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 slope of the tangent, we differentiate y with respect to x:

$y = \frac{3}{(5-3x)^2}$

Using the quotient rule:

$y' = \frac{(0 \cdot (5-3x)^2) - (3 \cdot 2(5-3x)(-3))}{(5-3x)^4} = \frac{18(5-3x)}{(5-3x)^4} = \frac{18}{(5-3x)^3}$

Now substituting x = 2 gives:

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

The slope of the tangent is -18.

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

$m_{normal} = -\frac{1}{-18} = \frac{1}{18}$

Using the point-slope form of the equation:

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

we substitute the coordinates of P (2, 3):

$y - 3 = \frac{1}{18}(x - 2)$

Multiplying through by 18 to eliminate the fraction results in:

$18(y - 3) = x - 2$

This simplifies to:

$x - 18y + 56 = 0$

Thus, the equation in the required form ax + by + c = 0 is:

$1x - 18y + 56 = 0$ with a = 1, b = -18, and c = 56.