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Begin by inspecting the equation $$z^{2}+4 y^{2}-x^{2}=1$$ to determine what type of surface it represents by comparing its form to well-known surface equations. Notice it has all three variables squared, which indicates it is a quadric surface. Furthermore, the coefficients of the squared terms are different and one squared term is negative, which means it is a hyperboloid.

Now we need to determine if it is a one-sheet hyperboloid or a two-sheet hyperboloid by looking at the coefficients of the squared terms. If the equation has more positive squared terms than negative ones (i.e., two positive and one negative), it is a one-sheet hyperboloid. In this case, the equation has one positive and two negative squared terms, meaning it is a two-sheet hyperboloid.

For a two-sheet hyperboloid, the equation can be written as: $$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = -1$$ Comparing this with our given equation, we find the parameters to be: $$a^2 = 1, b^2 = 4, \text{and } c^2 = 1$$. Therefore, $$a = 1, b = 2, \text{and } c = 1$$.

To visualize the two-sheet hyperboloid, imagine two hourglass-shaped surfaces along the coordinate axes which touch each other at the origin but do not intersect. The \(a\) parameter controls the extent of the surface along the \(x\)-axis, the \(b\) parameter controls the extent along the \(y\)-axis, and the \(c\) parameter controls the extent along the \(z\)-axis. In our case, the surface is wider along the \(y\)-axis due to the value of \(b=2\). These surfaces are infinite, since the hyperboloid extends indefinitely along the \(x\)-axis in both directions, and the same goes for the other axes. In conclusion, the equation $$z^{2}+4 y^{2}-x^{2}=1$$ defines a two-sheet hyperboloid with parameters $$a = 1, b = 2, \text{and } c = 1$$, and can be described as two hourglass-shaped surfaces along the coordinate axes that touch at the origin and extend infinitely along the axes.