The point P is the point on the curve $x = 2 an \left( y + \frac{\pi}{12} \right)$ with y-coordinate $\frac{\pi}{4}$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 6

We need to differentiate the curve to find the slope of the tangent:

$\frac{dx}{dy} = 2 \sec^2 \left( y + \frac{\pi}{12} \right)$

Now substituting $y = \frac{\pi}{4}$ into the derivative:

$\frac{dx}{dy} = 2 \sec^2 \left( \frac{\pi}{4} + \frac{\pi}{12} \right) = 2 \sec^2 \left( \frac{\pi}{3} \right)$

Since $\sec \left( \frac{\pi}{3} \right) = 2$:

$\frac{dx}{dy} = 2 \cdot 2^2 = 8$

The slope of the tangent at point P is the reciprocal of this:

$m_{tangent} = \frac{1}{8}$.

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

$$m_{normal} = -\frac{1}{m_{tangent}} = -8.$

Using the point-slope form of the equation of a line:

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

Substituting the known values $y_1 = \frac{\pi}{4}$, $x_1 = 2\sqrt{3}$, and $m = -8$:

$y - \frac{\pi}{4} = -8 (x - 2\sqrt{3})$

This simplifies to the equation of the normal:

$y = -8x + 16\sqrt{3} + \frac{\pi}{4}.$