91Ó°ÊÓ

Parameter elimination involves removing the parameter \( t \) from our parametric equations to express the relationship directly between \( x \) and \( y \). For the given equations, \( x = \frac{1}{9}(18 - t^2) \) and \( y = \frac{1}{3}t \), our goal is to connect \( x \) and \( y \) without involving \( t \). To do this, we first solve the equation for \( t \): - From \( y = \frac{1}{3}t \), we find \( t = 3y \).We then substitute this expression into the equation for \( x \): - Replace \( t \) in \( x = \frac{1}{9}(18 - t^2) \) with \( 3y \), giving us \( x = \frac{1}{9}(18 - (3y)^2) \).Finally, simplify this expression: - This simplifies to \( x = 2 - y^2 \), and we end up with a relationship that shows \( x \) as a function of \( y \) and not \( t \). This is useful because it reveals the underlying geometric shape represented by our parametric equations—a parabola.

With the equation \( x = 2 - y^2 \), we can now plot the parabola in a way that makes sense conceptually. This equation is a standard form that depicts a parabola opening horizontally to the left. Let's break down how to properly plot it:

• The vertex is at \((2, 0)\), which is noteworthy because it's derived from the constant term 2 when \( y = 0 \).
• As \( y \) increases or decreases, \( x \) decreases because \( y^2 \) is being subtracted. The symmetry of \( y^2 \) means as \( y \) moves away from zero, both in the positive and negative directions, it creates a mirrored effect on both parts of the parabola.
• Critical points to calculate and plot include \((1, -1)\) and \((1, 1)\), derived from substituting values of \( t \), \( -3 \, \text{and} \, 3 \), into the parametric equations.

An important note: remember the path plotted represents those values of \( y \) where the equation holds true given the constraints of \( t \). In this case, as \( t \geq -3 \), \( y = \frac{1}{3}t \) implies \( y \) stretches from \(-\infty\) to \(\infty\). This determines the length and position of the parabola plotted on a grid.

Curve orientation refers to the direction in which the curve is traversed as the parameter \( t \) increases. In this exercise, it's integral to determine not just the shape of the graph but also the direction it moves.With the parametric equations, as \( t \) increases:

• The starting point at \( t = -3 \) is \( (1, -1) \), after calculations of substituting \( t \) values.
• Moving towards \( t = 0 \) leads to \( (2, 0) \), as \( x \) is at its maximum (vertex of the parabola).
• Increasing \( t \) further to \( t = 3 \) reaches \( (1, 1) \).

This shows the curve is traversed from the rightmost point towards the left as \( t \) increases. This orientation, leftward, is important for understanding the dynamic nature of parametric graphs.Identifying this orientation helps clarify how particular points are linked in the graph, further aiding comprehension of the parametric equations at a deeper level.