曲线 $C_1$ 的参数方程为 $x = 3\sec^2 \theta$, $y = 2\tan \theta$, $0 \le \theta < \frac{\pi}{2}$。 $L_1$ 和 $L_2$ 是曲线 $C_1$ 的切线。

1. Find the Cartesian equation by eliminating the parameter $\theta$.

$x = 3\sec^2 \theta \implies \sec^2 \theta = \frac{x}{3}$
$y = 2\tan \theta \implies \tan \theta = \frac{y}{2}$

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

2. Determine the domain and range of the curve based on the given interval for $\theta$.

Given $0 \le \theta < \frac{\pi}{2}$:
When $\theta = 0$, $x = 3\sec^2(0) = 3$ and $y = 2\tan(0) = 0$.
As $\theta \to \frac{\pi}{2}^-$, $\sec \theta \to \infty$ and $\tan \theta \to \infty$, which implies $x \to \infty$ and $y \to \infty$.
Thus, the curve exists for $x \ge 3$ and $y \ge 0$.

3. State the coordinates of the endpoint and sketch the curve.

The equation $y^2 = \frac{4}{3}(x - 3)$ represents a parabola opening to the right with its vertex at $(3, 0)$.
Since $y \ge 0$, the curve $C_1$ is only the top half of this parabola.
The endpoint corresponds to $\theta = 0$, which is $(3, 0)$.

1. Find the derivatives of $x$ and $y$ with respect to $\theta$.

Given $x = 3\sec^2 \theta$,
$\frac{dx}{d\theta} = 3(2\sec \theta)(\sec \theta \tan \theta) = 6\sec^2 \theta \tan \theta$

Given $y = 2\tan \theta$,
$\frac{dy}{d\theta} = 2\sec^2 \theta$

2. Apply the chain rule for parametric equations to find $\frac{dy}{dx}$ and simplify.

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
$\frac{dy}{dx} = \frac{2\sec^2 \theta}{6\sec^2 \theta \tan \theta}$
$\frac{dy}{dx} = \frac{1}{3\tan \theta}$

1. Deduce the parameter value $\theta$ for a tangent parallel to the $y$-axis

Since $L_1$ is parallel to the $y$-axis, it is a vertical line and its gradient $\frac{dy}{dx}$ is undefined. Therefore, the denominator of $\frac{dy}{dx}$ must be zero.
$3\tan \theta = 0$
$\tan \theta = 0$
Given $0 \le \theta < \frac{\pi}{2}$, we have:
$\theta = 0$

2. Calculate the corresponding $x$-coordinate and state the equation of the line

Substitute $\theta = 0$ into the parametric equation for $x$:
$x = 3\sec^2(0)$
$x = 3(1)^2 = 3$
Since the tangent line is vertical, its equation is of the form $x = c$. Thus, the equation of $L_1$ is:
$x = 3$

1. Use the $y$-coordinate of the given point to find the value of $\tan \theta$.

$2\tan \theta = 1$
$\tan \theta = \frac{1}{2}$

2. Substitute the value of $\tan \theta$ into the expression for $\frac{dy}{dx}$ from part (b) to find the gradient of the tangent.

$\text{Gradient} = \frac{dy}{dx} = \frac{1}{3\tan \theta}$
$\text{Gradient} = \frac{1}{3\left(\frac{1}{2}\right)}$
$\text{Gradient} = \frac{2}{3}$

1. Determine $\tan \beta$ using the gradient of $L_2$ and the orientation of $L_1$.

From part (c)(i), $L_1$ is parallel to the $y$-axis (a vertical line).
From part (c)(ii), the tangent $L_2$ at $(3.75, 1)$ corresponds to $y = 2\tan\theta = 1 \implies \tan\theta = 0.5$. Its gradient is $m = \frac{1}{3\tan\theta} = \frac{1}{1.5} = \frac{2}{3}$.
Let the acute angle that $L_2$ makes with the horizontal axis be $\alpha$. Then $\tan \alpha = \frac{2}{3}$.
Since $L_1$ is vertical, the acute angle $\beta$ between $L_1$ and $L_2$ is $\frac{\pi}{2} - \alpha$.
$\tan \beta = \tan \left( \frac{\pi}{2} - \alpha \right) = \frac{1}{\tan \alpha} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

1. Simplify the parametric equations of $C_2$ to compare with $C_1$

For $C_2$, $y = 2\tan(\pi - \theta) = -2\tan \theta$.
The parametric equations for $C_2$ are therefore $x = 3\sec^2 \theta$, $y = -2\tan \theta$.

2. Identify the geometric transformation from the coordinate mapping

Comparing $C_1$ and $C_2$, for the same parameter $\theta$, the $x$-coordinates are identical while the $y$-coordinates are negatives of each other.
This maps any point $(x, y)$ on $C_1$ to $(x, -y)$ on $C_2$.
Therefore, the transformation is a reflection in the $x$-axis.