Problem 47 - College Algebra with Corequisite Support

Question

For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at (1,1) and (11,1) and one focus at (12,1) .

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{(x-6)^2}{25} - \frac{(y-1)^2}{11} = 1\).

Step 1: Identify the Center

The center of the hyperbola is the midpoint between the vertices. Given vertices are at (1,1) and (11,1).The midpoint (center) is calculated as:\[\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 + 11}{2}, \frac{1 + 1}{2}\right) = (6, 1)\]Thus, the center is at \((6, 1)\).

Step 2: Find the Distance to Vertices

The distance from the center to a vertex along the transverse axis is defined by \(a\). Given vertices are at (1,1) and (11,1), the distance \(2a\) is the distance between the two vertices along the x-axis:\[2a = 11 - 1 = 10 \Rightarrow a = 5\]

Step 3: Find the Distance to a Focus

The distance from the center to a focus along the transverse axis is defined by \(c\). Given one focus at (12,1), the distance from the center (6,1) is:\[c = |12 - 6| = 6\]

Step 4: Use the Relationship Between a, b, and c

In a hyperbola, we have the relationship \(c^2 = a^2 + b^2\). We know \(a = 5\) and \(c = 6\), so we find \(b\):\[c^2 = a^2 + b^2 \Rightarrow 6^2 = 5^2 + b^2 \Rightarrow 36 = 25 + b^2 \Rightarrow b^2 = 11\]

Step 5: Write the Standard Form Equation

For a hyperbola centered at \((h, k)\) with a transverse axis along the x-axis, the equation is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Substituting the known values: \(h = 6\), \(k = 1\), \(a^2 = 25\), \(b^2 = 11\):\[\frac{(x-6)^2}{25} - \frac{(y-1)^2}{11} = 1\]

Key Concepts

Center of Hyperbola

To find the center of a hyperbola, you need to determine the midpoint between its vertices. In this exercise, the vertices are given at coordinates \((1,1)\) and \((11,1)\). The center can be calculated by finding the midpoint between these two points using the formula:

• Midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Substituting the given coordinates:

• \(x_1 = 1\), \(x_2 = 11\)
• \(y_1 = 1\), \(y_2 = 1\)

Following the formula, the center of the hyperbola is calculated as \((6, 1)\). This midpoint provides a crucial part of the hyperbola's equation and helps define its structure in the coordinate plane.

Distance Between Vertices

The distance between the vertices of a hyperbola provides useful information, particularly in finding \(a\), which is half of this distance. In the horizontal direction, it is along the transverse axis. From the exercise, the vertices are at \((1,1)\) and \((11,1)\).

• The formula to find the entire distance between these points is: \(2a = |x_2 - x_1|\).

Given:

• \(x_2 = 11\), \(x_1 = 1\)

The calculation is straightforward:

• \(2a = 11 - 1 = 10\)
• Thus, \(a = \frac{10}{2} = 5\)

Knowing \(a\) is significant because it aids in structuring the hyperbola's equation and describing its size.

Relationship Between a, b, and c

The relationship between \(a\), \(b\), and \(c\) in a hyperbola is given by the equation \(c^2 = a^2 + b^2\). This equation helps identify the hyperbola's curvature. With \(a\) and \(c\) already established from previous calculations, we can easily determine \(b^2\).

• Given: \(a = 5\) and \(c = 6\)
• Calculate \(b^2\) using the relationship: \(6^2 = 5^2 + b^2\)
• The calculations follow: \(36 = 25 + b^2\)
• \(b^2 = 36 - 25 = 11\)

This relationship is integral to completing the standard form of the hyperbola's equation and describes its geometric structure further.

Standard Form Equation

The standard form equation of a hyperbola centered at \((h, k)\) is critical for describing its shape. When the transverse axis is horizontal, the equation is:

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

From this exercise, the values for \(h\), \(k\), \(a^2\), and \(b^2\) are:

• \(h = 6\), \(k = 1\)
• \(a^2 = 25\), \(b^2 = 11\)

Inserting these into the equation gives:

• \( \frac{(x-6)^2}{25} - \frac{(y-1)^2}{11} = 1\)

This equation encapsulates all the significant features of the hyperbola such as the center location, the direction of the transverse axis, and the extent of its vertices and foci.