The point P lies on the curve with equation $y = ext{ln}(rac{1}{3}x)$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 5

To find the slope of the tangent at point P, we need to differentiate the curve:

$\frac{dy}{dx} = \frac{1}{x} \cdot \frac{d}{dx}\bigg(\frac{1}{3}x\bigg) = \frac{1}{3x}.$

Substituting $x = 3$:

$\frac{dy}{dx} = \frac{1}{3(3)} = \frac{1}{9}.$

Therefore, the slope of the tangent at point P is $\frac{1}{9}$.

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

$m = -\frac{1}{\frac{1}{9}} = -9.$

Thus, the slope of the normal at point P is -9.

Using the point-slope form of the line equation:

$y - y_1 = m(x - x_1),$ where $(x_1, y_1) = (3, 0)$ and $m = -9$, we get:

$y - 0 = -9(x - 3).$

Simplifying gives:

$y = -9x + 27.$

This can be expressed in the form $y = ax + b$ where $a = -9$ and $b = 27$.