Figure 2 shows a sketch of part of the curve C with parametric equations $$ x = 1 - \frac{1}{2}t, \quad y = 2t - 1 $$ The curve crosses the y-axis at the point A and crosses the x-axis at the point B - Edexcel - A-Level Maths Pure - Question 17 - 2013 - Paper 1

To find when the curve crosses the x-axis (point B), we set $y = 0$:

1. Setting $y = 0$ in the parametric equation: $0 = 2t - 1 \Rightarrow 2t = 1 \Rightarrow t = \frac{1}{2}$

2. Now substituting $t = \frac{1}{2}$ into the $x$ equation: $x = 1 - \frac{1}{2}(\frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4}$

3. Therefore, the x-coordinate of point B is $\frac{3}{4}$.

To find the equation of the normal to the curve at point A, we first need to determine the slope of the tangent line at A:

1. The derivative rac{dy}{dx} can be found using: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

2. From the parametric equations:

   • $\frac{dy}{dt} = 2$
   • $\frac{dx}{dt} = -\frac{1}{2}$

3. Therefore, $\frac{dy}{dx} = \frac{2}{-\frac{1}{2}} = -4$

4. The slope of the normal line, $m_{normal}$, is the negative reciprocal of the tangent slope: $m_{normal} = \frac{1}{4}$

5. Using point-slope form of the line equation: $y - 3 = \frac{1}{4}(x - 0) \Rightarrow y = \frac{1}{4}x + 3$

6. Thus, the equation of the normal line at point A is: $y = \frac{1}{4}x + 3$.

To find the area of region R bounded by the curve, the x-axis, and the line $x = -1$:

1. Setting up the integral for area:

   The area A can be expressed as: $A = \int_{-1}^{0} (y) \, dx$

2. Express $y$ in terms of $t$ and change the limits according to $x = 1 - \frac{1}{2}t$:

   $x = 1 - \frac{1}{2}t \Rightarrow t = 2(1 - x)$ for $x$ limits from -1 to 0:

   When $x = -1$, $t = 4$ and when $x = 0$, $t = 0$.

3. Therefore, $A = \int_{4}^{0} (2t - 1) \left(-\frac{1}{2} \right) dt$ $A = \frac{1}{2} \int_{0}^{4} (2t - 1) dt$

4. Evaluating the integral:

   $\int (2t - 1) dt = t^2 - t$

5. Evaluating from 0 to 4: $= [4^2 - 4] - [0^2 - 0] = 16 - 4 = 12$

6. Therefore, the area of region R is: $A = \frac{1}{2} (12) = 6.$

Thus, the exact area of R is $6$.