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 $C_1$.

$x = \frac{2}{t-1} \implies t - 1 = \frac{2}{x} \implies t = \frac{x+2}{x}$
Substituting this expression for $t$ into the equation for $y$:
$y = \frac{4}{t+1} = \frac{4}{\frac{x+2}{x} + 1} = \frac{4}{\frac{2x+2}{x}}$
$y = \frac{2x}{x+1} = 2 - \frac{2}{x+1}$

2. Identify asymptotes and hidden domain restrictions.

From $y = 2 - \frac{2}{x+1}$, the curve is a hyperbola with a vertical asymptote at $x = -1$ and a horizontal asymptote at $y = 2$.
Since $t \in \mathbb{R}$, $t-1$ is finite, so $x = \frac{2}{t-1} \neq 0$.
As $t \to \pm \infty$, $x \to 0$ and $y \to 0$. Therefore, the curve approaches the origin but does not touch it, creating a hole at $(0,0)$.

3. Describe the features to be sketched on the diagram.

Sketch the two branches of the hyperbola for $x < -1$ and $x > -1$.
Draw and label the dotted lines for the asymptotes $x = -1$ and $y = 2$.
Mark an open circle at the origin $(0,0)$ to explicitly show the point of discontinuity (the curve has no $x$- or $y$-intercepts).

1. Relate the parametric equations of $C_2$ to those of $C_1$.

Let the coordinates of $C_1$ be $(x_1, y_1)$ and the coordinates of $C_2$ be $(x_2, y_2)$.
$x_2 = \frac{1}{1-t} = -\frac{1}{2} \left(\frac{2}{t-1}\right) = -\frac{1}{2} x_1$
$y_2 = \frac{4}{t+1} = y_1$

2. Identify the geometrical transformations corresponding to the coordinate mapping.

The coordinate mapping $(x, y) \mapsto (-\frac{1}{2}x, y)$ is equivalent to replacing $x$ with $-2x$ in the Cartesian equation.
This corresponds to:

1. A scaling (or stretch) parallel to the $x$-axis by scale factor $\frac{1}{2}$.
2. A reflection in the $y$-axis.

(Note: As the transformations act on the same axis linearly, they can be applied in either order.)

1. Find the Cartesian equation of $C_3$ (the reciprocal function of $C_1$).

From the parametric equations of $C_1$, we have $t - 1 = \frac{2}{x} \implies t = \frac{2+x}{x}$.
Substituting into $y$:
$y = \frac{4}{\left(\frac{2+x}{x}\right) + 1} = \frac{4x}{2+2x} = \frac{2x}{x+1}$
The reciprocal function $C_3$ is given by $y = \frac{1}{f(x)}$:
$y = \frac{x+1}{2x} = \frac{1}{2} + \frac{1}{2x}$

2. Determine the asymptotes and points of intersection with the axes.

For the curve $y = \frac{1}{2} + \frac{1}{2x}$:

• As $x \to \pm\infty$, $y \to \frac{1}{2}$. Thus, the horizontal asymptote is $y = \frac{1}{2}$.
• As $x \to 0$, $y \to \pm\infty$. Thus, the vertical asymptote is $x = 0$ (the $y$-axis).

To find the intersection with the axes, set $y = 0$:
$0 = \frac{x+1}{2x} \implies x + 1 = 0 \implies x = -1$
The point of intersection with the $x$-axis is $(-1, 0)$. Since $x=0$ is a vertical asymptote, there is no $y$-intercept.