Problem 14 - 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. \(9 y^{2}-4 x^{2}=1\)

Step-by-Step Solution

Answer

Vertices: (0, ±1/3), Foci: (0, ±√13/6), Asymptotes: y = ±2/3x.

Step 1: Recognize Standard Form of Hyperbola

The standard form of a hyperbola is given by \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for a vertical hyperbola, and \(-\frac{y^2}{b^2} + \frac{x^2}{a^2} = 1\) for a horizontal hyperbola. The given equation is \(9y^{2}-4x^{2}=1\) and can be transformed to the form \(\frac{y^2}{\frac{1}{9}} - \frac{x^2}{\frac{1}{4}} = 1\), indicating a vertical hyperbola.

Step 2: Identify Vertices

For a vertical hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), vertices are at \((0, \pm a)\). Here, \(a^2 = \frac{1}{9}\), so \(a = \frac{1}{3}\). Therefore, the vertices are \((0, \frac{1}{3})\) and \((0, -\frac{1}{3})\).

Step 3: Identify Foci

The foci of a vertical hyperbola are given by \((0, \pm c)\), where \(c^2 = a^2 + b^2\). Here, \(a^2 = \frac{1}{9}\) and \(b^2 = \frac{1}{4}\). Calculating \(c^2 = \frac{1}{9} + \frac{1}{4} = \frac{4 + 9}{36} = \frac{13}{36}\), so \(c = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6}\). The foci are \((0, \frac{\sqrt{13}}{6})\) and \((0, -\frac{\sqrt{13}}{6})\).

Step 4: Write Equations of Asymptotes

For a vertical hyperbola, the equations of the asymptotes are given by \(y = \pm \frac{a}{b}x\). Substituting the values, \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\), the equations become \(y = \pm \frac{\frac{1}{3}}{\frac{1}{2}}x = \pm \frac{2}{3}x\). Thus, the equations of the asymptotes are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Key Concepts

Standard Form

In hyperbola equations, identifying the correct standard form is essential to interpret the other elements like vertices and foci. The standard form of a hyperbola depends on its orientation:

• Vertical hyperbolas take the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
• Horizontal hyperbolas look like \(-\frac{y^2}{b^2} + \frac{x^2}{a^2} = 1\).

For example, the given equation \(9y^2 - 4x^2 = 1\) can be transformed by division to match the vertical hyperbola's form: \(\frac{y^2}{\frac{1}{9}} - \frac{x^2}{\frac{1}{4}} = 1\).
This demonstrates that the hyperbola in our exercise is vertical. Recognizing the standard form helps to categorize the hyperbola and aids in calculating other characteristics, guiding the way to solve related questions effectively.

Vertices

Vertices of a hyperbola are crucial points that help define its shape. For a vertical hyperbola in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices lie on the y-axis, specifically at \((0, \pm a)\).

In our example, since \(a^2 = \frac{1}{9}\), the value of \(a\) becomes \(\frac{1}{3}\). Thus, the vertices are located at \( (0, \frac{1}{3}) \) and \( (0, -\frac{1}{3}) \).

Vertices are significant as they indicate the points where the hyperbola intersects its transverse axis. They also mark the closest approach of the two branches of the hyperbola.

Foci

The foci of a hyperbola are another set of critical components. These points, located within each branch, are integral to the hyperbola's definition. In a vertical hyperbola, they are found at \((0, \pm c)\), where \(c^2 = a^2 + b^2\).
Using our given equation's values, where \(a^2 = \frac{1}{9}\) and \(b^2 = \frac{1}{4}\), we calculate:

• \(c^2 = \frac{1}{9} + \frac{1}{4} = \frac{13}{36}\).
• \(c = \frac{\sqrt{13}}{6}\).

Thus, the foci are \((0, \frac{\sqrt{13}}{6})\) and \((0, -\frac{\sqrt{13}}{6})\).

The foci are important because the differences in distances from any point on the hyperbola to the foci remain constant. This unique property distinguishes hyperbolas from other conic sections.

Asymptotes

Asymptotes for a hyperbola guide the branches of the curve, showing the ultimate direction they head as they extend. For vertical hyperbolas, the equations of asymptotes are \(y = \pm \frac{a}{b}x\).

Substituting our values of \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\) solves to:

• \(y = \pm \frac{\frac{1}{3}}{\frac{1}{2}}x = \pm \frac{2}{3}x\).

Therefore, the equations describing the asymptotes are \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).

Asymptotes are lines that the hyperbola never touches but approaches infinitely close. They provide a framework that helps sketch the hyperbola's shape accurately.