Figure 3 shows the curve C with parametric equations $x = 8 ext{cos} t, y = 4 ext{sin} 2t, 0 < t eq rac{2 ext{π}}{3}.$ The point P lies on C and has coordinates (4, 2√3) - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 7

To find the equation of the line $l$ which is normal to the curve $C$ at point $P(4, 2 ext{√}3)$:

1. Calculate derivatives:
   • Differentiate $x = 8 ext{cos}t$ and $y = 4 ext{sin}(2t)$ with respect to $t$.
   • At t = rac{ ext{π}}{3}, use $dx/dt = -8 ext{sin}t$ and $dy/dt = 8 ext{cos}(2t)$.
   • Calculate $dx/dt$ and $dy/dt$; substitute to find the slope of the tangent line at point P.
2. Since the normal line is perpendicular to the tangent, its slope will be the negative reciprocal of the tangent slope:
   • If slope of tangent at P is $m$, then slope of normal $l$ is -rac{1}{m}. Use this to get the equation in point-slope form:
   $y - 2 ext{√}3 = -m(x - 4)$

To show the area $A$ is given by the integral:

1. The area formula for parametric equations is: A = rac{1}{2} \int_{a}^{b} y \frac{dx}{dt} dt where $y = 4 ext{sin}(2t)$ and rac{dx}{dt} = -8 ext{sin}t.
2. Identify the limits for $t$ based on intersection of the curve with the line $x=4$.
3. Substitute and simplify:

   • Evaluate $A = \int_{[0, 2 ext{π}/3]} (4 ext{sin}(2t))(-8 ext{sin}t) ext{dt}$. This gives:

     $A = \int (64 ext{sin}t ext{cos}t) ext{dt}$.

To solve for the area $A$:

1. Compute the integral using a substitution u = 2t ext{, thus } ext{dt} = rac{1}{2} ext{du}, changing limits accordingly.
   • The area integral simplifies to:
   • $A = 32 \int ext{sin}(u) ext{cos}(u) ext{du}$
2. Using the identity $ext{sin}(2u) = 2 ext{sin}(u) ext{cos}(u)$, further simplify.
3. Finally, evaluate and express in the form $a + b ext{√}3$ to find constants $a$ and $b$.