Analysis frequency of the DFT?
I will provide two interpretations of the DFT samples $X[k]$;

The first interpretation relies on the concept of an inner product, $\langle x(t), \phi_k(t) \rangle $, between the signal $x(t)$ and a base function $\phi_k(t)$ which measures and quantifies the amount of the base function $\phi_k(t)$ in the signal $x(t)$. This operation is essentially performing an analysis of $x(t)$.

When this interpretation is applied to the DFT, those discrete functions $\phi_k[n] = e^{j\frac{2 \pi}{N} k n}$ become the base functions (as complex exponentials of a specific frequency) of an orthonormal basis and those values $X[k]$ become the inner products between $x[n]$ and the complex exponentials; computing the amount of that particular base function in the signal $x[n]$. Note that the analysis is performed in the discrete-time sampled frequency $\omega_k$ and its equivalent continuous-time frequency $\Omega_k$ can be found from the  sampling operation and frequency normalization, which provides the second interpretation below.

Consider a continuous-time bandlimited signal $x_\text{c}(t)$ which is sampled by $f_\text{s}$ samples per second, or equivalently by a sampling period of $T_\text{s} \triangleq \tfrac{1}{f_\text{s}}$ seconds per sample, without aliasing, to produce the discrete-time samples $x[n]$ of length $N$;

$$x[n] = x_\text{c}(n T_\text{s})$$ 

An $N$-point DFT of this sequence be defined as:

$$X[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} n k } ~~~ , ~~~\text{for} ~~~ k=0,1,...,N-1$$

These DFT values, $X[k]$, can also be considered as the samples of the DTFT $X(e^{j \omega})$ associated with $x[n]$ as;

$$X[k] = X(e^{j \omega}) \bigg|_{\omega_k = \frac{2 \pi}{N}k}$$

The DTFT itself is a frequency scaled and repeated version of the original continuous time signal's continuous Fourier transform $X_\text{c}(j\Omega)$ as

$$X(e^{j \omega}) = \frac{1}{T_\text{s}} X_\text{c}\left(j \tfrac{\omega}{T_\text{s}} \right)$$

in the base period $-\pi \le \omega &lt; \pi$ 

Therefore the corresponding continuous-time frequency (in Hz.) that a given DFT sample $X[k]$ resides can be found by the relation 

$$f_k = \frac{f_\text{s}}{N} k$$

for $ -\tfrac{N}{2} \le k &lt; \tfrac{N}{2}$. Note the use of negative $k$ values.

So your equation provides this frequency $f_k$ (saying $f_\text{a}$ to indicate its analysis nature) as an equivalent continuous-time frequency that's being analysed by the complex exponential base function $\phi_k[n]=e^{j\frac{2\pi}{N}k n}$ with a discrete-time frequency $\omega_k = \frac{2 \pi}{N} k$. 

Note that the value of $f_\text{a}$ is obtained by considering the sampling operation applied on the continuous time-signal $x_\text{c}(t)$ that's used to obtain the samples $x[n]$