Problem 48 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ (x+5)^{2}-16 y^{2}=16 $$

Step-by-Step Solution

Answer

The equation describes a hyperbola with center (-5, 0) and can be graphed using its standard form.

Step 1: Identify the Type of Conic

The given equation is \((x+5)^2 - 16y^2 = 16\). Notice that it resembles the standard form of a hyperbola equation, \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where the minus sign indicates it's a hyperbola oriented left-right.

Step 2: Rewrite the Equation in Standard Form

To transform the given equation to the standard form of a hyperbola, we divide every term by 16:\[\frac{(x+5)^2}{16} - \frac{16y^2}{16} = \frac{16}{16}\]Simplifying gives:\[\frac{(x+5)^2}{16} - \frac{y^2}{1} = 1\] This is now in standard form for a hyperbola, where \(a^2 = 16\), \(b^2 = 1\), \(h = -5\), and \(k = 0\).

Step 3: Identify Key Components of the Hyperbola

From the standard form equation \(\frac{(x+5)^2}{16} - \frac{y^2}{1} = 1\), we identify:- Center: \((-5, 0)\)- Transverse axis (horizontal) because \( (x-h)^2 \) term is first- \(a = 4\) since \(a^2 = 16\)- \(b = 1\) since \(b^2 = 1\)

Step 4: Graph the Hyperbola

Plot the center of the hyperbola at \((-5, 0)\). Draw a rectangle centered at the origin with width 2a = 8 and height 2b = 2, since \(a = 4\) and \(b = 1\). The asymptotes pass through the center with slopes \(\pm \frac{b}{a} = \pm \frac{1}{4}\). Draw the branches of the hyperbola approaching these asymptotes. The hyperbola opens left and right since the \(x\)-term appears first.

Key Concepts

Standard form of conic sections

Conic sections, including circles, parabolas, ellipses, and hyperbolas, have distinct equations in their standard forms. These equations help in identifying the type of conic and understanding its properties.

The equation needs to be in its standard form to analyze the components and graph it accurately. For hyperbolas, the standard form is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\).

• The negative sign in the standard form equation indicates the hyperbola's orientation.
• When the \((x-h)^2\) term appears first, the hyperbola opens left-right. Conversely, if \((y-k)^2\) term is first, it opens up-down.

Identifying the standard form allows students to easily extract important features such as the center and asymptotes.

Graphing hyperbolas

Graphing hyperbolas involves determining several key elements from their standard form, including the center, vertices, and asymptotes. These components are crucial for drawing an accurate graph.

For the hyperbola \(\frac{(x+5)^2}{16} - \frac{y^2}{1} = 1\), follow these steps:

• **Center**: The center is \((-5, 0)\), taken from \(h = -5\) and \(k = 0\).
• **Vertices**: Vertices are located \(a\) units away from the center along the transverse axis. Here, \(a = 4\), so vertices are at \((-9, 0)\) and \((-1, 0)\).
• **Asymptotes**: Asymptotes provide the directions that the hyperbola approaches but never touches. Given \(a = 4\) and \(b = 1\), the slopes of the asymptotes are \(\pm \frac{b}{a} = \pm \frac{1}{4}\).

Draw a rectangle with dimensions \(2a\) by \(2b\) centered at \((-5, 0)\), use the asymptote slopes to draw lines through the center, and sketch the hyperbola's curves approaching these lines.

Equation transformation

Transformation of an equation into its standard form is a crucial step for graphing and understanding hyperbolas. This process ensures correct identification of components like the center and asymptotes.

Take the original equation \((x+5)^2 - 16y^2 = 16\). To convert it into standard form, do the following:

• **Divide each term by 16**: This isolates the equation to equal one, a necessity for standard form.
• **Simplify**: The equation \(\frac{(x+5)^2}{16} - \frac{y^2}{1} = 1\) arises after simplification. This reveals parameters: \(a^2 = 16\) (hence \(a = 4\)) and \(b^2 = 1\) (hence \(b = 1\)).

This transformation helps not just in graphing, but also in finding various properties related to the hyperbola, leading to a clearer understanding of the equation's geometric representation.

Conic sections

Conic sections comprise a set of curves derived from slicing a right circular cone in different angles and positions. These include circles, ellipses, parabolas, and hyperbolas. Each is characterized by unique equations and properties.

Hyperbolas are particularly interesting because:

• **Orientation**: They open either horizontally or vertically depending on how the equation is structured.
• **Components**: They have two separate branches, asymptotes, a center, and vertices that define their space.
• **Standard form**: Adjusting an equation to reveal its type is pivotal for understanding and graphing it.

Understanding conic sections enriches students’ ability to visualize and interpret these shapes in various mathematical and real-world contexts.