Orthogonal trajectories are curves that intersect each member of a given family of curves at right angles. To find the orthogonal trajectories for the family of circles given by the equation x^2 + y^2 + 2fy - 1 = 0, we need to find the equation of the curves that intersect each circle orthogonally.

Let's proceed with finding the orthogonal trajectories step by step:

Step 1: Find the derivative of the given family of circles equation.
Differentiating the given equation with respect to x, we get:
2x + 2y(dy/dx) + 2f(dy/dx) = 0

Step 2: Find the slope of the tangent to the circles.
The slope of the tangent to a circle with equation x^2 + y^2 + 2fy - 1 = 0 can be determined by differentiating the equation implicitly with respect to x and solving for dy/dx. This gives us:
dy/dx = (-x - f) / (y + f)

Step 3: Find the slope of the orthogonal trajectories.
The slopes of orthogonal trajectories are negative reciprocals of the slopes of the circles. Therefore, the slope of the orthogonal trajectories is:
m = -1 / (dy/dx) = -1 / [(-x - f) / (y + f)] = (y + f) / (x + f)

Step 4: Find the differential equation of the orthogonal trajectories.
Using the slope-intercept form of a line, we can write the equation of the orthogonal trajectories as:
(y - y1) = m(x - x1)

Substituting the slope m and the point (x1, y1) into the equation, we get:
(y - y1) = [(y + f) / (x + f)](x - x1)

Step 5: Simplify the equation.
Expanding the equation, we have:
yx + fy - y1x - y1f = yx + yf + fx + f^2

Rearranging the terms, we get:
fy - y1f - y1x = yf + fx + f^2

Simplifying, we obtain:
(y - y1)x + (f - y1)f = 0

Comparing this equation with the given options, we can see that the correct answer is option 'C': x^2 + y^2 + 2gx - 1 = 0.

Therefore, the orthogonal trajectories for the family of circles x^2 + y^2 + 2fy - 1 = 0 are represented by the equation x^2 + y^2 + 2gx - 1 = 0.