A curve $C_1$ is defined parametrically by $x = \frac{2}{t-1}\quad\text{and}\quad y = \frac{4}{t+1},\qquad t \ne \pm 1.$

1. Eliminate the parameter $t$ to find the Cartesian equation of $C_1$.

From $x = \frac{2}{t-1}$, we make $t$ the subject:
$x(t-1) = 2$
$t-1 = \frac{2}{x}$
$t = \frac{2}{x} + 1 = \frac{x+2}{x}$
Substitute this expression for $t$ into the equation for $y$:
$y = \frac{4}{t+1}$
$y = \frac{4}{\frac{x+2}{x} + 1}$
$y = \frac{4}{\frac{2x+2}{x}}$
$y = \frac{4x}{2x+2} = \frac{2x}{x+1}$

2. Identify the asymptotes of the curve.

To clearly see the asymptotes, perform algebraic division or rewrite the Cartesian expression:
$y = \frac{2(x+1) - 2}{x+1} = 2 - \frac{2}{x+1}$
From this standard form, the vertical asymptote is $x = -1$.
The horizontal asymptote is $y = 2$.

3. Determine any excluded points due to the nature of the parameter $t$ and sketch the graph.

Since the parameter $t$ is finite (i.e. $t \in \mathbb{R} \setminus \{-1, 1\}$), we check the bounds.
As $t \to \pm\infty$, $x = \frac{2}{t-1} \to 0$. However, there is no finite real value of $t$ that yields $x = 0$.
When $x = 0$, $y = \frac{2(0)}{0+1} = 0$.
Therefore, the point $(0,0)$ is not part of the curve and must be represented as a hole (hollow circle).

Sketching features:

• Draw branches of the rectangular hyperbola in the upper-left quadrant and lower-right quadrant relative to the asymptotes $(-1, 2)$.
• Include a hollow circle at the origin $(0,0)$.

1. Compare the parametric equations of $C_1$ and $C_2$ to find the relationship between their coordinates for the same parameter $t$.

Let the coordinates of a point on $C_1$ be $(x, y)$ and the corresponding point on $C_2$ be $(x', y')$.
Comparing the $x$-coordinates:
$x' = \frac{1}{1-t} = -\frac{1}{2}\left(\frac{2}{t-1}\right) = -\frac{1}{2}x$
Comparing the $y$-coordinates:
$y' = \frac{4}{t+1} = y$
Therefore, the point $(x, y)$ on $C_1$ maps to the point $\left(-\frac{1}{2}x, y\right)$ on $C_2$.

2. Interpret the coordinate mapping $(x, y) \mapsto (-\frac{1}{2}x, y)$ as a sequence of geometrical transformations.

The overall transformation $x' = -\frac{1}{2}x$ and $y' = y$ (equivalent to replacing $x$ with $-2x$ in the Cartesian equation $y = \mathrm{f}(x)$ to get $y = \mathrm{f}(-2x)$) can be broken down into two distinct operations on the $x$-coordinate:

1. $x \mapsto \frac{1}{2}x$, which is a scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$.
2. $x \mapsto -x$, which is a reflection in the $y$-axis.

Since both transformations affect only the $x$-axis with constant factors, their order of application does not matter.

1. Find the Cartesian equation of $C_3$

Since $C_3$ is the reciprocal function of $C_1$, we take the reciprocal of the $y$-coordinates while $x$ remains the same. The parametric equations for $C_3$ are $x = \frac{2}{t-1}$ and $y = \frac{t+1}{4}$.
Eliminating $t$ by rearranging the $x$ equation gives $t = \frac{2}{x} + 1$.
Substituting this into $y$:
$y = \frac{\left(\frac{2}{x} + 1\right) + 1}{4} = \frac{\frac{2}{x} + 2}{4} = \frac{1}{2x} + \frac{1}{2}$

2. Determine the equations of the asymptotes

For the curve $y = \frac{1}{2x} + \frac{1}{2}$:
As $x \to 0$, $y \to \pm \infty$, which gives a vertical asymptote at $x = 0$ (the $y$-axis).
As $x \to \pm \infty$, $y \to \frac{1}{2}$, which gives a horizontal asymptote at $y = \frac{1}{2}$.

3. Determine any points of intersection with the axes, accounting for parametric restrictions

For the $y$-intercept, $x = 0$ is an asymptote, so the curve does not cross the $y$-axis.
For the $x$-intercept, set $y = 0$:
$\frac{1}{2x} + \frac{1}{2} = 0 \implies \frac{1}{2x} = -\frac{1}{2} \implies x = -1$
However, substituting $x = -1$ back into the parameter gives $t = \frac{2}{-1} + 1 = -1$. The question explicitly states $t \ne \pm 1$, so this point is excluded from the domain.
Consequently, there is a "hole" at $(-1, 0)$ instead of an intercept, meaning $C_3$ has no points of intersection with the axes.

4. State the sketching requirements for $C_3$

Sketch the standard rectangular hyperbola shifted upwards by $\frac{1}{2}$.

• Draw the vertical asymptote $x=0$ and the horizontal asymptote $y=\frac{1}{2}$.
• Draw the two branches of the curve in the top-right and bottom-left regions defined by the asymptotes.
• Explicitly draw an empty circle (a hole) at $(-1, 0)$ to denote the excluded point.