The point P lies on the curve with equation $x = (4y - extrm{sin}(2y))^2$ Given that P has $(x,y)$ coordinates $(p, rac{ u}{2})$, where p is a constant, a) find the exact value of p - Edexcel - A-Level Maths Pure - Question 6 - 2015 - Paper 3

To find the coordinates of point A where the tangent cuts the y-axis, perform the following steps:

1. Differentiate the curve equation: Starting from: $x = (4y - \textrm{sin}(2y))^2$ Using implicit differentiation, we have:

   $\frac{dx}{dy} = 2(4y - \textrm{sin}(2y))(4 - 2\cos(2y))$

   Evaluate this derivative at P where y = \frac{\pi}{2}:

   $\frac{dx}{dy} \Big|_{y = \frac{\pi}{2}} = 2(2\pi)(4 - 2\cos(\pi)) = 2(2\pi)(4 + 2) = 12\pi$

   The slope of the tangent is therefore: $m = \frac{dy}{dx} = \frac{1}{12\pi}$

2. Equation of the tangent line at P: The tangent line equation in point-slope form is: $y - \frac{\pi}{2} = m(x - 4\pi^2)$ Substitute m: $y - \frac{\pi}{2} = \frac{1}{12\pi}(x - 4\pi^2)$

3. Finding the y-coordinate when x = 0: Set x = 0:

   $y - \frac{\pi}{2} = \frac{1}{12\pi}(0 - 4\pi^2)$

   Simplifying this gives: $y - \frac{\pi}{2} = -\frac{4\pi}{12} = -\frac{\pi}{3}$

   So, $y = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6}$

4. Coordinates of A: Thus the coordinates of A are: $A(0, \frac{\pi}{6})$