A curve C has equation y = \frac{3}{(5-3x)^2}, \quad x \neq \frac{5}{3} The point P on C has x-coordinate 2 - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 5

First, we differentiate y with respect to x:

$$y' = \frac{d}{dx}\left(\frac{3}{(5-3x)^2}\right) = 3 \cdot \frac{d}{dx}((5-3x)^{-2}) = 3 \cdot (-2)(5-3x)^{-3}(-3) = \frac{18}{(5-3x)^3}.$$

Now let's find the slope at point P (x = 2):

$$y'(2) = \frac{18}{(5-3(2))^3} = \frac{18}{(5-6)^3} = \frac{18}{(-1)^3} = -18.$$

The slope of the normal line is the negative reciprocal of the slope of the tangent:

$\text{slope of normal} = -\frac{1}{-18} = \frac{1}{18}.$

Using the point (2, 3) with the slope of normal, we apply the point-slope form:

$y - y_1 = m(x - x_1)\Rightarrow y - 3 = \frac{1}{18}(x - 2).$

Multiply through by 18 to eliminate the fraction:

$18(y - 3) = x - 2\Rightarrow 18y - 54 = x - 2\Rightarrow x - 18y + 52 = 0.$

This gives us the normal in the desired form with a = 1, b = -18, and c = 52.