Question

Eliminate the parameter and sketch the graphs. $$ x=2 t^{2}, \quad y=t^{4}+1 $$

Short answer

The curve is a parabola: \( y = \frac{x^2}{4} + 1 \). Vertex: (0, 1).

Step-by-step solution

Express t in terms of x

Given the equation for x: \( x = 2t^2 \). Solve for \( t^2 \) by dividing both sides by 2: \( t^2 = \frac{x}{2} \).

Express y in terms of x

Now, express \( y \) in terms of \( x \). From Step 1, \( t^2 = \frac{x}{2} \). Hence, \( t^4 = \left( \frac{x}{2} \right)^2 = \frac{x^2}{4} \). Substitute this into the equation for \( y \): \( y = t^4 + 1 = \frac{x^2}{4} + 1 \).

Final Parametric Equation

The parametric equations \( x = 2t^2 \) and \( y = t^4 + 1 \) are now expressed as a single non-parametric equation: \( y = \frac{x^2}{4} + 1 \).

Sketch the Graph

The equation \( y = \frac{x^2}{4} + 1 \) is a parabola that opens upwards. The vertex of this parabola is at \( (0, 1) \). It is symmetrical about the y-axis since it is not dependent on the sign of \( t \). Sketch this parabola on a coordinate plane.

Key concepts

graph sketching

When you are tasked with graph sketching, it helps to break the process down into manageable steps. At its core, graph sketching involves translating equations into visual representations on a coordinate plane. Here, the main objective is to take the final non-parametric equation and use it to outline the graph's shape and position.

For the equation derived from the parametric ones, such as \( y = \frac{x^2}{4} + 1 \), we need to recognize it as a parabola. Observing the equation, notice that it is in the standard form of a parabola \( y = ax^2 + c \). This indicates it will have a U-shape and opens upwards due to the positive coefficient of \( x^2 \).

• Determine key features: This includes identifying the vertex, axis of symmetry, and direction. In this case, the vertex is at \( (0, 1) \) and the parabola is symmetrical about the y-axis.
• Plot points: Select a few values of \( x \) to find corresponding \( y \), confirming curve shape.
• Draw the curve: With the vertex and additional points, sketch the smooth U-shaped curve.

With these steps, the graph should effectively represent the equation. Always check symmetry, and ensure that fundamental features like vertex position and direction are clearly shown.

eliminating the parameter

Eliminating the parameter is a key process in converting parametric equations into a single equation relating \( x \) and \( y \). The goal is to remove the parameter, often represented as \( t \), to simplify analysis and graphing.

Steps involved in eliminating a parameter typically include:

• Solving one parametric equation for the parameter \( t \). As seen with \( x = 2t^2 \), it is rearranged to find \( t^2 = \frac{x}{2} \).
• Substituting into the other equation. Use \( t^2 = \frac{x}{2} \) in \( y = t^4 + 1 \) to express \( y \) directly in terms of \( x \), leading to \( y = \frac{x^2}{4} + 1 \).

This approach transforms two parametric expressions into a single, comprehensible equation. It simplifies the task of graph sketching and helps in understanding the relationship between the variables \( x \) and \( y \). Elimination of the parameter unveils the geometric nature of the curve more directly.

parabola

Parabolas are a familiar type of conic section, recognized for their unique open curve shape. The equation \( y = \frac{x^2}{4} + 1 \) forms a parabola, and it's advantageous to understand its characteristics:

Firstly, parabolas' shapes are defined by their coefficients and the presence of \( x^2 \). A positive coefficient ensures the curve opens upwards. In this exercise, \( \frac{1}{4} \) is positive, resulting in upward opening.

Secondly, the vertex, \( (0, 1) \), is critical for understanding the entire structure of the parabola, being its lowest point when opening upwards. The vertex provides the point of symmetry and is found by examining constants in the equation.

Further properties include the axis of symmetry, the imaginary line through the vertex that splits the parabola into mirror images. This axis is the y-axis here, since \( x \) terms have no horizontal displacement. Understanding

• shape,
• direction,
• vertex,
• axis of symmetry,

of a parabola elucidates how it behaves and can guide accurate graph sketching, ensuring clear visualization of given equations.