Don't Fall for the Parametric Trap: Hidden Restrictions

Problem Statement

A curve is defined by the parametric equations:

$$x = \frac{t-2}{t+3}, \quad y = \frac{t+2}{t-3}$$

where $t \in \mathbb{R}$ and $t \neq -3, 3$.

Find the Cartesian equation of the curve and state its complete domain and range.

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Hints

• Step 1: Isolate $t$ in the equation for $x$. Be mindful of algebraic steps that introduce hidden restrictions (e.g., dividing by an expression containing $x$).
• Step 2: Substitute the resulting expression for $t$ into the equation for $y$ and simplify to find the Cartesian equation.
• Step 3: The domain and range cannot be determined solely by looking at the final Cartesian equation's asymptotes. You must evaluate the behavior of the original parametric functions as $t \to \pm \infty$ to identify "missing" coordinate pairs.

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Solution

First, express $t$ in terms of $x$:

$$x = \frac{t-2}{t+3}$$ $$x(t+3) = t-2$$ $$xt + 3x = t-2$$ $$xt - t = -3x - 2$$ $$t(x-1) = -(3x+2)$$ $$t = \frac{3x+2}{1-x}$$

Note: For this expression to be defined, $x \neq 1$.

Next, substitute this expression for $t$ into the parametric equation for $y$:

$$y = \frac{t+2}{t-3}$$ $$y = \frac{\frac{3x+2}{1-x} + 2}{\frac{3x+2}{1-x} - 3}$$

Multiply the numerator and the denominator by $(1-x)$ to simplify:

$$y = \frac{(3x+2) + 2(1-x)}{(3x+2) - 3(1-x)}$$ $$y = \frac{3x + 2 + 2 - 2x}{3x + 2 - 3 + 3x}$$ $$y = \frac{x+4}{6x-1}$$

Determining the Domain:

From the denominator of the Cartesian equation, we have a vertical asymptote where $6x - 1 = 0$, meaning $x \neq \frac{1}{6}$.

Additionally, we found earlier that $x \neq 1$ because there is no finite real value of $t$ that maps to $x = 1$. (As $t \to \pm \infty$, $x \to 1$).

Domain: $x \in \mathbb{R}, x \neq \frac{1}{6}, x \neq 1$

Determining the Range:

From the Cartesian equation $y = \frac{x+4}{6x-1}$, the horizontal asymptote dictates $y \neq \frac{1}{6}$.

We must also find the $y$-value corresponding to the missing domain point $x = 1$. Substituting $x = 1$ into our Cartesian equation:

$$y = \frac{1+4}{6(1)-1} = \frac{5}{5} = 1$$

Because $x$ can never equal $1$, $y$ can never equal $1$.

Range: $y \in \mathbb{R}, y \neq \frac{1}{6}, y \neq 1$

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Takeaways

When eliminating a parameter to form a Cartesian equation, algebraic manipulation routinely masks essential domain constraints. The final equation $y = \frac{x+4}{6x-1}$ is merely a continuous rational curve, but the original parametric system only generates a subset of that curve.

Tracking limits is just as important as finding algebraic asymptotes. While the condition $t \neq 3$ maps neatly to the Cartesian vertical asymptote $x \neq \frac{1}{6}$, the behavior of the system as $t \to \pm \infty$ creates a "hole" at $(1, 1)$. Because infinity is a limit rather than an evaluable number, the parametric equations will approach but never actually reach the coordinate $(1, 1)$. Evaluating asymptotes strictly on the final Cartesian expression will consistently fail to identify these unmapped terminal points.