ and/or $Q$ for frequency warping effects (like you get with the blinear transform), but if you paid big bucks for something that gets you a 2nd-order IIR filter, it doesn't matter if you implement it with any Direct Form or transposed Direct Form or Lattice or Normalized Ladder or Hal Chamberlin's State-Variable or Andrew Simpson's modeling of linear analog with trapezoidal integration, eventually you get to 5 coefficients and they can be mapped to these 5 independent parameters.  it's all the same.  whether or not you paid money for a license or not.  the math is stronger than any claims made by whomever you are licensing from.

Just FYI, i solved where the true peak or valley frequency is when there is a $tilt$ that is not zero.  the frequency where the peak or valley has been nudged over by the tilt is:

$$ \omega_\text{peak} \ = \ \omega_0 \ 
\sqrt{ \frac{Q^2 \left(R - \frac{1}{R}\right)}{G_\text{boost}^2 - R + 2 Q^2 (R - 1)} + \\ \sqrt{\frac{G_\text{boost}^2 - \frac{1}{R} + 2 Q^2 \left(\frac{1}{R} - 1\right)}{G_\text{boost}^2 - R + 2 Q^2 (R - 1)} + \frac{Q^4 \left(R - \frac{1}{R}\right)^2}{\left(G_\text{boost}^2 - R + 2 Q^2 (R - 1)\right)^2} } } $$

you can see that when $tilt = 0$, then $R=1$ and consequently $\omega_\text{peak} = \omega_0$.   the peak gain $G_\text{boost}$ might also have to be adjusted a little and that has yet to be worked out.  a good first guess would be $G_\text{boost} \leftarrow \frac{G_\text{boost}}{\sqrt{R}}$ or maybe $G_\text{boost} \leftarrow G_\text{boost}-(\sqrt{R}-1)$.