The point P lies on the curve with equation $y = ext{ln} \left( \frac{1}{3} x \right)$ - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 5

Differentiate the function $y = \text{ln} \left( \frac{1}{3} x \right)$ using the chain and quotient rule:

$\frac{dy}{dx} = \frac{1}{\frac{1}{3} x} \cdot \frac{1}{3} = \frac{1}{x}.$

Substitute $x = 3$ into the derivative to find the slope of the tangent line at point P:

$\frac{dy}{dx} \bigg|_{x=3} = \frac{1}{3}.$

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

$m_{normal} = -\frac{1}{\frac{1}{3}} = -3.$

Using the point-slope form of the line, the equation of the normal line at point P $(3, 0)$ is:

$y - 0 = -3(x - 3) \Rightarrow y = -3x + 9.$
Thus, the equation of the normal line is $y = -3x + 9$, where $a = -3$ and $b = 9$.