曲线 $C$ 由以下参数方程定义：
$x = 2\cos\theta + 1, \quad y = \sin\theta, \quad \text{where } -\pi \le \theta \le \pi.$

1. Convert the parametric equations to a Cartesian equation to identify the curve's shape.

From $x = 2\cos\theta + 1$, we have $\cos\theta = \frac{x-1}{2}$.
Given $y = \sin\theta$.
Using the identity $\cos^2\theta + \sin^2\theta = 1$:
$\left(\frac{x-1}{2}\right)^2 + y^2 = 1$
$\frac{(x-1)^2}{4} + y^2 = 1$
This is the equation of an ellipse centred at $(1, 0)$, with a horizontal semi-axis of $2$ and a vertical semi-axis of $1$. The parameter range $-\pi \le \theta \le \pi$ traces the full ellipse exactly once.

2. Find the points of intersection with the $x$-axis.

Set $y = 0$:
$\sin\theta = 0 \implies \theta = -\pi, 0, \pi$
When $\theta = 0$, $x = 2(1) + 1 = 3$.
When $\theta = \pm\pi$, $x = 2(-1) + 1 = -1$.
The points of intersection with the $x$-axis are $(3, 0)$ and $(-1, 0)$.

3. Find the points of intersection with the $y$-axis.

Set $x = 0$:
$2\cos\theta + 1 = 0 \implies \cos\theta = -\frac{1}{2}$
For $-\pi \le \theta \le \pi$, $\theta = -\frac{2\pi}{3}$ or $\frac{2\pi}{3}$.
When $\theta = \frac{2\pi}{3}$, $y = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
When $\theta = -\frac{2\pi}{3}$, $y = \sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.
The points of intersection with the $y$-axis are $\left(0, \frac{\sqrt{3}}{2}\right)$ and $\left(0, -\frac{\sqrt{3}}{2}\right)$.

4. Sketch the curve.

(Draw an ellipse matching the identified properties)

• Centre at $(1,0)$
• Vertices at $(-1,0)$ and $(3,0)$
• Highest and lowest points at $(1,1)$ and $(1,-1)$ respectively
• Intercepts correctly labelled.

1. Set up the integral for the total area enclosed by the curve using symmetry.

\text{The curve is symmetric about the } x \text{-axis. The top half is above the } x \text{-axis for } 0 \le \theta \le \pi \text{.}
\text{Area} = 2 \int_{-1}^{3} y , dx

2. Substitute the parametric equations and adjust the limits of integration.

x = 2\cos\theta + 1 \implies \frac{dx}{d\theta} = -2\sin\theta
\text{When } x = -1, \theta = \pi \text{; when } x = 3, \theta = 0.
\text{Area} = 2 \int_{\pi}^{0} (\sin\theta) (-2\sin\theta) , d\theta
= 4 \int_{0}^{\pi} \sin^2\theta , d\theta

3. Apply the double-angle formula to prepare the trigonometric expression for integration.

\text{Area} = 4 \int_{0}^{\pi} \left( \frac{1 - \cos 2\theta}{2} \right) , d\theta
= 2 \int_{0}^{\pi} (1 - \cos 2\theta) , d\theta

4. Evaluate the definite integral to find the exact area.

\text{Area} = 2 \left[ \theta - \frac{1}{2}\sin 2\theta \right]_{0}^{\pi}
= 2 \left( \left(\pi - \frac{1}{2}\sin 2\pi\right) - (0 - 0) \right)
= 2\pi

1. Find the gradient of the normal to the curve using parametric differentiation.

$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta, \quad \frac{\mathrm{d}y}{\mathrm{d}\theta} = \cos\theta$
$\text{Gradient of tangent, } \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\cos\theta}{-2\sin\theta} = -\frac{1}{2\tan\theta}$
$\text{Gradient of normal, } m_N = -\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} = 2\tan\theta$

2. Formulate the equation of the normal to the curve at the given point.

$y - \sin\theta = 2\tan\theta (x - (2\cos\theta + 1))$

3. Find the length $OQ$ by determining the $x$-intercept of the normal.

Substitute $y = 0$ to find the coordinates of $Q$:
$-\sin\theta = \frac{2\sin\theta}{\cos\theta} (x - 2\cos\theta - 1)$
Since $0 < \theta < \frac{\pi}{2}$, $\sin\theta \neq 0$, so we divide both sides by $\sin\theta$:
$-1 = \frac{2}{\cos\theta} (x - 2\cos\theta - 1)$
$-\frac{1}{2}\cos\theta = x - 2\cos\theta - 1$
$x_Q = \frac{3}{2}\cos\theta + 1$
Since $\cos\theta > 0$ for $0 < \theta < \frac{\pi}{2}$, the length $OQ = x_Q = \frac{3\cos\theta + 2}{2}$.

4. Find the length $OR$ by determining the $y$-intercept of the normal.

Substitute $x = 0$ to find the coordinates of $R$:
$y - \sin\theta = 2\tan\theta (0 - 2\cos\theta - 1)$
$y = \sin\theta - 4\sin\theta - 2\tan\theta$
$y_R = -3\sin\theta - 2\tan\theta$
Since $0 < \theta < \frac{\pi}{2}$, $\sin\theta > 0$ and $\tan\theta > 0$, meaning $y_R < 0$.
Thus, the length $OR = |y_R| = 3\sin\theta + 2\tan\theta$.

5. Calculate the area of triangle $OQR$ and simplify the trigonometric expression.

$\text{Area} = \frac{1}{2} \times OQ \times OR$
$\text{Area} = \frac{1}{2} \left( \frac{3\cos\theta + 2}{2} \right) \left( 3\sin\theta + 2\frac{\sin\theta}{\cos\theta} \right)$
$= \frac{1}{4} (3\cos\theta + 2) \left[ \frac{\sin\theta(3\cos\theta + 2)}{\cos\theta} \right]$
$= \frac{1}{4}\tan\theta (3\cos\theta + 2)^2$