Figure 3 shows the curve C with parametric equations x = 8 cos t, y = 4 sin 2t, 0 ≤ t ≤ π/2 - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 7

To find the equation of the line l that is normal to curve C at P:

1. Compute the derivatives:

   • From the equations: $\frac{dx}{dt} = -8 \sin t$ $\frac{dy}{dt} = 8 \cos 2t$.

2. At point P, substituting t = \frac{\pi}{3}:

   • $\frac{dx}{dt} = -8 \sin(\frac{\pi}{3}) = -8 \cdot \frac{\sqrt{3}}{2} = -4\sqrt{3}$
   • $\frac{dy}{dt} = 8 \cos(\frac{2\pi}{3}) = 8 \cdot -\frac{1}{2} = -4.$

3. Thus, the slope of the tangent line at P is: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4}{-4\sqrt{3}} = \frac{1}{\sqrt{3}}.$

4. The slope of the normal line is: $m = -\sqrt{3}.$

5. The equation of the normal line l in point-slope form: $y - 2\sqrt{3} = -\sqrt{3}(x - 4).$

The area R bounded by curve C, the x-axis, and the line x = 4 can be calculated by integrating with respect to t:

Using the area formula: $A = \int_{a}^{b} y \, dx = \int_{t_1}^{t_2} y(t) \frac{dx}{dt} \, dt$, where $y = 4 \sin 2t \text{ and } \frac{dx}{dt} = -8 \sin t$.

Thus:

$A = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (4 \sin 2t)(-8 \sin t) \, dt = -32 \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \sin 2t \sin t \, dt.$

Using the identity $\sin 2t = 2 \sin t \cos t$:

$A = -64 \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \sin^2 t \, \cos t \, dt.$

Rearranging gives: $A = \int_{\frac{5}{3}}^{4} \left( 64 \sin^{2} t / \cos t \right) \, dt.$

To find the area using the integral:

$A = \int_{\frac{5}{3}}^{4} \left( 64 \sin^{2} t / \cos t \right) \, dt.$

Using substitution where ( u = \sin t ) and adjusting limits:

1. When t = \frac{5}{3}, u = \sin(\frac{5}{3})
2. When t = 4, u = \sin(4).

Calculate the integral, evaluating to get the final area, expressed in the form a + b√3:

1. Collect constants and simplify:
2. Final expressions for a, b:
3. Evaluate to conclude with area in the required form.