Problem 5 - Algebra and Trigonometry with Analytic Geometry

Question

\(x=4 t^{2}-5, \quad y=2 t+3 ; \quad t\) in \(\mathbb{R}\)

Step-by-Step Solution

Answer

The relation between \(x\) and \(y\) is \(x = y^2 - 6y + 4\).

Step 1: Understanding the Functions

The equations provide parametric equations for a curve. These equations express both \(x\) and \(y\) in terms of a third variable, \(t\). Here: \[ x = 4t^2 - 5 \]and \[ y = 2t + 3 \]Our task is to find the relation between \(x\) and \(y\) by eliminating the parameter \(t\).

Step 2: Expressing t from y

To eliminate \(t\), we need to express it in terms of one of the given equations. Let's solve for \(t\) using the equation for \(y\):\[ y = 2t + 3 \]Subtract 3 from both sides:\[ y - 3 = 2t \]Divide by 2:\[ t = \frac{y - 3}{2} \]

Step 3: Substituting t in x's equation

Now that \(t\) is expressed in terms of \(y\), substitute \(t = \frac{y - 3}{2}\) into the \(x\) equation:\[ x = 4t^2 - 5 \]Replace \(t\):\[ x = 4\left(\frac{y - 3}{2}\right)^2 - 5 \]

Step 4: Simplifying the expression

We simplify the expression obtained:\[ x = 4\left(\frac{y - 3}{2}\right)^2 - 5 \]First, expand the squared term:\[ x = 4\left(\frac{(y - 3)^2}{4}\right) - 5 \]Simplify:\[ x = (y - 3)^2 - 5 \]

Step 5: Expanding and Rearranging

Expand the square and remove the parentheses:\[ x = (y - 3)^2 - 5 \]This becomes:\[ x = y^2 - 6y + 9 - 5 \]Simplify:\[ x = y^2 - 6y + 4 \]

Key Concepts

Eliminating the Parameter

When dealing with parametric equations, we often want to find the relationship between the dependent variables without the parameter. In this case, the parameter is the variable, \( t \). By eliminating \( t \), we can derive a Cartesian equation, a relation solely between \( x \) and \( y \).We start with the equation \( y = 2t + 3 \). To eliminate \( t \), rearrange for \( t \) by subtracting 3 from \( y \) and dividing by 2:

• \( y - 3 = 2t \)
• \( t = \frac{y - 3}{2} \)

The next step is to substitute this expression for \( t \) into the \( x \) equation \( x = 4t^2 - 5 \). By replacing \( t \), we start bridging our parametric form to the Cartesian form. Keep in mind, the goal is always to simplify until you see the elimination of the parameter. This paves the way for directly relating \( x \) and \( y \).

Quadratic Relations

After eliminating the parameter from the parametric equations, one might notice the appearance of a quadratic relation. In our exercise, the expression becomes:\[ x = (y - 3)^2 - 5 \]Quadratic relations show curves that are parabolas. Here, you simplify further to illustrate:

• Expand to get: \( x = (y - 3)^2 - 5 \)
• Extend it as \( x = y^2 - 6y + 9 - 5 \)
• It simplifies to \( x = y^2 - 6y + 4 \)

This relation is a quadratic equation in terms of \( y \), signifying that the initial parametric equations form a parabola in the Cartesian plane. Understanding the properties of a parabolic curve is vital, as these curves are common in physics practical scenarios, such as projectile motion.

Parametric to Cartesian Conversion

Converting parametric equations to Cartesian equations is essential for better visualization of graphs and understanding their properties directly. This conversion makes analysing the geometric shape more straightforward since it eliminates the extra parameter that can complicate calculations.For this exercise, the goal was to derive the equation \( x = y^2 - 6y + 4 \) from the original parametric pair:- \( x = 4t^2 - 5 \)- \( y = 2t + 3 \)Using the derived Cartesian equation, one can now easily plot or analyse the parabolic curve without dealing with the extraneous parameter \( t \), making it much simpler to investigate key properties such as:

• Vertex of the parabola
• Axis of symmetry
• Directional trends

These insights are not as easily extracted in parametric form. By converting, it simplifies investigation and facilitates comparisons with other Cartesian formulas.