The curve C has equation $y = \frac{x^{3}(x - 6) + 4}{x}, \quad x > 0$ - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 1

To determine if the tangents at points P and Q are parallel, we need to find the gradients of the curve at these points.

First, we differentiate the curve: $y = \frac{x^{3}(x - 6) + 4}{x} = x^{2}(x - 6) + \frac{4}{x}$

Using the quotient rule: $\frac{dy}{dx} = \frac{d}{dx} \left( x^{2}(x - 6) + 4x^{-1} \right)$ $= \frac{d}{dx}(x^{3} - 6x^{2}) + \frac{d}{dx}(4x^{-1})$ $= 3x^{2} - 12x - 4x^{-2}$

Evaluating at P (x = 1): $\frac{dy}{dx} \bigg|_{x=1} = 3(1)^{2} - 12(1) - 4(1)^{-2} = 3 - 12 - 4 = -13$

Evaluating at Q (x = 2): $\frac{dy}{dx} \bigg|_{x=2} = 3(2)^{2} - 12(2) - 4(2)^{-2} = 12 - 24 - 1 = -13$

Both gradients are equal: $\therefore \text{the tangents at P and Q are parallel.}$

The gradient of the tangent at point P is -13. The gradient of the normal is the negative reciprocal: $m = -\frac{1}{13}$

Using point P(1, -1) in the point-gradient form of the line: $y - (-1) = -\frac{1}{13}(x - 1)$ $y + 1 = -\frac{1}{13}x + \frac{1}{13}$

ightarrow 13y + 13 = -x + 1$$ $$\rightarrow x + 13y + 12 = 0$$ Thus, the normal equation is: $$\text{in the form } ax + by + c = 0,\ a = 1, b = 13, c = 12.$$