Problem 34 - Algebra and Trigonometry

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 conic is a hyperbola centered at (-2, 1). Graph it using a graphing device.

Step 1: Rearrange the Equation

Begin by rearranging the given equation to separate terms associated with the same variable:\[x^2 + 4x - 4y^2 + 8y = 0\]

Step 2: Complete the Square for x Terms

For the terms involving x, complete the square:1. Rewrite: \(x^2 + 4x\).2. Half the coefficient of x: \(\frac{4}{2} = 2\).3. Square it: \(2^2 = 4\).4. Add and subtract this inside the equation: \(x^2 + 4x + 4 - 4\).The x terms now become:\[(x+2)^2 - 4\]

Step 3: Complete the Square for y Terms

For the terms involving y, complete the square:1. Rewrite: \(-4(y^2 - 2y)\).2. Half the coefficient of y: \(\frac{-2}{2} = -1\).3. Square it: \((-1)^2 = 1\).4. Add and subtract inside the original equation: \(-4(y^2 - 2y + 1 - 1)\).The y terms become:\[-4(y-1)^2 + 4\]

Step 4: Write the Equation in Standard Form

Combine the newly arranged terms from completing the square:\[(x+2)^2 - 4(y-1)^2 - 4 + 4 = 0\]Which simplifies to:\[(x+2)^2 - 4(y-1)^2 = 0\]

Step 5: Recognize as a Conic Section

Rewrite the equation to standard form:\[(x+2)^2 = 4(y-1)^2\]Recognize this represents a hyperbola.

Step 6: Graph the Conic Section

Using a graphing device, graph the hyperbola given by the equation:The graph shows a hyperbola centered at \((-2, 1)\) with directrices parallel to the y-axis.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that can be imagined as the path created by the intersection of a plane and two opposite-facing cones. It has two separate curves, known as branches, which appear as mirrored images along their axes. The general form of a hyperbolic equation is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] where

• \((h, k)\) represents the center of the hyperbola,
• \(a\) and \(b\) define the distances from the center to the vertices and co-vertices, respectively.

The hyperbola in our example, \[ (x+2)^2 - 4(y-1)^2 = 0 \] was rearranged into its standard form, allowing for easier identification as a hyperbola. When graphed, the branches of a hyperbola open away from each other. The line through the center and perpendicular to the directrices is known as the transverse axis, while the conjugate axis is perpendicular to it. Hyperbolas are unique because they often describe phenomena in physics, such as satellite paths and certain reflective properties related to light and sound.

Completing the Square

Completing the square is an algebraic technique used to simplify quadratic expressions by converting them into a perfect square trinomial. This method is fundamental in obtaining the standard form of a conic section. It involves a few systematic steps:

• First, isolate the terms involving one variable, such as the \(x\) terms.
• Take half of the linear coefficient (the term with \(x\)), square it, and add and subtract this square within the equation.

For example, starting with the expression \(x^2 + 4x\), we:

Step-by-Step

• Divide \(4\) by \(2\), giving \(2\).
• Square \(2\) to get \(4\).
• Add and subtract \(4\) in the equation, transforming it to \((x + 2)^2 - 4\).

A similar approach is taken with the \(y\)-terms. Completing the square is an invaluable tool in algebra for solving quadratics, recognizing conic sections like hyperbolas, and transforming equations into forms that are easier to analyze and graph.

Standard Form of a Conic Section

Understanding the standard form of conic sections is crucial to accurately recognize and graph shapes like circles, ellipses, parabolas, and hyperbolas. Each conic has its own distinctive representation:

• Circles: \((x-h)^2 + (y-k)^2 = r^2\)
• Ellipses: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• Parabolas: \(y = ax^2 + bx + c\)
• Hyperbolas: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)

For hyperbolas, the standard form helps illustrate not only the position and orientation of the hyperbola but also key features like its center \((h, k)\), its vertices, and its asymptotes. In the analyzed exercise, \[ (x+2)^2 = 4(y-1)^2 \] could be rearranged and interpreted into standard form, guiding us to identify its center at \((-2, 1)\). Converting equations to their standard form simplifies the graphing process and greatly aids in understanding the conic's properties.