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 4 - 2010 - Paper 5

To find the derivative, use the quotient rule:

If ( y = \frac{3}{(5-3x)^2} ), then using the derivative formula, we have: $y' = -\frac{(5-3x)^2 \cdot 0 - 3 \cdot 2(5-3x)(-3)}{(5-3x)^4} = \frac{18(5-3x)}{(5-3x)^4} = \frac{18}{(5-3x)^3}.$

At point P (x = 2): $y' \bigg|_{x=2} = \frac{18}{(5 - 3 \cdot 2)^3} = \frac{18}{(-1)^3} = -18.$

Thus, the slope of the tangent line at P is -18.

The slope of the normal line is the negative reciprocal of the tangent slope. Thus: $m_{normal} = -\frac{1}{-18} = \frac{1}{18}.$

Using point-slope form: $y - y_1 = m(x - x_1)$ Substituting in point P(2, 3) and the slope of the normal line: $y - 3 = \frac{1}{18}(x - 2).$

Multiplying through by 18 to eliminate the fraction gives: $18(y - 3) = x - 2,$ which simplifies to: $x - 18y + 56 = 0.$

Thus, the equation of the normal line in the form ax + by + c = 0 is: $1x - 18y + 56 = 0.$