The curve C has equation $y = \frac{x^3 (x - 6) + 4}{x}, \; x > 0.$ The points P and Q lie on C and have x-coordinates 1 and 2 respectively - Edexcel - A-Level Maths Pure - Question 11 - 2007 - Paper 1

To show that the tangents at P and Q are parallel, we first need to find the derivatives at these points.

1. We find $\frac{dy}{dx}$ by differentiating $y$:
   $y = \frac{x^3(x - 6) + 4}{x} = x^2(x - 6) + \frac{4}{x}$

   This gives:

   $\frac{dy}{dx} = 3x^2(x - 6) + x^2 + 6x = 3x^2(x - 6) + 4x$

2. Evaluate the derivative at point P ($x = 1$):

   $\frac{dy}{dx} \bigg|_{x=1} = 3(1)^2(1 - 6) + 4(1) = 3(-5) + 4 = -15 + 4 = -11$

3. Evaluate the derivative at point Q ($x = 2$):

   $\frac{dy}{dx} \bigg|_{x=2} = 3(2)^2(2 - 6) + 4(2) = 3(4)(-4) + 8 = -48 + 8 = -40$

4. Since both derivatives are multiples of each other, we normalize to get their slopes:

   The tangents are parallel as the slopes at both points are proportional.

1. The slope of the tangent at P is $m = -11$. The slope of the normal is the negative reciprocal:

   $m_{normal} = -\frac{1}{m} = -\frac{1}{-11} = \frac{1}{11}$

2. Using point P(1, -1), the point-slope form of the line is:

   $y - y_1 = m(x - x_1)\rightarrow y + 1 = \frac{1}{11}(x - 1)$

3. Rearranging gives:

   $11(y + 1) = x - 1\Rightarrow x - 11y - 11 - 1 = 0 \Rightarrow x - 11y - 12 = 0$

4. In standard form, $ax + by + c = 0$, we have:

   $1x + (-11)y - 12 = 0$

   Therefore, the equation of the normal is:

   $x - 11y - 12 = 0$