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

To find the equation of the tangent at point P, we first need to find the derivative of C:

$\frac{dy}{dx} = 2x^2 - 4x + 8$

Next, we evaluate the derivative at x = 3:

$\frac{dy}{dx} \bigg|_{x=3} = 2(3)^2 - 4(3) + 8 = 18 - 12 + 8 = 14$

Thus, the slope m of the tangent at P is m = 14.

Now, using the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

Substituting P(3, 0):

$y - 0 = 14(x - 3)$

Expanding this gives us:

$y = 14x - 42$

So, the equation of the tangent at P is:

$y = 14x - 42$

Since the tangent at Q is parallel to the tangent at P, it must have the same slope of 14. Therefore, the equation of the tangent at Q can also be expressed in point-slope form as follows:

$y - y_Q = 14(x - x_Q)$

Now, we substitute this into the original curve C equation:

Solving for $y_Q$ leads us to:

$y = 14x - 14x_Q + y_Q$

Next, we set this equal to the original curve equation to solve for the coordinates of Q. The quadratic can be simplified leading us to find the exact values for $x_Q$ and $y_Q$.

After solving the quadratic form, we will identify the coordinates of point Q that also satisfy curve C, while maintaining the specified slope of the tangent.