Question

The focus of the parabola \( y^2 = 16x \) is

• A. \( (4, 0) \)
• B. \( (0, 4) \)
• C. \( (2, 0) \)
• D. \( (0, 2) \)

Hint:

For standard parabolas of the form \( y^2 = 4ax \), the focus is always at \( (a, 0) \) and the directrix is the vertical line \( x = -a \).

Answer 1

The Correct Option is A

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 $(4, 0)$.

Answer 2

Rather than matching the equation to a memorized standard form, we can derive the focus directly from the defining property of a parabola: every point on the curve is equidistant from the focus and from the directrix.

Since \( y^2 = 16x \) opens along the positive \( x \)-axis, let the focus be the point \( (h, 0) \) and the directrix be the vertical line \( x = -h \), for some positive value \( h \). For any point \( (x, y) \) on the parabola, the distance to the focus must equal the distance to the directrix:

\[ \sqrt{(x - h)^2 + y^2} = x + h \]

Squaring both sides:

\[ (x - h)^2 + y^2 = (x + h)^2 \]\[ x^2 - 2hx + h^2 + y^2 = x^2 + 2hx + h^2 \]\[ y^2 = 4hx \]

Comparing this derived relation \( y^2 = 4hx \) with the given equation \( y^2 = 16x \), we get \( 4h = 16 \), so \( h = 4 \).

We can verify this with an actual point on the curve. Taking \( x = 4 \), we get \( y^2 = 16(4) = 64 \), so \( y = 8 \), giving the point \( (4, 8) \). The distance from \( (4, 8) \) to the focus \( (4, 0) \) is \( \sqrt{(4-4)^2 + 8^2} = 8 \), and the distance from \( (4, 8) \) to the directrix \( x = -4 \) is \( 4 - (-4) = 8 \). Both distances match, confirming the focus.

Therefore, the correct answer is \( (4, 0) \).