Problem 28 - College Algebra

Question

Find the equations of the asymptotes for each hyperbola. \(\frac{(y-3)^{2}}{3^{2}}-\frac{(x+5)^{2}}{6^{2}}=1\)

Step-by-Step Solution

Answer

The asymptotes are \(y = \frac{1}{2}x + 5.5\) and \(y = -\frac{1}{2}x + 0.5\).

Step 1: Recognize the Standard Form

The given equation is the standard form of a hyperbola. It is written as \( \frac{(y-k)^{2}}{b^{2}} - \frac{(x-h)^{2}}{a^{2}} = 1 \). Comparing this with the given equation \( \frac{(y-3)^{2}}{3^{2}} - \frac{(x+5)^{2}}{6^{2}} = 1 \), we identify the center \((h, k)\) as \((-5, 3)\), \(b = 3\), and \(a = 6\).

Step 2: Identify the Orientation

In the given equation, the \((y-k)^{2}\) term comes first, indicating that this is a vertically oriented hyperbola. For a vertical hyperbola, the asymptotes are determined using the slope formula \(\frac{b}{a}\).

Step 3: Calculate the Slopes of the Asymptotes

The slopes of the asymptotes for a vertical hyperbola are \( \pm \frac{b}{a} \). Substitute the values of \(b\) and \(a\): \( \pm \frac{3}{6} = \pm \frac{1}{2} \).

Step 4: Write the Equations of the Asymptotes

Using the center \((-5, 3)\) and the slopes \( \pm \frac{1}{2} \), write the equations of the asymptotes. The general form is \(y - k = \pm m(x - h)\). The equations are: \(y - 3 = \frac{1}{2}(x + 5)\) and \(y - 3 = -\frac{1}{2}(x + 5)\).

Step 5: Simplify the Equations of the Asymptotes

Simplify the equations from the previous step. \(y = \frac{1}{2}x + 2.5 + 3 = \frac{1}{2}x + 5.5\) and \(y = -\frac{1}{2}x - 2.5 + 3 = -\frac{1}{2}x + 0.5\).

Key Concepts

Asymptotes

Asymptotes are invisible lines that a curve approaches but never actually reaches. In hyperbolas, asymptotes play a crucial role in describing the shape and orientation of the graph. They show the direction in which the hyperbola opens and help us understand its behavior at infinity.

• An important characteristic of hyperbolas is that their branches get infinitely close to the asymptotes.
• For the hyperbola described by its equation, the asymptotes intersect at the center of the hyperbola.

In this hyperbola, with center \((-5, 3)\), the asymptotes are linear equations that provide a guiding framework for sketching the curve accurately. Knowing how to find and use these equations simplifies the graphing process significantly.

Standard Form of a Hyperbola

The standard form of a hyperbola is a mathematical way to express its equation, contributing to easier identification of key features like the center and axes. The formula is usually written as \( \frac{(y-k)^{2}}{b^{2}} - \frac{(x-h)^{2}}{a^{2}} = 1 \), where:

• \(h, k\) is the center of the hyperbola.
• \(a\) and \(b\) indicate the distances from the center to the vertices and co-vertices, respectively.

When a hyperbola is given in standard form, it's straightforward to determine these parameters and further find the equations of the asymptotes. Understanding this form is the first step in analyzing the hyperbola's properties, such as the type of hyperbola (horizontal or vertical) and the orientation of its axes.

Vertical Hyperbola

A vertical hyperbola is one where the hyperbola opens upwards and downwards, dictated by the \( (y-k)^2 \) term appearing first in the equation. This is a key indicator in classifying the orientation of the hyperbola.

• In a vertical hyperbola, the axes are vertical and touch the top and bottom of each curve branch.
• The transverse axis, which is the line segment between the vertices, runs vertically.

For our hyperbola \( \frac{(y-3)^{2}}{3^{2}} - \frac{(x+5)^{2}}{6^{2}} = 1 \), the vertical orientation stems from the \( (y-3)^2 \) term taking precedence. Recognizing this pattern helps immensely when graphing, as it determines how the hyperbolic branches extend away from the center.

Slope of Asymptotes

The slope of the asymptotes in a hyperbola depends on the relation between \(a\) and \(b\), which are part of the standard form of the equation. For a vertical hyperbola, the slopes are given by \(+\frac{b}{a}\) and \(-\frac{b}{a}\).

• For our given hyperbola, the slope calculation becomes \(+\frac{3}{6}\) and \(-\frac{3}{6}\), simplifying to \(+\frac{1}{2}\) and \(-\frac{1}{2}\).
• These slopes indicate the steepness and direction of the asymptotes around the center, \( (-5, 3) \).

Once the slopes are known, writing the equations of the asymptotes becomes straightforward, as you can plug the slope into the slope-intercept form for a line. This crucial step reveals the invisible boundaries within which the hyperbola exists.