To eliminate the parameter $t$ from the given parametric equations:

$x = e^{-7t}, \quad y = e^{7t}$

we can start by solving each equation for $t$ and then equating them.

Step-by-Step Solution:

1. Solve for $t$ from each equation:

   • For $x = e^{-7t}$: $\ln(x) = -7t \Rightarrow t = -\frac{\ln(x)}{7}$

   • For $y = e^{7t}$: $\ln(y) = 7t \Rightarrow t = \frac{\ln(y)}{7}$

2. Equate the expressions for $t$:

   Since both expressions represent $t$, we can set them equal to each other: $-\frac{\ln(x)}{7} = \frac{\ln(y)}{7}$

3. Simplify the equation:

   Multiply both sides by 7 to eliminate the denominator: $-\ln(x) = \ln(y)$

4. Rewrite in terms of $x$ and $y$:

   This simplifies to: $\ln(y) = -\ln(x)$

5. Combine logarithmic terms:

   Using the property $\ln(a) = -\ln(b) \Rightarrow a = \frac{1}{b}$, we get: $y = \frac{1}{x}$

Final Answer:

The Cartesian equation of the curve is: $y = \frac{1}{x}$

Would you like further details on this solution or have additional questions?

Here are 5 related questions for further exploration:

1. How would you graph the curve $y = \frac{1}{x}$ in the Cartesian plane?
2. What happens to the curve as $x$ approaches zero?
3. How would this solution change if the powers in $e^{-7t}$ and $e^{7t}$ were different?
4. What is the domain and range of the curve $y = \frac{1}{x}$?
5. How could we verify this solution using another method?

Tip: When dealing with exponential equations, logarithms are often helpful for isolating variables or eliminating parameters.