Problem 38 - Precalculus Mathematics for Calculus

Question

Use a graphing device to graph the conic. $$x^{2}-4 y^{2}+4 x+8 y=0$$

Step-by-Step Solution

Answer

The graph represents a hyperbola centered at (-2, 1), opening left and right along the x-axis.

Step 1: Rearrange the Equation

Start by grouping like terms in the equation.Given:\[ x^2 - 4y^2 + 4x + 8y = 0 \]Rearrange it as:\[ (x^2 + 4x) - (4y^2 - 8y) = 0 \]

Step 2: Complete the Square

Complete the square for both \(x\) and \(y\) terms separately.1. For \(x\), complete the square: - Take \(x^2 + 4x\), and add \(4\) to complete the square: \[ x^2 + 4x = (x+2)^2 - 4 \]2. For \(y\), complete the square: - Factor out a \(-4\) from \(-4y^2 + 8y\): \[ -4(y^2 - 2y) \] - Complete the square inside the parenthesis: \[ y^2 - 2y = (y-1)^2 - 1 \] - Factor back in the \(-4\): \[ -4((y-1)^2 - 1) = -4(y-1)^2 + 4 \]

Step 3: Substitute Completed Squares and Simplify

Substitute the completed squares back into the rearranged equation:\[ (x+2)^2 - 4 - 4((y-1)^2 - 1) = 0 \]Simplify:\[ (x+2)^2 - 4(y-1)^2 + 4 - 4 = 0 \]This simplifies to:\[ (x+2)^2 - 4(y-1)^2 = 0 \]

Step 4: Recognize the Conic Type

The equation \[ (x+2)^2 = 4(y-1)^2 \] can be rewritten as:\[ \frac{(x+2)^2}{(\sqrt{4})^2} - \frac{(y-1)^2}{(1)^2} = 0 \]It is now clear this is a hyperbola with the center at \((-2, 1)\), which opens left and right.

Step 5: Graph the Hyperbola using a Device

Using a graphing calculator or software, plot the hyperbola with the transverse axis aligned horizontally.- Center at \((-2, 1)\)- Vertices distance horizontally along the \(x\)-axis determined by solving \(x\) part, namely \(b=2\).- Since \(a^2 = 4\), real axis scale is adjusted accordingly, with \(a = 2\). Plot the branches of the hyperbola, ensuring they open along the \(x\)-axis with smooth curves extending both to left and right from the center.

Key Concepts

Hyperbola

A hyperbola is one type of conic section, which is formed when a plane intersects both nappes of a double cone. Unlike other conic sections, such as circles and ellipses, hyperbolas have two symmetrical parts called branches. These branches reflect across the hyperbola's center.

One way to identify a hyperbola is by its standard equation, which usually takes the format:

• \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \)

The orientation of a hyperbola can vary:

• If the \( x^2 \) term is positive, the hyperbola opens left and right.
• If the \( y^2 \) term is positive, it opens up and down.

When graphing hyperbolas, it's important to recognize these formulas, as they govern the shape and opening direction of the graph. Hyperbolas are defined by two key points: the vertices and foci, which play a crucial role in shaping its graph.

Completing the Square

Completing the square is a useful algebraic technique to simplify expressions, particularly quadratic expressions. This method allows you to reformulate a quadratic equation into a perfect square trinomial, which is easier to work with for graphing and further calculations.To complete the square for an expression like \(x^2 + bx\), follow these steps:

• Take half of the \(b\) coefficient (the number before \(x\)).
• Square that value.
• Add and subtract this squared value inside the quadratic expression.

This process converts related terms into a binomial square: \(x^2 + bx = (x + \/product{b}{2})^2 - {centref{b^2}{4}}\).When applied to conic sections, completing the square helps transform complicated expressions into easier-to-understand standard forms, especially for hyperbolas, by aligning them with a recognizable pattern.

Graphing Hyperbolas

Graphing hyperbolas involves several key steps to ensure accuracy and understanding. After rewriting the hyperbola equation into its standard form through completing the square, you can begin to visualize its structure.First, identify the:

• Center: The point \((h, k)\) serves as the middle of the hyperbola.
• Axes: The transverse axis is the main axis along which the hyperbola opens, either horizontally or vertically.
• Vertices: Located along the transverse axis and are \(a\) units away from the center.
• Asymptotes: These are the diagonal lines that the branches of the hyperbola approach but never intersect.

Using a graphing device simplifies plotting as it can handle the precise posting locations of these elements.Set the graph's scale based on values of \(a\) and \(b\), which dictate the axis lengths and hyperbola's shape. Graphing software or calculators can plot these efficiently and display a clear depiction of the hyperbola.

Center of Hyperbola

The center of a hyperbola is a pivotal point from which its axes and symmetry are defined. In its standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center is represented by the coordinates \((h, k)\).

The center is always situated midway between the hyperbola's vertices and serves as a balancing point for its branches and asymptotes.

Found by transforming and simplifying the initial equation using methods like completing the square, the center determines the hyperbola's position in the coordinate plane. In our example, the center was calculated to be at \((-2, 1)\) after completing the square and aligning the equation with its standard form.

Recognizing the center helps in evaluating other characteristics of the hyperbola, such as its orientation, direction, and dimensions, crucial for accurate graphing and interpretation.