The curve C has equation $x = 8y \tan 2y$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 5

To find the equation of the tangent at point P, we first need to find (\frac{dx}{dy}) at (y = \frac{\pi}{8}).

Differentiating the curve's equation:

$\frac{dx}{dy} = 8 \tan(2y) + 16y \sec^2(2y)$

Next, we evaluate this at (y = \frac{\pi}{8}):

$\frac{dx}{dy} \bigg|_{y = \frac{\pi}{8}} = 8\tan\left(\frac{\pi}{4}\right) + 16\left(\frac{\pi}{8}\right)\sec^2\left(\frac{\pi}{4}\right)$

Simplifying gives:

$\frac{dx}{dy} \bigg|_{y = \frac{\pi}{8}} = 8 + 16\left(\frac{\pi}{8}\right)\cdot 2 = 8 + 4\pi$

The slope of the tangent line is thus (m = 8 + 4\pi).

Using the point-slope form of the line:

$y - \frac{\pi}{8} = (8 + 4\pi)(x - \frac{\pi}{8})$

Rearranging into the form (ay = x + b):

1. Isolate y:
   $y = (8 + 4\pi)x - (8 + 4\pi)\frac{\pi}{8} + \frac{\pi}{8}$
2. Simplifying gives:
   $a = \frac{1}{8 + 4\pi}$
   and
   $b = \frac{\pi - \frac{(8 + 4\pi)\pi}{8}}{(8 + 4\pi)}$

Thus, the equation of the tangent line at P in the desired form is given, with a and b determinable in terms of (\pi).