Step 1: Understanding the Question:
This question belongs to the topic of Coordinate Geometry, specifically focusing on Conic Sections. We are tasked with identifying the coordinates of the "focus" of a given parabola. A focus is a fixed point used to define the curve; for a horizontal parabola opening to the right, this point lies on the axis of symmetry. Understanding the standard orientation of the equation $y^2 = 4ax$ is crucial to solving this problem quickly.
Step 2: Key Formulas and approach:
The standard equation for a parabola with its vertex at the origin and symmetric about the x-axis is $y^2 = 4ax$.
In this specific configuration:
1. The Focus is located at $(a, 0)$.
2. The Directrix is the line $x = -a$.
3. The Latus Rectum has a length of $4a$.
Our approach is to compare the given equation to the standard form, solve for the constant 'a', and then use the definition of the focus coordinates for this specific orientation.
Step 3: Detailed Explanation:
The equation provided in the problem is $y^2 = 16x$.
We compare this equation term-by-term with the standard parabola form $y^2 = 4ax$.
From this comparison, we can see that $4a$ must be equal to 16.
Solving for 'a', we divide both sides by 4: $a = \frac{16}{4} = 4$.
Because the equation is in the form $y^2 = \dots$, the parabola opens horizontally. Since the coefficient 16 is positive, it specifically opens to the right.
For such a parabola, the focus is defined by the coordinates $(a, 0)$.
Substituting our calculated value of $a = 4$, the coordinates become $(4, 0)$.
Step 4: Final Answer:
The focus of the parabola $y^2 = 16x$ is located at the point $(4, 0)$.