The point P lies on the curve with equation $x = (4y - ext{sin } 2y)^2$ Given that P has $(x, y)$ coordinates $igg( p, rac{ ext{π}}{2} igg)$, where $p$ is a constant, (a) find the exact value of $p$ - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 3

To find the coordinates of point A, we first determine the derivative of $x$ with respect to $y$.

From the function,

$x = (4y - ext{sin } 2y)^2$

Using the chain rule, we have:

rac{dx}{dy} = 2(4y - ext{sin } 2y) imes (4 - 2 ext{cos } 2y)

Now, substitute y = rac{ ext{π}}{2} to find the slope at point P:

• $4 - 2 ext{cos}( ext{π}) = 4 + 2 = 6$.
• Therefore, at y = rac{ ext{π}}{2}, $4y - ext{sin } 2y = 2 ext{π}$, leading to:

rac{dx}{dy} = 2(2 ext{π})(6) = 24 ext{π}

The equation of the tangent line at P can now be deduced:

Using the point-slope form, y - rac{ ext{π}}{2} = rac{24 ext{π}}{24} (x - 4 ext{π}^2)

Simplifying this equation leads to: y - rac{ ext{π}}{2} = ext{π} (x - 4 ext{π}^2)

To find the y-intercept, let $x = 0$:

y - rac{ ext{π}}{2} = ext{π} (0 - 4 ext{π}^2)

This yields: y = rac{ ext{π}}{2} - 4 ext{π}^3

Hence, the coordinates of point A are ig(0, rac{ ext{π}}{2} - 4 ext{π}^3ig).