The point p lies on the curve with eqn x = (4y - sin(2y)^2, given that p has coordinates (p,π/2), p is a constant, a) find the exact value of p; the tangent to the curve at P cuts the y-axis at A, b) use calculus to find the coordinates of A.

a) Sub y= π/2 into equation, hence x coordinate is 4π^2 b) to find equation of the tangent, differentiate the equation using the chain rule (wrt y) and then substitute the coordinates of p into the differentiated equation. Then use dy/dx = 1/(dx/dy) x = (4y - sin(2y)^2  dx/dy = (4y - sin(2y)2(4 - 2cos(2y)) dx/dy(4π^2,π/2) = 24π ; hence dy/dx = 1/24π This is now the gradient of the tangent. Using y =mx +c (used for linear equations), where m is the gradient, x and y are coordinates of a point and c is the y intercept. Substitute values we have into this, then rearrange and we have that c = π/3, which is the y intercept; coordinates of A = (0, π/3)