Problem 15 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The vertices are (−2, 2) and (4, 2), the foci are (−4, 2) and (6, 2), asymptotes are y=(4/3)x−(4/3)+2 and y=−(4/3)x+(4/3)+2.

Step 1: Rewriting the Given Hyperbola Equation

The given hyperbola equation is already in standard form: \( \frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1 \). This indicates that it is a horizontal hyperbola.

Step 2: Identifying the Center

The center of the hyperbola can be identified from the equation \((x-h)^2/a^2 - (y-k)^2/b^2 = 1 \) as \((h, k)\). Here, \(h = 1\) and \(k = 2\), so the center is \((1, 2)\).

Step 3: Finding Vertices

For a horizontal hyperbola, vertices are located \(a\) units from the center along the horizontal axis. Here, \(a = \sqrt{9} = 3\), so the vertices are \((1 \pm 3, 2)\), which are \((-2, 2)\) and \((4, 2)\).

Step 4: Calculating the Distances to Foci

The distance from the center to each focus is \(c\) where \(c = \sqrt{a^2 + b^2}\). Here, \(a^2 = 9\) and \(b^2 = 16\), so \(c = \sqrt{9 + 16} = \sqrt{25} = 5\).

Step 5: Determining the Foci

With the calculated \(c\), the foci are \((h \pm c, k)\). Substituting the values, the foci are \((1 \pm 5, 2)\), which gives us \((-4, 2)\) and \((6, 2)\).

Step 6: Writing Equations of Asymptotes

For a hyperbola in the form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the asymptotes are given by the equations \(y - k = \pm \frac{b}{a}(x - h)\). Substituting in the values gives the asymptotes as \(y - 2 = \pm \frac{4}{3}(x - 1)\) or \(y = \frac{4}{3}x - \frac{4}{3} + 2\) and \(y = -\frac{4}{3}x + \frac{4}{3} + 2\).

Key Concepts

Standard Form

The standard form of a hyperbola provides a clear and concise way to understand its properties. It allows for easy identification of key characteristics such as the center, vertices, and foci. For a horizontal hyperbola, the equation appears as:

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

The terms \((x-h)^2\) and \((y-k)^2\) indicate that the center of the hyperbola is located at point \((h, k)\).
This is the point from which the distances to vertices and foci are calculated.
The parameters \(a^2\) and \(b^2\) are crucial for determining the length of the axes. Here, the values \(a^2 = 9\) and \(b^2 = 16\) help us know more about the hyperbola's shape.
The given equation, \( \frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1 \), is already in standard form indicating that it is a horizontal hyperbola.

Vertices

Vertices are specific points on a hyperbola that lie along the axis of symmetry. They provide a visual cue to the shape and orientation of the hyperbola.
For horizontal hyperbolas like the given one, vertices are located along the horizontal axis.

• The distance from the center to each vertex is equal to \(a\).

In our equation, \(a = \sqrt{9} = 3\). Consequently, the vertices are calculated from the center (1, 2):

• \((1 + 3, 2) = (4, 2)\)
• \((1 - 3, 2) = (-2, 2)\)

These points identify the narrowest part of the hyperbola, as the structure of a hyperbola widens as it extends away from the vertices.

Foci

The foci are important points located along the same line as the vertices, but further outside the hyperbola's opening. They are vital for understanding the hyperbola's reflective properties.
The distance from the center

• to each focus \(c\) is computed using:

\[ c = \sqrt{a^2 + b^2} \] Substituting the values, we get \(c = \sqrt{9 + 16} = \sqrt{25} = 5\).
With the center at (1, 2), we find the foci:

• \((1 + 5, 2) = (6, 2)\)
• \((1 - 5, 2) = (-4, 2)\)

These points lie on the major axis and are crucial for properties related to the hyperbola's optical characteristics.

Asymptotes

Asymptotes are key linear guides for the shape and behavior of a hyperbola as it extends towards infinity. They serve to inform how wide and at what angle the hyperbola's arms open.
For hyperbolas, the asymptotes are not intersecting lines but rather lines toward which the hyperbola approaches as it moves further out.
The equations of the asymptotes for a horizontal hyperbola in standard form are given by:

• \[ y - k = \pm \frac{b}{a}(x - h) \]
Inserting our values, we find:
• \(y - 2 = \pm \frac{4}{3}(x-1)\)
These can be rewritten in slope-intercept form:
• \(y = \frac{4}{3}x - \frac{4}{3} + 2\)
• \(y = -\frac{4}{3}x + \frac{4}{3} + 2\)
The lines show how the hyperbola behaves at infinity and the diagonal orientation relative to the center.