Problem 34 - College Algebra

Question

Sketch a graph of the hyperbola, labeling vertices and foci. \(81 x^{2}-9 y^{2}=1\)

Step-by-Step Solution

Answer

The hyperbola is horizontal with vertices at \((\pm \frac{1}{9}, 0)\) and foci at \((\pm \frac{\sqrt{10}}{9}, 0)\).

Step 1: Identify the Standard Form

The given equation is \(81x^2 - 9y^2 = 1\). This can be written in the standard form of a hyperbola as \( \frac{x^2}{\frac{1}{81}} - \frac{y^2}{\frac{1}{9}} = 1 \), which is \( \frac{x^2}{\left(\frac{1}{9}\right)^2} - \frac{y^2}{1^2} = 1 \). This is a hyperbola oriented horizontally.

Step 2: Identify Vertices and Foci

For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are at \((\pm a, 0)\). Here \(a^2=\frac{1}{81}=\left(\frac{1}{9}\right)^2\) so \(a=\frac{1}{9}\). The foci are found using \(c = \sqrt{a^2 + b^2}\). With \(b^2=\frac{1}{9}=\left(\frac{1}{3}\right)^2\), calculate \(c\).

Step 3: Calculate Foci Distance

In this case, \(a = \frac{1}{9}\) and \(b = \frac{1}{3}\). Calculating \(c\), we have \(c = \sqrt{\left(\frac{1}{9}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{1}{81} + \frac{1}{9}} = \sqrt{\frac{1}{81} + \frac{9}{81}} = \sqrt{\frac{10}{81}} = \frac{\sqrt{10}}{9}\). Thus, the foci are at \( (\pm \frac{\sqrt{10}}{9}, 0)\).

Step 4: Sketch the Graph

Draw the coordinate axes. Plot the vertices at \( (\frac{1}{9}, 0)\) and \( (-\frac{1}{9}, 0)\). Mark the foci approximately at \((\pm \frac{\sqrt{10}}{9}, 0)\). Because the hyperbola is horizontal, sketch the two separate curves opening left and right from the vertices.

Key Concepts

Vertices of Hyperbola

Understanding the concept of the vertices in a hyperbola is a fundamental step when dealing with these unique shapes. The vertices of a hyperbola are the points where each curve (or branch) meets its closest point to the center of the hyperbola. These points are crucial as they help outline the hyperbola's shape and orientation. The standard form of a hyperbola guiding us here is given by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for a horizontal hyperbola or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\) for a vertical one.

For the exercise at hand, our hyperbola is oriented horizontally, meaning the vertices are positioned along the horizontal axis. The vertices can simply be found at the points \((\pm a, 0)\).

• Horizontal Hyperbola: \( (\pm a, 0) \)
• Vertical Hyperbola: \( (0, \pm b) \)

In the problem provided, we have \(a = \frac{1}{9}\) because the equation is transformed to \(\frac{x^2}{\left(\frac{1}{9}\right)^2} - \frac{y^2}{1^2} = 1\). Therefore, the vertices are located at \((\frac{1}{9}, 0)\) and \((-\frac{1}{9}, 0)\). These positions indicate how the curves open away horizontally from either side of the center located at the origin (0,0).

Foci of Hyperbola

The foci of a hyperbola are points from which the curves of the hyperbola seem to "come" and "go". They are significant because the hyperbola's definition—much like an ellipse—is governed by the distances between these points. For a standard hyperbola in the form of \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci are located on the horizontal axis, whereas for \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\), they align along the vertical axis.

To find the exact location of the foci, use the formula \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to each focus. For our specific hyperbola equation, we calculated \(a = \frac{1}{9}\) and \(b = \frac{1}{3}\). Using these values:

• \(c = \sqrt{\left(\frac{1}{9}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{1}{81} + \frac{9}{81}} = \sqrt{\frac{10}{81}} = \frac{\sqrt{10}}{9}\)
• Foci are located at \((\pm \frac{\sqrt{10}}{9}, 0)\)

This calculation positions the foci relative to the center and helps to better visualize the hyperbola's elongation and direction along the x-axis.

Graphing Hyperbolas

Graphing a hyperbola involves careful placement of its vertices and foci on a coordinate system to accurately portray its characteristic shape. A hyperbola graph consists of two separate curves called branches, and these curves never meet.

Here's a simple step-by-step method to sketch a hyperbola:

• Start by identifying the vertices, which in this problem are \((\frac{1}{9}, 0)\) and \((-\frac{1}{9}, 0)\).
• Next, mark the position of the foci. For our hyperbola's setting, this is at approximately \((\pm \frac{\sqrt{10}}{9}, 0)\).
• Determine the asymptotes of the hyperbola, which guide the overall shape, though they don't appear explicitly in our simplified graph. In a standard equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), these lines can be defined as \(y = \pm\frac{b}{a}x\).
• Draw the branches of the hyperbola opening left and right from each vertex. These should extend toward the direction of the foci but not touch them.

Keep in mind, hyperbolas are symmetric with respect to both their axes and their center, making them elegant and regular once plotted. To illustrate the hyperbola in the exercise, ensure the branches of the graph don't intersect and appear approximately equidistant from the asymptotes at any point.