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:
We need to determine the geometric coordinates of the focus for the given parabola equation \( y^2 = 16x \).


Step 2: Key Formula or Approach:
The standard equation of a rightward opening horizontal parabola is \( y^2 = 4ax \).
For this standard form, the focus is located at the coordinates \( (a, 0) \).


Step 3: Detailed Explanation:
Compare the given equation \( y^2 = 16x \) with the standard form \( y^2 = 4ax \).
Equating the coefficients of the \( x \) term:
\[ 4a = 16 \] \[ a = \frac{16}{4} = 4 \] Since the parabola is of the form \( y^2 = 4ax \), its focus is at \( (a, 0) \).
Substitute \( a = 4 \) to get the coordinates of the focus:
Focus = \( (4, 0) \).


Step 4: Final Answer:
The focus of the parabola is \( (4, 0) \).

Answer 2

The parabola \( y^2 = 16x \) opens to the right along the positive \( x \)-axis, since it is of the form \( y^2 = 4ax \) with a positive coefficient. Let's check each of the given coordinates against this orientation and the general focus relation where \( 4a \) equals the coefficient of \( x \).

1. \( (4, 0) \): This point lies on the positive \( x \)-axis, matching the axis of symmetry of \( y^2 = 16x \). Comparing with \( y^2 = 4ax \), we get \( 4a = 16 \), so \( a = 4 \), placing the focus exactly at \( (4, 0) \).
2. \( (0, 4) \): This point lies on the \( y \)-axis. A focus at \( (0, 4) \) would correspond to an upward-opening parabola of the form \( x^2 = 4ay \), which is a completely different orientation from \( y^2 = 16x \). This does not fit.
3. \( (2, 0) \): This lies on the correct axis, but it would require \( 4a = 8 \), giving a coefficient of \( x \) equal to \( 8 \), not \( 16 \). Since our equation clearly has coefficient \( 16 \), this value of \( a \) is too small.
4. \( (0, 2) \): Like option (B), this lies on the \( y \)-axis and would belong to a vertically opening parabola, not the horizontally opening one given here. This does not fit either.

Only \( (4, 0) \) is consistent both with the axis along which \( y^2 = 16x \) opens and with the coefficient \( 16 \) matching \( 4a \).

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