Problem 67 - A Graphical Approach to Precalculus with Limits

Question

Find an equation for each hyperbola. Vertices \((0,6)\) and \((0,-6) ;\) asymptotes \(y=\pm \frac{1}{2} x\)

Step-by-Step Solution

Answer

The equation is \(\frac{y^2}{36} - \frac{x^2}{144} = 1\).

Step 1: Identify the Orientation

Since the vertices are along the y-axis, the hyperbola is vertical. This means the standard form of the equation for the hyperbola is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).

Step 2: Calculate the Center and 'a' Value

The center of the hyperbola is midway between the vertices. So, the center \((h,k)\) is \((0,0)\). The distance from the center to each vertex (\(a\)) is 6 (the distance between \((0,0)\) and \((0,6)\)). Thus, \( a = 6 \).

Step 3: Determine the Slope of the Asymptotes

The asymptotes are given by \(y = \pm \frac{1}{2} x\), which implies the slope \(\pm \frac{1}{2}\) is \(\pm \frac{a}{b}\). Thus, \(\frac{6}{b} = \frac{1}{2}\).

Step 4: Solve for 'b'

From \(\frac{6}{b} = \frac{1}{2}\), solve for \(b\): \( b = 6 \times 2 = 12 \). So, \( b = 12 \).

Step 5: Write the Equation

Substitute the values of \(a\), \(b\), \(h\), and \(k\) into the standard form equation: \( \frac{(y-0)^2}{6^2} - \frac{(x-0)^2}{12^2} = 1 \). Simplify to get \( \frac{y^2}{36} - \frac{x^2}{144} = 1 \).

Key Concepts

Vertices

The vertices of a hyperbola are key points that define its shape and orientation. A hyperbola's vertices are located at the maximum distance from its center along its main axis. In this exercise, the vertices are given as

• (0, 6) and (0, -6)

Here, it's clear that the vertices are aligned along the y-axis. This means the hyperbola is vertical.
To calculate the distance from the center to each vertex, which is denoted as 'a', we measure the length from the center to either vertex. With both vertices at 6 units away from the center

• 0 at each extremity (0,6) and (0,-6)

The 'a' value is 6. This distance is crucial for forming the standard equation of the hyperbola.

Asymptotes

Asymptotes of a hyperbola are the lines that the hyperbola approaches but never actually reaches. They offer a helpful guide to the hyperbola's overall orientation in its plane. For our case, the asymptotes are given by the equations

• \( y = \pm \frac{1}{2} x \)

The slope of the asymptotes, \( \pm \frac{1}{2} \), is derived from the ratio \( \pm \frac{a}{b} \), where 'a' and 'b' are the distances from the center to a vertex and to a co-vertex, respectively.
By knowing the value of 'a' is 6, you can set up the equation \( \frac{6}{b} = \frac{1}{2} \), because the slope of the asymptotes must match. Solving for 'b', \( b \) turns out to be 12. These asymptotes help visually determine how 'spread out' the hyperbola is.

Equation

The equation of a hyperbola describes its complete shape and form. Since our hyperbola is vertical, the standard form of the equation is

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

Here, (h,k) is the center of the hyperbola, 'a' is the distance to the vertices along the vertical axis, and 'b' is the distance to the co-vertices along the horizontal axis.
For our hyperbola:

• 'a' is 6
• 'b' is 12
• The center is (0,0)

Substitute these values into the standard form to get

• \( \frac{y^2}{36} - \frac{x^2}{144} = 1 \)

This equation forms the complete mathematical representation of our hyperbola, telling us exactly its shape and orientation.

Center of hyperbola

The center of a hyperbola serves as the main reference point for all other measurements and calculations, like vertices and distances. It is fundamentally the midpoint between the vertices along the main axis.
In our case, since the vertices are (0,6) and (0,-6), the center is midway along the y-axis, exactly at

• (0,0)

This position is crucial, as it defines the other parameters of the equation. Locating the center is the first step in analyzing or sketching a hyperbola, as it helps determine the orientation and guides the equation used in its formulation.