Let `P(4,-4)` and `Q(9,6)` be two points on the parabola, `y^2=4x` and let X be any point on the are POQ of this parabola, where O is the vertex of this parabola, such that the area of `Delta PXQ` is maximum. Then this maximum area (in square units) is `(25k)/(4)`. The value of k is

Explanation

To solve the problem, we need to find the maximum area of triangle \( PXQ \) where \( P(4, -4) \) and \( Q(9, 6) \) are points on the parabola \( y^2 = 4x \) and \( O(0, 0) \) is the vertex of the parabola. The area of triangle \( PXQ \) can be expressed in terms of the coordinates of point \( X(x, y) \) on the parabola.

### Step 1: Area of Triangle Formula
The area \( A \) of triangle \( PXQ \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Here, \( P(4, -4) \), \( Q(9, 6) \), and \( X(x, y) \).

### Step 2: Substitute Coordinates
Substituting the coordinates of points \( P \), \( Q \), and \( X \):
\[ A = \frac{1}{2} \left| 4(6 - y) + 9(y + 4) + x(-4 - 6) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 24 - 4y + 9y + 36 - 10x \right| \]
\[ A = \frac{1}{2} \left| 60 + 5y - 10x \right| \]

### Step 3: Express \( x \) in terms of \( y \)
Since point \( X \) lies on the parabola \( y^2 = 4x \), we can express \( x \) as:
\[ x = \frac{y^2}{4} \]

### Step 4: Substitute \( x \) in Area Formula
Substituting \( x \) in the area formula:
\[ A = \frac{1}{2} \left| 60 + 5y - 10\left(\frac{y^2}{4}\right) \right| \]
\[ A = \frac{1}{2} \left| 60 + 5y - \frac{5y^2}{2} \right| \]
\[ A = \frac{1}{2} \left| -\frac{5y^2}{2} + 5y + 60 \right| \]

### Step 5: Simplify the Area Expression
To simplify:
\[ A = \frac{1}{4} \left| -5y^2 + 10y + 120 \right| \]

### Step 6: Find Critical Points
To maximize the area, we need to find the critical points by differentiating the area function with respect to \( y \) and setting it to zero:
\[ \frac{dA}{dy} = \frac{1}{4} \cdot (-10y + 10) = 0 \]
\[ -10y + 10 = 0 \implies y = 1 \]

### Step 7: Calculate Maximum Area
Substituting \( y = 1 \) back into the area formula:
\[ A = \frac{1}{4} \left| -5(1)^2 + 10(1) + 120 \right| \]
\[ A = \frac{1}{4} \left| -5 + 10 + 120 \right| = \frac{1}{4} \left| 125 \right| = \frac{125}{4} \]

### Step 8: Relate to Given Form
The problem states that the maximum area can be expressed as \( \frac{25k}{4} \). Thus, we set:
\[ \frac{125}{4} = \frac{25k}{4} \]
This implies:
\[ 125 = 25k \implies k = 5 \]

### Final Answer
The value of \( k \) is \( \boxed{5} \).