Problem 3 - Algebra and Trigonometry with Analytic Geometry

Question

\(x=t^{2}+1, \quad y=t^{2}-1 ; \quad-2 \leq t \leq 2\)

Step-by-Step Solution

Answer

The relationship is \( y = x - 2 \).

Step 1: Understand the Given Equations

We are given two parametric equations: 1. \( x = t^2 + 1 \) 2. \( y = t^2 - 1 \) The variable \( t \) ranges from -2 to 2. Our goal is to find the relationship between \( x \) and \( y \) independent of \( t \).

Step 2: Eliminate the Parameter

Notice that both equations involve \( t^2 \). Solve for \( t^2 \) in each equation:From \( x = t^2 + 1 \), it follows that \( t^2 = x - 1 \).From \( y = t^2 - 1 \), it follows that \( t^2 = y + 1 \).Since both expressions equal \( t^2 \), set them equal to each other:\[ x - 1 = y + 1 \].

Step 3: Simplify the Equation

Solve \( x - 1 = y + 1 \) for \( y \) in terms of \( x \):\[ x - 1 = y + 1 \]\[ x - 1 - 1 = y \]\[ y = x - 2 \]This equation represents the relationship between \( x \) and \( y \).

Step 4: Confirm the Validity of the Range

Verify if the equation \( y = x - 2 \) holds for the range of \( t \) given, which is \(-2 \leq t \leq 2\).Calculate the minimum and maximum values for \( x \) and \( y \):- For the maximum value \( t = 2 \), \( x = 2^2 + 1 = 5\) and \( y = 2^2 - 1 = 3 \).- For the minimum value \( t = -2 \), \( x = (-2)^2 + 1 = 5 \) and \( y = (-2)^2 - 1 = 3 \).Therefore, \( y = x - 2 \) holds true throughout the range.

Key Concepts

Elimination of Parameter

In parametric equations, each variable is expressed in terms of a third variable, called a parameter. One of the most important techniques is eliminating this parameter to find a direct relationship between the variables involved.
In the given problem, we start with the parametric equations:

• \( x = t^2 + 1 \)
• \( y = t^2 - 1 \)

Both equations contain the term \( t^2 \). The goal of eliminating the parameter is to express \(x\) and \(y\) directly without involving \(t\). By rearranging each equation to isolate \(t^2\), we find:

• \( t^2 = x - 1 \)
• \( t^2 = y + 1 \)

Since both are equal to \(t^2\), they can be set equal to each other: \[ x - 1 = y + 1 \]This equation provides a direct relationship between \(x\) and \(y\) without \(t\). It simplifies further to \( y = x - 2 \). This way, we eliminate the parameter effectively.

Relationship Between Variables

After eliminating the parameter \( t \), we discover the relational equation \( y = x - 2 \). This equation forms a straight line when plotted on a graph, showcasing a simple linear relationship between \( x \) and \( y \).

• For every unit increase or decrease in \( x \), \( y \) will change by the same amount, specifically \( y \) decreases by 2 when compared directly to \( x \).
• This linear relationship indicates that as \(x\) increases or decreases, \(y\) will always be two units less.

Understanding this simple relationship allows for easier calculations and predictions when interpreting these equations, as you know exactly how the two variables affect one another.

Range of Parameter

The range of the parameter refers to the limits within which the parameter \(t\) is defined. This is crucial because it affects the possible values of \(x\) and \(y\). In the given problem, \(t\) ranges from -2 to 2. Calculating the extreme values of \(x\) and \(y\) within this range ensures that the equation \(y = x - 2\) holds true.

• When \( t = 2 \), \( x = t^2 + 1 = 5 \) and \( y = t^2 - 1 = 3 \).
• Similarly, when \( t = -2 \), the calculations for \( x \) and \( y\) yield the same results: 5 and 3 respectively.

This shows that irrespective of whether \( t \) is at its maximum or minimum, the relational equation \( y = x - 2 \) remains valid. Thus, the range of the parameter confirms the consistency and applicability of the solution across all valid values of \(t\).