The curve C has equation $y = 9 - 4x - \frac{8}{x}, \quad x > 0.$ The point P on C has x-coordinate equal to 2 - Edexcel - A-Level Maths Pure - Question 11 - 2009 - Paper 1

The gradient of the normal is the negative reciprocal of the gradient of the tangent. Since the gradient of the tangent at P is -2, the gradient of the normal is:
$\text{Gradient of normal} = \frac{1}{2}.$
Using point-slope form again at point P (2, -3):
$y - (-3) = \frac{1}{2}(x - 2)$
which simplifies to:
$y + 3 = \frac{1}{2}x - 1 \Rightarrow y = \frac{1}{2}x - 4.$
Thus, the equation of the normal at P is:
$y = \frac{1}{2}x - 4.$

To find the area of triangle APB, we first identify the coordinates of points A and B.

1. To find point A, where the tangent intersects the x-axis (where $y = 0$):

   $0 = 1 - 2x \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}.$

   Thus, point A is (0.5, 0).

2. To find point B, where the normal intersects the x-axis:

   $0 = \frac{1}{2}x - 4 \Rightarrow \frac{1}{2}x = 4 \Rightarrow x = 8.$

   Thus, point B is (8, 0).

Now, we can calculate the area of triangle APB using the formula:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.$
The base AB is the distance between points A and B, which is $8 - 0.5 = 7.5$. The height is the y-coordinate of point P, which is -3.
Thus, the area is:
$\text{Area} = \frac{1}{2} \times 7.5 \times 3 = \frac{22.5}{2} = 11.25.$
Therefore, the area of triangle APB is 11.25.