The point P lies on the curve with equation $x = (4y - ext{sin}(2y))^2$ Given that P has $(x,y)$ coordinates $(p, rac{ ext{ }{2}})$, 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, we first need to find the derivative of x with respect to y at point P.

Using implicit differentiation:

1. Differentiate both sides:
   $\frac{dx}{dy} = 2(4y - \text{sin}(2y))(4 - 2\text{cos}(2y))$
2. Substitute y = \frac{\pi}{2} into the derivative: $\frac{dx}{dy} = 2(2\pi)(4 - 2\text{cos}(\pi)) = 2(2\pi)(4 + 2) = 2(2\pi)(6) = 24\pi$

The slope or gradient of the tangent line at point P is \frac{dy}{dx} = \frac{1}{24\pi}.

Using point-slope form, we get the equation of the tangent:

$y - \frac{\pi}{2} = \frac{1}{24\pi}(x - 4\pi^2)$

Setting x = 0 to find the y-intercept (point A):

$y - \frac{\pi}{2} = \frac{1}{24\pi}(0 - 4\pi^2)$ This simplifies to:

$y - \frac{\pi}{2} = -\frac{1}{6} \to y = \frac{\pi}{2} - \frac{1}{6}$ Converting to a common denominator: $y = \frac{3\pi - 1}{6}$

Thus, the coordinates of A are:

$A(0, \frac{3\pi - 1}{6})$