To solve the problem, we need to find the maximum area of triangle \( \Delta PXQ \) where \( P(4, -4) \), \( Q(9, 6) \), and \( X(x, y) \) is any point on the arc \( POQ \) of the parabola defined by \( y^2 = 4x \).
### Step-by-Step Solution:
1. **Identify the coordinates of points**:
- Point \( P \) is \( (4, -4) \).
- Point \( Q \) is \( (9, 6) \).
- The vertex \( O \) of the parabola \( y^2 = 4x \) is at \( (0, 0) \).
2. **Area of Triangle Formula**:
The area \( A \) of triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
For points \( P(4, -4) \), \( Q(9, 6) \), and \( X(x, y) \):
\[
A = \frac{1}{2} \left| 4(6 - y) + 9(y + 4) + x(-4 - 6) \right|
\]
3. **Substituting the coordinates**:
\[
A = \frac{1}{2} \left| 24 - 4y + 9y + 36 - 10x \right|
\]
Simplifying gives:
\[
A = \frac{1}{2} \left| 60 + 5y - 10x \right|
\]
4. **Express \( y \) in terms of \( x \)**:
Since \( y^2 = 4x \), we can express \( y \) as:
\[
y = \sqrt{4x} = 2\sqrt{x}
\]
Substitute \( y \) into the area formula:
\[
A = \frac{1}{2} \left| 60 + 5(2\sqrt{x}) - 10x \right| = \frac{1}{2} \left| 60 + 10\sqrt{x} - 10x \right|
\]
5. **Maximize the area**:
To maximize the area, we need to maximize the expression \( 60 + 10\sqrt{x} - 10x \). Let:
\[
f(x) = 60 + 10\sqrt{x} - 10x
\]
Differentiate \( f(x) \):
\[
f'(x) = \frac{5}{\sqrt{x}} - 10
\]
Set the derivative to zero to find critical points:
\[
\frac{5}{\sqrt{x}} - 10 = 0 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4}
\]
6. **Second derivative test**:
Calculate the second derivative:
\[
f''(x) = -\frac{5}{2x^{3/2}}
\]
Since \( f''(x) < 0 \), this indicates a maximum at \( x = \frac{1}{4} \).
7. **Calculate maximum area**:
Substitute \( x = \frac{1}{4} \) back into \( f(x) \):
\[
f\left(\frac{1}{4}\right) = 60 + 10\left(\frac{1}{2}\right) - 10\left(\frac{1}{4}\right) = 60 + 5 - 2.5 = 62.5
\]
Therefore, the area \( A \) is:
\[
A = \frac{1}{2} \times 62.5 = 31.25
\]
8. **Expressing in the required form**:
The maximum area is given as \( \frac{25k}{4} \). Thus, we set:
\[
31.25 = \frac{25k}{4} \implies k = \frac{31.25 \times 4}{25} = 5
\]
### Final Answer:
The value of \( k \) is \( 5 \).
