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 1 - 2008 - Paper 1

The gradient of the normal is the negative reciprocal of the tangent's slope. The slope of the tangent is -2, therefore, the slope of the normal is:

$\frac{1}{2}$

Using point-slope form again, we have:

$y - y_1 = m(x - x_1)\Rightarrow y + 3 = \frac{1}{2}(x - 2)$

Thus, $y + 3 = \frac{1}{2}x - 1 \Rightarrow y = \frac{1}{2}x - 4$

The equation of the normal is $y = \frac{1}{2}x - 4$.

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

1. The tangent line intersects the x-axis where $y = 0$:

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

   Thus, point A is $(\frac{1}{2}, 0)$.

2. The normal line intersects the x-axis where $y = 0$:

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

   Thus, point B is $(8, 0)$.

Now we have the coordinates of points A $(\frac{1}{2}, 0)$, B $(8, 0)$, and P $(2, -3)$.

To find the area of triangle APB, we use the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Base = distance between A and B: $\text{Base} = 8 - \frac{1}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}$

Height = y-coordinate of P = 3.

Thus, $\text{Area} = \frac{1}{2} \times \frac{15}{2} \times 3 = \frac{45}{4} = 11.25$

Therefore, the area of triangle APB is 11.25.