Depolarization factors of a drude metal plasmonic spheroid A practice question asks for the depolarization factors $ L_{i=x,y,z} $ of a plasmonic spheroid made of a drude metal having the same resonance as the SPP resonance frequency.
The answer turns out to be $ L_x = L_y = \frac{1}{4} ; L_z = \frac{1}{2} $. I'm at a loss at how this was obtained. How can we infer those figures only knowing whats given?
 A: Polarizability for a spheroidal particle of permittivity $\epsilon_2$ placed in a medium of permittivity $\epsilon_1$ due to electric field in direction $j$ is given by:
$$ \alpha_j = 4 \pi \epsilon_0 R_x R_y R_z \cfrac{\epsilon_1 - \epsilon_2}{3 \epsilon_2 + 3 L_j (\epsilon_1 - \epsilon_2)}  $$
where $R_x$, $R_y$, $R_z$ are the lengths of axes along x, y, z.
Since we are interested in the resonance, we want the denominator to become zero, which gives the resonant condition:
$$ 3 \epsilon_2 + 3 L_j (\epsilon_1 - \epsilon_2) = 0 $$
which leads to the relation
$$ \epsilon_1 = \cfrac{L_j - 1}{L_j} \epsilon_2 $$
and the resonant frequency:
$$ \omega_{0j} = \cfrac{\Omega_P}{\sqrt{\epsilon_{1} - \epsilon_2 \left(1-\cfrac{1}{L_j}\right)}} $$
Now we want it to be equal to SPP resonance which is given by:
$$ \omega_{SPP} = \cfrac{\Omega_P}{\sqrt{\epsilon_{1} + \epsilon_2}} $$
Comparing the two expressions gives us:
$$ -\epsilon_2 \left(1-\frac{1}{L_j}\right) = \epsilon_2 $$
and we get $ L_j = 1/2 $.
Since the answer is $L_z = \frac{1}{2}$ it means it was assumed that the incident field is along axis z. 
The three factors $L_x$, $L_y$, $L_z$ always sum up to 1. And since the question asks about a spheroid it means that two of the axes have the same length and the two other factors are equal, therefore $L_x$ = $L_y$ = $\frac{1}{4}$.
