## 3 Results  
Let us turn to the asymptotic behavior at high k. If one of the eigenvalues of the index of refraction operator \(\hat{N}\) becomes infinite, then \(\text{det}\hat{N}=n_{1}n_{2}n_{3}n_{4}\to\infty\) also diverges, since it is the product of all the eigenvalues \(n_{i}\). If the elements of the matrix \(\hat{M}\) are finite, then, according to the expression for matrix \(\overline{m}\) (Equation (8)), whose elements compose \(\hat{N}\) (Equation (9)), such divergence is only possible for waves propagating in directions \(\hat{\mathbf{k}}\) such that  
\[\Delta_{\infty}=\text{det}\left( \begin{array}{cc}\overline{\epsilon}_{33}& \overline{X}_{33}\\ \overline{Y}_{33}&\overline{\mu}_{33}\end{array} \right)=0. \tag{13}\]  
Considering Equation (6) and that \(\mathbf{q}_{3}=\hat{\mathbf{k}}\), Equation (13) can be rewritten as an equation of a quartic conical surface  
\[h(\mathbf{k})=0, \tag{14}\]  
where \(h(\mathbf{k})=\left(\mathbf{k}^{T}\hat{e}\mathbf{k}\right)\left(\mathbf{k}^{T}\hat{\mu}\mathbf{k} \right)-\left(\mathbf{k}^{T}\hat{X}\mathbf{k}\right)\left(\mathbf{k}^{T}\hat{Y}\mathbf{k}\right)\) for a generic bianisotropic medium without an assumption of reciprocity. Direct comparison shows that function \(h(\mathbf{k})\) of Equation (14) is identical to Equation (2). Note that functions \(a(\mathbf{k})=\left(\mathbf{k}^{T}\hat{a}\mathbf{k}\right)=\sum_{ij}a_{ij}k_{i}k_{j}\), where \(\hat{a}=\hat{e}\), \(\hat{\mu}\), \(\hat{X}\), \(\hat{Y}\), are quadratic forms on k-space. If \(\Delta_{\infty}\to 0\), then the longitudinal components of fields \(E_{z}\) and \(H_{z}\) are much greater than the transverse components, so \(\mathbf{E}=E_{z}\mathbf{\hat{k}}\), \(\mathbf{H}=H_{z}\mathbf{\hat{k}}\). Consequently, [PERSON]'s equations can be written as  
\[\mathbf{k}\cdot\mathbf{D}=\left(\mathbf{k}^{T}\mathbf{\hat{\varepsilon}}\mathbf{k}\right)\!E_{z}+ \left(\mathbf{k}^{T}\hat{X}\mathbf{k}\right)\!H_{z}=0, \tag{15}\]  
\[\mathbf{k}\cdot\mathbf{B}=\left(\mathbf{k}^{T}\mathbf{\hat{\mu}}\mathbf{k}\right)\!H_{z}+\left(\bm {k}^{T}\hat{Y}\mathbf{k}\right)\!E_{z}=0. \tag{16}\]  
Equations (15) and (16) have a nonzero solution for the longitudinal fields \((E_{z},H_{z})\) only if Equation (14) is met. Note that from \(\mathbf{k}\times\mathbf{E}=k_{0}\mathbf{B}\) and \(\mathbf{k}\times\mathbf{H}=-k_{0}\mathbf{D}\), it follows that for HKPs \(\mathbf{D}=\mathbf{B}=0\), since HKP fields are purely longitudinal \(\mathbf{E}=E_{z}\mathbf{\hat{k}}\), \(\mathbf{H}=H_{z}\mathbf{\hat{k}}\). An important characteristic of an HKP is its longitudinal impedance, which we introduce using Equations (15) and (16):  
\[Z_{l}=\frac{E_{z}}{H_{z}}=-\frac{\left(k^{T}\hat{X}k\right)}{\left(k^{T}\mathbf{ \hat{\varepsilon}}k\right)}=-\frac{\left(k^{T}\hat{\mu}k\right)}{\left(k^{T} \hat{Y}k\right)} \tag{17}\]  
For reciprocal materials, \(\left(\mathbf{k}^{T}\hat{Y}\mathbf{k}\right)=-\left(\mathbf{k}^{T}\hat{X}\mathbf{k}\right)\) and Equation (14) breaks into two solution branches:  
\[\sqrt{\left(\mathbf{k}^{T}\mathbf{\hat{\varepsilon}}\mathbf{k}\right)\!\left(\mathbf{k}^{T} \hat{\mu}\mathbf{k}\right)}\pm i\!\left(\mathbf{k}^{T}\hat{X}\mathbf{k}\right)=0, \tag{18}\]  
corresponding to impedances  
\[Z_{l}=\pm i\ \sqrt{\frac{\left(\mathbf{k}^{T}\hat{\mu}k\right)}{\left(\mathbf{k}^{T} \hat{\varepsilon}\mathbf{k}\right)}} \tag{19}\]  
The conventional condition of hyperbolicity requires the principle values of the dielectric permittivity tensor \(\hat{\varepsilon}\) to be of different signs [2; 3; 4]. Indeed, for such materials, \(h(\mathbf{k})=\left(\mathbf{k}^{T}\hat{\varepsilon}\mathbf{k}\right)\) can only be zero if principle values of \(\hat{\varepsilon}\) have different signs. According to Equations (17) and (19), the longitudinal impedance for the HKP in these materials is \(Z_{l}^{-1}=0\).