The curve C has equation $x = 8y \tan (2y)$ The point P has coordinates $ \left(\frac{\pi}{8}, \frac{\pi}{8}\right)$ (a) Verify that P lies on C - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 5

To find the equation of the tangent to C at the point P, we first need to calculate the derivative of $x$ with respect to $y$:

1. Differentiate the equation $x = 8y \tan(2y)$ with respect to $y$: $\frac{dx}{dy} = 8 \tan(2y) + 8y \cdot 2 \sec^2(2y) = 8 \tan(2y) + 16y \sec^2(2y)$

2. Evaluate the derivative at $y = \frac{\pi}{8}$:

   • Start by calculating $\tan(\frac{\pi}{4})$ and $\sec(\frac{\pi}{4})$: $\tan(\frac{\pi}{4}) = 1, \quad \sec(\frac{\pi}{4}) = \sqrt{2}$
   • Substitute to find: $\frac{dx}{dy} = 8(1) + 16 \cdot \frac{\pi}{8} \cdot 2 = 8 + 4\pi$

3. The slope of the tangent line is given by the derivative evaluated at point P: $\text{slope} = \frac{dx}{dy} = 8 + 4\pi$

4. Use the point-slope form of a line: $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = \left(\frac{\pi}{8}, \frac{\pi}{8}\right)$ and $m = 8 + 4\pi$: $y - \frac{\pi}{8} = (8 + 4\pi)\left(x - \frac{\pi}{8}\right)$

5. Rearranging this to the form $ay = x + b$, we identify the values of $a$ and $b$ in terms of $\\pi$:

   • Rearranging gives the final equation of the tangent line at P.