曲线 $C$ 由参数方程定义 $$\begin{aligned} x &= 4\sin t, \\ y &= b\cos t - 1, \end{aligned}$$ 其中 $0 \le t \le \pi$ 且 $1 < b < 4$。

1. Eliminate the parameter $t$ to determine the Cartesian equation of the curve.

$x = 4\sin t \implies \sin t = \frac{x}{4}$
$y = b\cos t - 1 \implies \cos t = \frac{y+1}{b}$
Using the Pythagorean identity $\sin^2 t + \cos^2 t = 1$, we get:
$\left(\frac{x}{4}\right)^2 + \left(\frac{y+1}{b}\right)^2 = 1$
This represents an ellipse centered at $(0, -1)$.

2. Determine the restricted domain for $x$ based on the range of $t$.

Given the parameter range $0 \le t \le \pi$, we know that $\sin t \ge 0$.
Therefore, $x = 4\sin t \ge 0$.
This confirms that $C$ is exactly the right half of the ellipse.

3. Find the exact coordinates where the curve meets the $y$-axis.

The curve meets the $y$-axis when $x = 0$.
$4\sin t = 0 \implies t = 0 \text{ or } t = \pi$
When $t = 0$, $y = b\cos(0) - 1 = b - 1$.
When $t = \pi$, $y = b\cos(\pi) - 1 = -b - 1$.
Since $b > 1$, $b - 1 > 0$ and $-b - 1 < 0$.
The curve intersects the $y$-axis at $(0, b-1)$ and $(0, -b-1)$.

4. Describe the required sketch of the curve $C$.

Draw a smooth, C-shaped semi-ellipse in the quadrant $x \ge 0$. The curve starts at the upper $y$-intercept $(0, b-1)$ (which is above the $x$-axis), curves outwards to reach a maximum $x$-coordinate at the point $(4, -1)$, and curves back inwards to terminate at the lower $y$-intercept $(0, -b-1)$ (which is below the $x$-axis).

1. Differentiate the parametric equations to find $\frac{dy}{dx}$.

$\frac{dx}{dt} = 4\cos t$
$\frac{dy}{dt} = -b\sin t$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-b\sin t}{4\cos t} = -\frac{b}{4}\tan t$

2. Equate the gradient to the gradient of the given line to find the value of $t$.

The gradient of the line $y = \frac{1}{4}bx$ is $\frac{b}{4}$.
Since the tangent is parallel to this line, its gradient must be equal:
$-\frac{b}{4}\tan t = \frac{b}{4}$
$\tan t = -1$
Given the domain $0 \le t \le \pi$, the only solution is:
$t = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

3. Find the coordinates of the point of tangency.

Substitute $t = \frac{3\pi}{4}$ into the parametric equations:
$x = 4\sin\left(\frac{3\pi}{4}\right) = 4\left(\frac{\sqrt{2}}{2}\right) = 2\sqrt{2}$
$y = b\cos\left(\frac{3\pi}{4}\right) - 1 = b\left(-\frac{\sqrt{2}}{2}\right) - 1 = -\frac{\sqrt{2}}{2}b - 1$

The coordinates of the point are $\left(2\sqrt{2}, -\frac{\sqrt{2}}{2}b - 1\right)$.

4. Formulate the equation of the tangent and simplify it to the required form.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - \left(-\frac{\sqrt{2}}{2}b - 1\right) = \frac{b}{4}\left(x - 2\sqrt{2}\right)$
$y + \frac{\sqrt{2}}{2}b + 1 = \frac{b}{4}x - \frac{\sqrt{2}}{2}b$
$y = \frac{bx}{4} - \frac{\sqrt{2}}{2}b - \frac{\sqrt{2}}{2}b - 1$
$y = \frac{bx}{4} - \sqrt{2}b - 1$ (Shown)

1. State the equations of the curve and the tangent for $b = 2$, and find the $x$-coordinate of the point of tangency.

Given $b=2$, the parametric equations of $C$ are $x = 4\sin t$ and $y = 2\cos t - 1$.
The equation of the tangent from (b) is $y = \frac{1}{2}x - 2\sqrt{2} - 1$.
To find the point of tangency, we equate the gradient of the curve to the gradient of the tangent (which is $\frac{1}{2}$):
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2\sin t}{4\cos t} = -\frac{1}{2}\tan t$
$-\frac{1}{2}\tan t = \frac{1}{2} \implies \tan t = -1$
Since $0 \le t \le \pi$, $t = \frac{3\pi}{4}$.
At $t = \frac{3\pi}{4}$, the $x$-coordinate is $x = 4\sin\left(\frac{3\pi}{4}\right) = 2\sqrt{2}$.

2. Identify the limits of integration for the bounded region.

The bounded region lies between the $y$-axis ($x=0$) and the point of tangency ($x=2\sqrt{2}$).
For the curve $C$, at $x=0$, $4\sin t = 0 \implies t=0 \text{ or } t=\pi$.
The tangency point has $y = 2\cos\left(\frac{3\pi}{4}\right) - 1 = -\sqrt{2}-1$, which is on the lower segment of $C$. Thus, at $x=0$, we take the corresponding lower point where $t=\pi$ (which yields $y=-3$).
The required area is the difference between the integral of the curve and the integral of the tangent over $x \in [0, 2\sqrt{2}]$:
$\text{Area} = \int_{0}^{2\sqrt{2}} y_{\text{curve}} \, \mathrm{d}x - \int_{0}^{2\sqrt{2}} y_{\text{tangent}} \, \mathrm{d}x$

3. Evaluate the integral of the area under the curve using parametric integration.

Substituting $x = 4\sin t$ and $\mathrm{d}x = 4\cos t \, \mathrm{d}t$, and using limits $t=\pi$ to $t=\frac{3\pi}{4}$:
$\int_{0}^{2\sqrt{2}} y \, \mathrm{d}x = \int_{\pi}^{\frac{3\pi}{4}} (2\cos t - 1)(4\cos t) \, \mathrm{d}t$
$= \int_{\pi}^{\frac{3\pi}{4}} (8\cos^2 t - 4\cos t) \, \mathrm{d}t$
Using the double-angle formula $\cos 2t = 2\cos^2 t - 1$:
$= \int_{\pi}^{\frac{3\pi}{4}} (4\cos 2t + 4 - 4\cos t) \, \mathrm{d}t$
$= \left[ 2\sin 2t + 4t - 4\sin t \right]_{\pi}^{\frac{3\pi}{4}}$
$= \left( 2\sin\frac{3\pi}{2} + 4\left(\frac{3\pi}{4}\right) - 4\sin\frac{3\pi}{4} \right) - \left( 2\sin 2\pi + 4\pi - 4\sin\pi \right)$
$= \left( -2 + 3\pi - 4\left(\frac{\sqrt{2}}{2}\right) \right) - (4\pi)$
$= -\pi - 2 - 2\sqrt{2}$

4. Evaluate the integral of the area under the tangent.

$\int_{0}^{2\sqrt{2}} \left( \frac{1}{2}x - 2\sqrt{2} - 1 \right) \mathrm{d}x = \left[ \frac{1}{4}x^2 - (2\sqrt{2} + 1)x \right]_{0}^{2\sqrt{2}}$
$= \frac{1}{4}(2\sqrt{2})^2 - (2\sqrt{2} + 1)(2\sqrt{2})$
$= 2 - (8 + 2\sqrt{2}) = -6 - 2\sqrt{2}$

5. Calculate the exact bounded area.

$\text{Area} = \int_{0}^{2\sqrt{2}} y_{\text{curve}} \, \mathrm{d}x - \int_{0}^{2\sqrt{2}} y_{\text{tangent}} \, \mathrm{d}x$
$\text{Area} = (-\pi - 2 - 2\sqrt{2}) - (-6 - 2\sqrt{2})$
$= -\pi - 2 - 2\sqrt{2} + 6 + 2\sqrt{2}$
$= 4 - \pi$