The parametric equations of a curve C are $x = 3\sec t$ and $y = -2\tan t$, where $0 < t < \frac{\pi}{2}$. Find the cartesian equation of the curve, stating any restriction on the values of $x$ and $y$. Sketch the curve C, indicating the equation of the asymptote if any. A curve $C_1$ has parametric equations $x = 2\sin t + 1, \quad y = 2\cos 3t + 4\sin t, \quad -\frac{\pi}{3} \le t \le \frac{\pi}{3}.$ The tangent to the curve $C_1$ at the point $(1,2)$ is given by $L$ as shown in the diagram below. The line $L$ cuts the line $x = 2$ at $(2,4)$. Show that the area bounded by the curve $C_1$, the line $L$ and the line $x = 2$ is given by $a - \displaystyle\int_{0}^{\theta} 4\cos 3t\cos t + 8\sin t\cos t\;\mathrm{d}t$, where $a$ and $\theta$ are constants to be determined in exact form. Hence find the exact value of this area.

1. Express the trigonometric functions in terms of $x$ and $y$.

Given $x = 3\sec t$ and $y = -2\tan t$, we have:
$\sec t = \frac{x}{3}$
$\tan t = -\frac{y}{2}$

2. Use a trigonometric identity to eliminate the parameter $t$.

Using the identity $\sec^2 t - \tan^2 t = 1$:
$\left(\frac{x}{3}\right)^2 - \left(-\frac{y}{2}\right)^2 = 1$
$\frac{x^2}{9} - \frac{y^2}{4} = 1$

3. Determine the restrictions on $x$ and $y$ based on the given domain for $t$.

Given the domain $0 < t < \frac{\pi}{2}$:
$\sec t > 1 \implies \frac{x}{3} > 1 \implies x > 3$
$\tan t > 0 \implies -\frac{y}{2} > 0 \implies y < 0$

1. State the asymptotes of the hyperbola based on the cartesian equation.

From part (a)(i), the cartesian equation is $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
The asymptotes of this hyperbola are given by $\frac{x^2}{9} - \frac{y^2}{4} = 0$, which simplifies to $y = \pm \frac{2}{3}x$.

2. Select the relevant branch and asymptote, and describe the sketch.

Based on the parameter $0 < t < \frac{\pi}{2}$, we have $x > 3$ and $y < 0$.
Therefore, the curve lies entirely in the 4th quadrant, and the relevant asymptote is $y = -\frac{2}{3}x$.
The sketch consists of the line $y = -\frac{2}{3}x$ drawn as a dotted line, and a solid curve that begins at $(3,0)$ with an open (unfilled) circle, moving downwards to the right while approaching the asymptote.

1. Calculate the area under the line $L$ between $x=1$ and $x=2$.

The line $L$ passes through $(1, 2)$ and $(2, 4)$.
The area under $L$ from $x=1$ to $x=2$ is the area of a trapezium:
$\text{Area under } L = \frac{1}{2}(2 + 4)(2 - 1) = 3$

2. Find the limits of the parameter $t$ corresponding to $x=1$ and $x=2$.

For the lower limit $x = 1$:
$2\sin t + 1 = 1 \implies \sin t = 0 \implies t = 0$
For the upper limit $x = 2$:
$2\sin t + 1 = 2 \implies \sin t = \frac{1}{2} \implies t = \frac{\pi}{6}$

3. Express the area under the curve $C_1$ in terms of the parameter $t$.

The area under $C_1$ from $x=1$ to $x=2$ is given by $\int_{1}^{2} y \,\mathrm{d}x$.
Differentiating $x = 2\sin t + 1$ with respect to $t$ gives $\frac{\mathrm{d}x}{\mathrm{d}t} = 2\cos t$.
Substituting $y$ and $\frac{\mathrm{d}x}{\mathrm{d}t}$ into the integral:
$\int_{0}^{\frac{\pi}{6}} (2\cos 3t + 4\sin t)(2\cos t) \,\mathrm{d}t = \int_{0}^{\frac{\pi}{6}} (4\cos 3t\cos t + 8\sin t\cos t) \,\mathrm{d}t$

4. Formulate the required bounded area and deduce the values of $a$ and $\theta$.

The required bounded area is the difference between the area under $L$ and the area under $C_1$:
$\text{Bounded Area} = 3 - \int_{0}^{\frac{\pi}{6}} (4\cos 3t\cos t + 8\sin t\cos t) \,\mathrm{d}t$
Comparing this with the given expression $a - \int_{0}^{\theta} (4\cos 3t\cos t + 8\sin t\cos t) \,\mathrm{d}t$, we find:
$a = 3 \quad \text{and} \quad \theta = \frac{\pi}{6}$

5. Use trigonometric identities to rewrite the integrand into an integrable form.

Using the factor formula $2\cos A \cos B = \cos(A+B) + \cos(A-B)$ and double-angle formula $2\sin A \cos A = \sin 2A$:
$4\cos 3t\cos t = 2(\cos 4t + \cos 2t) = 2\cos 4t + 2\cos 2t$
$8\sin t\cos t = 4\sin 2t$
So the integral simplifies to:
$\int_{0}^{\frac{\pi}{6}} (2\cos 4t + 2\cos 2t + 4\sin 2t) \,\mathrm{d}t$

6. Evaluate the definite integral exactly.

Integrating term by term:
$I = \left[ \frac{1}{2}\sin 4t + \sin 2t - 2\cos 2t \right]_{0}^{\frac{\pi}{6}}$
Substitute the upper limit $t = \frac{\pi}{6}$:
$\frac{1}{2}\sin\left(\frac{2\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) - 2\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} - 2\left(\frac{1}{2}\right) = \frac{3\sqrt{3}}{4} - 1$
Substitute the lower limit $t = 0$:
$\frac{1}{2}\sin(0) + \sin(0) - 2\cos(0) = -2$
Subtracting the evaluations:
$I = \left(\frac{3\sqrt{3}}{4} - 1\right) - (-2) = \frac{3\sqrt{3}}{4} + 1$

7. Find the exact final value of the bounded area.

Subtract the evaluated integral from $a$:
$\text{Area} = 3 - \left( \frac{3\sqrt{3}}{4} + 1 \right) = 2 - \frac{3\sqrt{3}}{4}$