Let `P(4,-4)` and `Q(9,6)` be points on the parabola `y^(2)=4a(x-b)`. Let R be a point on the arc of the parabola between P and Q. Then the area of `DeltaPQR` is largest when.

Options

A. `anglePRQ=(pi)/(2)`
B. the point R is `(4,4)`
C. the point R is `((1)/(4),1)`
D. None of these

Explanation

To find the point R on the parabola \( y^2 = 4a(x-b) \) that maximizes the area of triangle PQR, where P(4, -4) and Q(9, 6) are given points, we can follow these steps:

### Step 1: Identify the parabola parameters
Given the points P(4, -4) and Q(9, 6), we need to find the values of \( a \) and \( b \) for the parabola equation \( y^2 = 4a(x-b) \).

### Step 2: Substitute point P into the parabola equation
Substituting point P(4, -4) into the parabola equation:
\[ (-4)^2 = 4a(4-b) \]
This simplifies to:
\[ 16 = 4a(4-b) \]
Dividing both sides by 4:
\[ 4 = a(4-b) \quad \text{(Equation 1)} \]

### Step 3: Substitute point Q into the parabola equation
Now substituting point Q(9, 6) into the parabola equation:
\[ 6^2 = 4a(9-b) \]
This simplifies to:
\[ 36 = 4a(9-b) \]
Dividing both sides by 4:
\[ 9 = a(9-b) \quad \text{(Equation 2)} \]

### Step 4: Solve the equations for \( a \) and \( b \)
From Equation 1:
\[ a(4-b) = 4 \quad \Rightarrow \quad a = \frac{4}{4-b} \]
Substituting \( a \) into Equation 2:
\[ 9 = \frac{4}{4-b}(9-b) \]
Cross-multiplying gives:
\[ 9(4-b) = 4(9-b) \]
Expanding both sides:
\[ 36 - 9b = 36 - 4b \]
Rearranging gives:
\[ -9b + 4b = 0 \quad \Rightarrow \quad -5b = 0 \quad \Rightarrow \quad b = 0 \]

### Step 5: Find \( a \)
Substituting \( b = 0 \) back into Equation 1:
\[ 4 = a(4-0) \quad \Rightarrow \quad 4 = 4a \quad \Rightarrow \quad a = 1 \]

### Step 6: Write the equation of the parabola
The parabola is now:
\[ y^2 = 4x \]

### Step 7: General point on the parabola
The general point \( R \) on the parabola can be expressed as:
\[ R(t) = (t^2, 2t) \]

### Step 8: Area of triangle PQR
The area \( A \) of triangle PQR can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting \( P(4, -4) \), \( Q(9, 6) \), and \( R(t) = (t^2, 2t) \):
\[ A = \frac{1}{2} \left| 4(6 - 2t) + 9(2t + 4) + t^2(-4 - 6) \right| \]
Simplifying this:
\[ = \frac{1}{2} \left| 24 - 8t + 18t + 36 - 10t^2 \right| \]
\[ = \frac{1}{2} \left| -10t^2 + 10t + 60 \right| \]
\[ = 5 \left| -t^2 + t + 6 \right| \]

### Step 9: Maximize the area
To maximize the area, we need to find the maximum of the quadratic function \( -t^2 + t + 6 \). The vertex of a parabola given by \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \):
\[ t = -\frac{1}{2(-1)} = \frac{1}{2} \]

### Step 10: Find coordinates of point R
Substituting \( t = \frac{1}{2} \) into the general point \( R(t) \):
\[ R\left(\frac{1}{2}\right) = \left(\left(\frac{1}{2}\right)^2, 2 \cdot \frac{1}{2}\right) = \left(\frac{1}{4}, 1\right) \]

### Conclusion
Thus, the area of triangle PQR is largest when \( R \) is at the point \( \left(\frac{1}{4}, 1\right) \).