If the graphs of `x^(2) = 4(y+9)` and `x + ky = 6` intersect on the x-axis , then k =

Options

A. 0
B. 6
C. `-6`
D. any real number

Explanation

To solve the problem, we need to determine the value of \( k \) such that the graphs of the equations \( x^2 = 4(y + 9) \) and \( x + ky = 6 \) intersect on the x-axis.

### Step-by-Step Solution:

1. **Understand the Intersection on the X-axis**:
- The intersection of the two graphs on the x-axis means that the y-coordinate at the point of intersection is 0. Therefore, we can denote the point of intersection as \( (a, 0) \).

2. **Substitute \( y = 0 \) in the First Equation**:
- The first equation is \( x^2 = 4(y + 9) \).
- Substituting \( y = 0 \):
\[ x^2 = 4(0 + 9) \implies x^2 = 36 \implies x = 6 \text{ or } x = -6 \]
- Thus, the points of intersection on the x-axis are \( (6, 0) \) and \( (-6, 0) \).

3. **Substitute \( y = 0 \) in the Second Equation**:
- The second equation is \( x + ky = 6 \).
- Substituting \( y = 0 \):
\[ x + k(0) = 6 \implies x = 6 \]
- This means that for the point \( (6, 0) \), the equation holds true regardless of the value of \( k \).

4. **Check the Other Point**:
- Now, we check the second point \( (-6, 0) \):
\[ -6 + k(0) = 6 \implies -6 = 6 \]
- This is not true, hence \( (-6, 0) \) does not satisfy the second equation.

5. **Conclusion on the Value of \( k \)**:
- Since the point \( (6, 0) \) satisfies the second equation for any value of \( k \), and the point \( (-6, 0) \) does not, we conclude that the graphs intersect on the x-axis at \( (6, 0) \) for any real number \( k \).

### Final Answer:
Thus, \( k \) can be any real number.