Problem 31 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Vertices \((0,\pm 8),\) asymptotes \(y=\pm 2 x\)

Step-by-Step Solution

Answer

The hyperbola equation is \(\frac{y^2}{64} - \frac{x^2}{16} = 1\).

Step 1: Determine the Center and Orientation

The vertices of the hyperbola are given as \((0, \pm 8)\). Since the vertices have the same x-coordinate, the hyperbola is vertical and centered at \((0, 0)\). The standard form of a vertical hyperbola is \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where \((h, k)\) is the center.

Step 2: Identify 'a' from the Vertices

The distance from the center to each vertex is the value of \(a\). For the vertices \((0, \pm 8)\), we have \(a = 8\). Since \((h, k)\) is \((0, 0)\), the equation takes the form \(\frac{y^2}{64} - \frac{x^2}{b^2} = 1\).

Step 3: Utilize Asymptotes to Find 'b'

The equations of the asymptotes for a vertical hyperbola are given by \( y = \pm \frac{a}{b} x \). Given the asymptotes are \(y = \pm 2x\), it follows that \(\frac{a}{b} = 2\). Substitute \(a = 8\) to get \(\frac{8}{b} = 2\). Solving for \(b\), we find \(b = 4\).

Step 4: Construct the Hyperbola Equation

Now that we have \(a = 8\) and \(b = 4\), substitute these into the standard form of the vertical hyperbola: \[\frac{y^2}{64} - \frac{x^2}{16} = 1.\]

Key Concepts

Vertices of Hyperbola

In the study of hyperbolas, understanding the vertices is essential. The vertices are the points on the hyperbola that lie closest to the center. They define the structure of the hyperbola and help in determining the other essential parameters. Given the problem, we have vertices at coordinates

• (0, 8) and
• (0, -8)

These points are aligned along the y-axis, indicating the type of hyperbola we are dealing with. The center of the hyperbola is at the midpoint between these vertices. Therefore, in this problem, the center is at (0,0). This informs us that the distance from the center to a vertex, labeled as 'a', is important. For this exercise, 'a' equals 8. This information sets the stage for constructing an accurate hyperbola equation.

Asymptotes of Hyperbola

The asymptotes of a hyperbola provide a lot of information about its shape and orientation. They are diagonal lines that the hyperbola approaches but never touches, guiding the opening of the curves. In a vertical hyperbola, such as in our exercise, the standard equations for asymptotes are

• y = +\( \frac{a}{b} \)x and
• y = -\( \frac{a}{b} \)x

In this example, the asymptotes are given by y = ±2x. This configuration helps determine the relationship between 'a' and 'b'. We have \( \frac{a}{b} = 2 \). Knowing 'a' is 8, this helps us solve for 'b' to find it equals 4. Thus, these asymptotes help in fine-tuning the parameters that define the hyperbola's equation.

Standard Form of Hyperbola

The standard form of a hyperbola's equation is crucial for its accurate representation. A hyperbola centered at (h, k) is defined by two possible equations

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

Since our hyperbola is vertical and centered at (0,0), we use the second form. For this specific hyperbola, we know 'a' equals 8 and 'b' equals 4. Substituting these values, we get the equation: \( \frac{y^2}{64} - \frac{x^2}{16} = 1 \). This equation accurately describes the shape and orientation of the hyperbola with respect to its center.

Vertical Hyperbola

Understanding the orientation of a hyperbola is important in classifying its equation. A vertical hyperbola opens in the up-and-down direction, parallel to the y-axis. The main indicator is the placement of 'a' under \((y-k)^2\) in its standard form equation. This affects the asymptotes and the orientation of the entire graph. In our exercise, the vertices (0,8) and (0,-8) confirm the vertical nature, as they lie along the y-axis. The asymptotes y = ±2x are steep, maintaining the hyperbola's vertical curve direction. Besides, the vertical nature indicates that any central movements would be along the y-axis only. Hence, understanding these insights aids in visualizing how the hyperbola behaves and ensures proper application of its properties.