A tangent and a normal are drawn at the point P(8, 8) on the parabola `y^(2)=8x` which cuts the axis of the parabola at the points A and B respectively. If the centre of the circle through P, A and B is C, then the sum of `sin(anglePCB)` and `cot(anglePCB)` is equal to

Answer

To solve the problem, we need to find the sum of \( \sin(\angle PCB) \) and \( \cot(\angle PCB) \) where \( P \), \( A \), and \( B \) are points on the parabola \( y^2 = 8x \).

### Step 1: Find the coordinates of points \( A \) and \( B \)

1. **Equation of the Parabola**: The given parabola is \( y^2 = 8x \).
2. **Point P**: The point \( P \) is given as \( (8, 8) \).

### Step 2: Find the equation of the tangent at point \( P(8, 8) \)

1. **Slope of the Tangent**: The derivative of \( y^2 = 8x \) gives us:
\[ 2y \frac{dy}{dx} = 8 \implies \frac{dy}{dx} = \frac{4}{y} \]
At \( P(8, 8) \):
\[ \text{slope} = \frac{4}{8} = \frac{1}{2} \]

2. **Equation of the Tangent**: Using the point-slope form:
\[ y - 8 = \frac{1}{2}(x - 8) \]
Rearranging gives:
\[ y = \frac{1}{2}x + 4 \]

### Step 3: Find the intersection of the tangent with the x-axis (point \( A \))

1. **Set \( y = 0 \) in the tangent equation**:
\[ 0 = \frac{1}{2}x + 4 \implies \frac{1}{2}x = -4 \implies x = -8 \]
Thus, \( A = (-8, 0) \).

### Step 4: Find the equation of the normal at point \( P(8, 8) \)

1. **Slope of the Normal**: The slope of the normal is the negative reciprocal of the tangent slope:
\[ \text{slope of normal} = -2 \]

2. **Equation of the Normal**: Using the point-slope form:
\[ y - 8 = -2(x - 8) \]
Rearranging gives:
\[ y = -2x + 24 \]

### Step 5: Find the intersection of the normal with the x-axis (point \( B \))

1. **Set \( y = 0 \) in the normal equation**:
\[ 0 = -2x + 24 \implies 2x = 24 \implies x = 12 \]
Thus, \( B = (12, 0) \).

### Step 6: Find the coordinates of the center \( C \) of the circle through points \( P \), \( A \), and \( B \)

1. **Midpoint of \( AB \)**:
\[ C = \left( \frac{-8 + 12}{2}, 0 \right) = \left( \frac{4}{2}, 0 \right) = (2, 0) \]

### Step 7: Find \( \sin(\angle PCB) \) and \( \cot(\angle PCB) \)

1. **Finding \( \tan(\angle PCB) \)**:
- Slope of line \( PB \):
\[ \text{slope of } PB = \frac{0 - 8}{12 - 8} = \frac{-8}{4} = -2 \]
- Slope of line \( PC \):
\[ \text{slope of } PC = \frac{0 - 8}{2 - 8} = \frac{-8}{-6} = \frac{4}{3} \]

2. **Using the slopes to find \( \tan(\angle PCB) \)**:
\[ \tan(\angle PCB) = \left| \frac{-2 - \frac{4}{3}}{1 + (-2) \cdot \frac{4}{3}} \right| = \left| \frac{-\frac{6}{3} - \frac{4}{3}}{1 - \frac{8}{3}} \right| = \left| \frac{-\frac{10}{3}}{-\frac{5}{3}} \right| = 2 \]

3. **Finding \( \sin(\angle PCB) \)**:
\[ \sin(\angle PCB) = \frac{4}{5} \]

4. **Finding \( \cot(\angle PCB) \)**:
\[ \cot(\angle PCB) = \frac{3}{4} \]

### Step 8: Calculate the final result

1. **Sum of \( \sin(\angle PCB) \) and \( \cot(\angle PCB) \)**:
\[ \sin(\angle PCB) + \cot(\angle PCB) = \frac{4}{5} + \frac{3}{4} \]
Finding a common denominator (20):
\[ = \frac{16}{20} + \frac{15}{20} = \frac{31}{20} \]

### Final Answer:
The sum of \( \sin(\angle PCB) \) and \( \cot(\angle PCB) \) is \( \frac{31}{20} \).