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

Substituting $y = \frac{\pi}{4}$ into the equation:

$x = 2 \tan\left(\frac{\frac{\pi}{4} + \frac{\pi}{12}}{2}\right)$

First, calculate the angle:

$\frac{\frac{\pi}{4} + \frac{\pi}{12}}{2} = \frac{3\pi + \pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$

Then find:

$\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$

Now, substituting back gives:

$x = 2 \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

To find the slope of the tangent line at point P, differentiate the curve with respect to y:

$\frac{dx}{dy} = 2 \sec^2\left(\frac{y + \frac{\pi}{12}}{2}\right) \cdot \frac{1}{2} = \sec^2\left(\frac{y + \frac{\pi}{12}}{2}\right)$

Evaluate this at $y = \frac{\pi}{4}$:

$\sec^2(\frac{\frac{\pi}{4} + \frac{\pi}{12}}{2}) = \sec^2(\frac{\pi}{6}) = 4$

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

If slope of tangent $m = 4$, then slope of normal $m_{normal} = -\frac{1}{4}$.

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

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

Substituting $m = -\frac{1}{4}$, $x_1 = \frac{2}{\sqrt{3}}$, and $y_1 = \frac{\pi}{4}$ gives:

$y - \frac{\pi}{4} = -\frac{1}{4}\left(x - \frac{2}{\sqrt{3}}\right)$

This is the equation of the normal at point P.