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

To find the equation of the tangent to C at point P, we must determine the derivative $\frac{dx}{dy}$ at P:

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

Substituting $y = \frac{\pi}{8}$ gives:

$\frac{dx}{dy} = 8 \tan\left( \frac{\pi}{4} \right) + 16 \cdot \frac{\pi}{8} \sec^2\left( \frac{\pi}{4} \right)\n$

This results in:

$\frac{dx}{dy} = 8 + 2\pi$

The slope of the tangent line at point P is the inverse of rac{dx}{dy}, so:

$m = \frac{1}{8 + 2\pi}$

Using the point-slope form of the line, the equation of the tangent at P $\left( \frac{\pi}{8}, \frac{\pi}{8} \right)$ is:

$$\left( y - \frac{\pi}{8} \right) = m \left( x - \frac{\pi}{8} \right)$$

Substituting $m$:

$$\left( y - \frac{\pi}{8} \right) = \frac{1}{8 + 2\pi} \left( x - \frac{\pi}{8} \right)$$

Rearranging to the form $ay = x + b$ gives:

$y = \frac{1}{8 + 2\pi} x + b$

where the value of $b$ can be calculated from substituting in the point P to finalize the equation.