Problem 32 - Algebra and Trigonometry

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices \((0, \pm 6),\) asymptotes \(y=\pm \frac{1}{3} x\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{y^2}{36} - \frac{x^2}{324} = 1\).

Step 1: Identify the Center

The vertices of the hyperbola are given at \((0, \pm 6)\). Since the vertices are vertically aligned, the center of the hyperbola should be located at \((0,0)\). This is obtained by averaging the vertex coordinates, \((0,6)\) and \((0,-6)\), resulting in the center \((0,0)\).

Step 2: Determine the Orientation

The vertices are at \((0, \pm 6)\), indicating that the hyperbola opens vertically, which is characteristic of a vertical hyperbola.

Step 3: Find 'a'

The distance from the center to each vertex (\(a\)) is 6 (since vertices are at \((0, \pm 6)\)). Therefore, \(a = 6\).

Step 4: Use Asymptotes to Determine 'b'

The equations of the asymptotes for a vertical hyperbola centered at the origin are \(y = \pm \frac{a}{b}x\). Here, the given asymptotes are \(y = \pm \frac{1}{3}x\), so we have \(\frac{a}{b} = \frac{1}{3}\). Substituting \(a = 6\), we get \(\frac{6}{b} = \frac{1}{3}\). Solving for \(b\) gives \(b = 18\).

Step 5: Write the Equation

The standard form of a vertical hyperbola centered at \((0,0)\) is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Substitute \(a = 6\) and \(b = 18\) into the equation to obtain \(\frac{y^2}{36} - \frac{x^2}{324} = 1\).

Key Concepts

Vertices of Hyperbola

A hyperbola is a beautiful and complex structure in mathematics, and understanding its vertices is essential. The vertices are the points where the hyperbola is closest to or farthest from its center. In this case, the vertices given are

• (0, 6)
• (0, -6)

.These points are directly aligned vertically, hinting at the vertical orientation of the hyperbola. Since the vertices are found at these coordinates, the center of the hyperbola is located exactly halfway between them. You can calculate the center by averaging the y-values of the vertices (6 and -6), leading to the center

• (0, 0)

. The distance from the center to each vertex is what we call \( a \). Here, the distance is 6, which means \( a = 6 \) for this hyperbola. This distance helps determine the size of the hyperbola along the axis in which it opens.

Asymptotes of Hyperbola

Asymptotes are lines that approach the hyperbola but never touch it, creating the general direction in which the branches open. For this hyperbola, the asymptotes are given as:

• \( y = \frac{1}{3}x \)
• \( y = -\frac{1}{3}x \)

Since these are linear equations of a hyperbola centered at the origin, they take the form \( y = \pm \frac{a}{b}x \). The slope of the asymptotes, \( \frac{1}{3} \), allows us to express a relationship between \( a \) and \( b \) in the hyperbola's equation. With \( a \) already determined as 6, substituting it into the relation \( \frac{a}{b} = \frac{1}{3} \) helps us find \( b \). Solving \( \frac{6}{b} = \frac{1}{3} \) gives \( b = 18 \). Understanding these calculations is key to grasping how the asymptotes set the limits for where the hyperbola bends around.

Equation of Hyperbola

The equation of a hyperbola defines its precise shape and orientation. For a vertical hyperbola centered at the origin, the standard form is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Now that we know both \( a \) and \( b \), we can write the specific equation for this hyperbola.First, substitute \( a = 6 \) and \( b = 18 \) into the standard equation. This results in:

• \( \frac{y^2}{36} - \frac{x^2}{324} = 1 \)

This equation represents our hyperbola, describing the curves created by points satisfying the equation property. This form ensures that the hyperbola opens vertically, as indicated by the vertices. The use of \( a^2 \) and \( b^2 \) in the denominators corresponds directly to the previously calculated values, locking in the dimensions and orientation perfectly.