The curve C has equation $$y=\frac{1}{2}x^3-9x^2+\frac{8}{x}+30, \quad x>0$$ (a) Find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 2

We need to substitute $x = 4$ into the equation of the curve:

$y = \frac{1}{2}(4)^3 - 9(4)^2 + \frac{8}{4} + 30$

Calculating this gives:

$y = \frac{1}{2}(64) - 9(16) + 2 + 30 = 32 - 144 + 2 + 30 = -8$

Thus, the point $P(4, -8)$ lies on $C$.

First, we find the gradient of the tangent at $x = 4$:

$\frac{dy}{dx}|_{x=4} = \frac{3}{2}(4)^2 - 18(4) - \frac{8}{(4)^2} = 24 - 72 - \frac{1}{2} = -48.5$

The gradient of the normal is the negative reciprocal:

$\text{Gradient of normal} = \frac{1}{48.5}$

Using the point-slope form of the line equation:

$y - (-8) = \frac{1}{48.5}(x - 4)$

Rearranging, we can convert to the form $ax + by + c = 0$. Multiplying through by 48.5 to eliminate the fraction gives:

$48.5y + 8*48.5 = x - 4$ After a series of steps, we arrive at the final form:

$7y - 2x + 64 = 0$