Question

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=e^{-t}, \quad y=e^{2 t}-1 $$

Short answer

The graph is a rectangular hyperbola opening towards the right and upwards. The rectangular form of the given parametric equations is \(x^2y = x^2 - 1\).

Step-by-step solution

Plotting Key Points

The first step involves identifying and plotting key points from the equations. The value of \(x\) diminishes as \(t\) increases from \(-\infty\) to \(+\infty\). The value of \(y\), however, increases from -1 to +\(\infty\) as \(t\) increases. When \(t = 0\), \(x = 1\) and \(y = 0\). These values mark an important point on the curve (1,0).

Sketching the Curve

The curve is sketched connecting these points and considering the behaviour discussed in the step above. The curve starts from \(+\infty\) on y-axis and goes to the point (1,0), then continues towards \(+\infty\) as \(t\) increases. The general shape of the curve is a rectangular hyperbola.

Converting to Rectangular Equation

To convert, we use our first equation to solve for \(t\) in terms of \(x\) and replace it in the equation for \(y\). Taking the natural log on both sides of the first equation, we get \(t = -\ln{x}\). Substituting this in the second equation, it becomes \(y = e^{2(-\ln{x})}-1 = x^{-2} - 1\)

Simplifying the Rectangular Equation

The last equation can be simplified to get the rectangular equation of the curve. Thus, the rectangular equation becomes \(x^2y = x^2 - 1\)