The curve C has equation $y = \frac{1}{3} x^3 - 4x^2 + 8x + 3$ - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

To find the equation of the tangent to C at the point P:

1. Differentiate the equation of C: $\frac{dy}{dx} = \frac{1}{3}(3x^2) - 8x + 8 = x^2 - 8x + 8.$

2. Evaluate the derivative at $x = 3$: $\frac{dy}{dx} \Bigg|_{x=3} = 3^2 - 8(3) + 8 = 9 - 24 + 8 = -7.$ Thus, the slope of the tangent at P is $m = -7$.

3. Using the point-slope form of the equation of a line: $y - y_1 = m(x - x_1),$ substitute $(x_1, y_1) = (3, 0)$: $y - 0 = -7(x - 3)$ simplifying, $y = -7x + 21.$ Hence, the tangent line at P is given by: $y = -7x + 21.$

Let the coordinates of point Q be $(x, y)$. Since the tangent to C at Q is parallel to the tangent at P, it must have the same slope, which is $-7$.

1. Differentiate the curve C again to find critical points: $\frac{dy}{dx} = x^2 - 8x + 8.$

2. Set the derivative equal to -7 to find $x$:

   x^2 - 8x + 15 = 0.$$

3. Factor this equation: $(x - 3)(x - 5) = 0,$ thus $x = 3$ or $x = 5.$ Since we already found point P at $x = 3$, we take $x = 5$.

4. Substitute $x = 5$ back into the equation of C to find $y$:

   $$= \frac{1}{3}(125) - 4(25) + 40 + 3 \ = \frac{125}{3} - 100 + 40 + 3 \ = \frac{125}{3} - 60 + \frac{9}{3} \ = \frac{125 - 180 + 9}{3} \ = \frac{-46}{3}.$$