Problem 9 - Precalculus: Functions and Graphs

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$16 x^{2}-36 y^{2}=1$$

Step-by-Step Solution

Answer

Vertices: \((\pm \frac{1}{4}, 0)\); Foci: \((\pm \frac{\sqrt{13}}{12}, 0)\); Asymptotes: \(y = \pm \frac{2}{3}x\).

Step 1: Write the Equation in Standard Form

The given equation is \(16x^2 - 36y^2 = 1\). To convert it into the standard form of a hyperbola, divide every term by 1:\[\frac{16x^2}{1} - \frac{36y^2}{1} = 1\]Then simplify by dividing through by 16:\[\frac{x^2}{\frac{1}{16}} - \frac{y^2}{\frac{1}{36}} = 1\]This simplifies to the standard form:\[\frac{x^2}{(\frac{1}{4})^2} - \frac{y^2}{(\frac{1}{6})^2} = 1\]

Step 2: Identify the Center and Identify "a" and "b"

For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the center is at \((0, 0)\), \(a = \frac{1}{4}\), and \(b = \frac{1}{6}\).

Step 3: Find the Vertices

The vertices for a hyperbola of this form are located at \((\pm a, 0)\). Therefore, the vertices are \(\left(\frac{1}{4}, 0\right)\) and \(-\left(\frac{1}{4}, 0\right)\).

Step 4: Calculate the Foci

The foci of the hyperbola are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). Calculate \(c\):\[c = \sqrt{(\frac{1}{4})^2 + (\frac{1}{6})^2} = \sqrt{\frac{1}{16} + \frac{1}{36}} = \sqrt{\frac{9}{144} + \frac{4}{144}} = \sqrt{\frac{13}{144}} = \frac{\sqrt{13}}{12}\]So the foci are located at \(\left( \pm \frac{\sqrt{13}}{12}, 0 \right)\).

Step 5: Derive the Equations of the Asymptotes

The asymptotes of a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are defined by the lines \(y = \pm \frac{b}{a}x\):\[y = \pm \left(\frac{1/6}{1/4}\right)x = \pm \frac{2}{3}x\]Thus, the equations of the asymptotes are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Step 6: Sketch the Graph

To sketch the hyperbola, plot the vertices at \(\left(\frac{1}{4}, 0\right)\) and \(-\left(\frac{1}{4}, 0\right)\), plot the foci at \(\left( \frac{\sqrt{13}}{12}, 0 \right)\) and \(-\left( \frac{\sqrt{13}}{12}, 0 \right)\), and draw the asymptotes \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\). The hyperbola will open horizontally along the x-axis between and beyond the vertices, approaching the asymptotes but never intersecting them.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are critical for understanding its shape and position. In this equation, we need to find these pivotal points. Vertices act as the points closest or farthest apart along the transverse axis, defining the "width" of a hyperbola.

With the standard form of a horizontal hyperbola being \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices can be easily found. Here, the equation is \(\frac{x^2}{(\frac{1}{4})^2} - \frac{y^2}{(\frac{1}{6})^2} = 1\).

• The value of \(a\), derived from \(a^2 = (\frac{1}{4})^2\), is \(\frac{1}{4}\).
• The vertices lie on the x-axis at \((\pm a, 0)\).
• Thus, the hyperbola’s vertices are at \((\frac{1}{4}, 0)\) and \((-\frac{1}{4}, 0)\).

Foci of Hyperbola

The foci, though invisible on a basic graph, play a role in the definition of a hyperbola. They determine the precise shape and how "stretched" or "compressed" it looks.

To locate the foci, calculate \(c\) using \(c = \sqrt{a^2 + b^2}\). This formula provides the distance from the center to each focus point.

• For this hyperbola, \(a = \frac{1}{4}\) and \(b = \frac{1}{6}\).
• Therefore, \(c = \sqrt{(\frac{1}{4})^2 + (\frac{1}{6})^2} = \sqrt{\frac{1}{16} + \frac{1}{36}}\).
• Solving gives \(c = \frac{\sqrt{13}}{12}\).
• The foci are at \((\pm \frac{\sqrt{13}}{12}, 0)\).

Asymptotes of Hyperbola

Asymptotes are the invisible lines that the hyperbola will approach, but never meet. They provide valuable guidelines for correctly sketching the graph of a hyperbola.

For hyperbolas structured as \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are given by the equations \(y = \pm \frac{b}{a}x\).

• For our example, \(a = \frac{1}{4}\) and \(b = \frac{1}{6}\).
• This leads to \(y = \pm \frac{1/6}{1/4}x = \pm \frac{2}{3}x\).
• Thus, the asymptotes are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Graphing Hyperbolas

Graphing a hyperbola involves plotting all its features: vertices, foci, and asymptotes. This process helps visualize the hyperbola's dynamic path as it extends indefinitely.

Begin by plotting the vertices and foci on a coordinate plane.

• Mark the vertices at \((\frac{1}{4}, 0)\) and \((-\frac{1}{4}, 0)\).
• Then, place the foci at \((\frac{\sqrt{13}}{12}, 0)\) and \((-\frac{\sqrt{13}}{12}, 0)\).
• Draw the asymptotes, \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\), as faint dotted lines.

This setup acts as a guide.
Now sketch the hyperbola, ensuring it approaches but never crosses the asymptotes and appears open toward the x-axis.

Standard Form of Hyperbola

The standard form of a hyperbola is key to identifying its features and subsequently manipulating it for graphing. In standard form, the hyperbola equation reveals aspects such as direction and dimensions through constants.

To convert the equation into standard form, each term must be divided by the constant originally on the right-hand side.
For our given equation, the process is:

• Start with \(16x^2 - 36y^2 = 1\).
• Each term should be divided to isolate \(1\) on the right: \(\frac{16x^2}{1} - \frac{36y^2}{1} = 1\).
• Simplify further to \(\frac{x^2}{(\frac{1}{4})^2} - \frac{y^2}{(\frac{1}{6})^2} = 1\).

After this transformation:

• The equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is now in standard form.
• This will signify a horizontally opening hyperbola due to the position of \(x\) in the formula.