Problem 28 - College Algebra

Question

For the following exercises, 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 \( y = \frac{1}{2}x + \frac{11}{2} \) and \( y = -\frac{1}{2}x + \frac{1}{2} \).

Step 1: Identify the Standard Form of Hyperbola

The given equation is \( \frac{(y-3)^{2}}{3^{2}} - \frac{(x+5)^{2}}{6^{2}} = 1 \). This is in the form \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), which represents a vertically-oriented hyperbola.

Step 2: Determine the Center of the Hyperbola

The center \((h, k)\) of the hyperbola can be found from the equation. Here, \(h = -5\) and \(k = 3\). So, the center is \((-5, 3)\).

Step 3: Find the Equation of Asymptotes

For a vertically-oriented hyperbola of the form \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), the asymptotes are given by the equations \( y = k \pm \frac{a}{b}(x-h) \).

Step 4: Plug in the Values for Asymptote Equations

In this hyperbola, \(a = 3\) and \(b = 6\). Therefore, \( \frac{a}{b} = \frac{3}{6} = \frac{1}{2} \). The equations for the asymptotes become:\[ y - 3 = \pm \frac{1}{2} (x + 5) \]

Step 5: Simplify the Asymptote Equations

Solve the equations \( y - 3 = \frac{1}{2}(x + 5) \) and \( y - 3 = -\frac{1}{2}(x + 5) \) by distributing and simplifying.The first simplifies to \( y = \frac{1}{2}x + \frac{5}{2} + 3 \). The second to \( y = -\frac{1}{2}x - \frac{5}{2} + 3 \).Further simplifying gives the equations:\[ y = \frac{1}{2}x + \frac{11}{2} \] and\[ y = -\frac{1}{2}x + \frac{1}{2} \].

Key Concepts

Asymptotes

Asymptotes are crucial components in understanding the behavior of hyperbolas. These are lines that the curve approaches but never actually reaches. They give insight into the "opening direction" and orientation of the hyperbola.
For hyperbolas, the equations of asymptotes differ based on whether the hyperbola is oriented horizontally or vertically. In our case, the hyperbola is vertically oriented.

For a vertically-oriented hyperbola, the equations of the asymptotes are derived from the standard form:

• The formula is given by: \( y = k \pm \frac{a}{b}(x-h) \)
• Here \( h \) and \( k \) are the coordinates of the center.
• \( a \) and \( b \) are taken from the denominators of \( y^2 \) and \( x^2 \) terms respectively.

By identifying these elements from the specific equation and plugging them into the formula, you can easily find the asymptotes for any hyperbola.

Standard Form

The standard form of a hyperbola is vital to identifying its characteristics. It provides clarity on the orientation, center, and length of the transverse and conjugate axes.
In the case of a vertically-oriented hyperbola, the standard form is expressed as:

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

This equation tells us that the hyperbola opens vertically. The terms \((y-k)^2\) and \( (x-h)^2 \) emphasize the parabola's focus along the y-axis.
Recognizing the standard form allows us to extract the necessary components needed for further calculations, such as the equation of the asymptotes and the center of the hyperbola.

Center of Hyperbola

The center of a hyperbola is a pivotal point that lies directly between the vertices and is the intersection point of the asymptotes.
In the equation \( \frac{(y-3)^2}{3^2} - \frac{(x+5)^2}{6^2} = 1 \), the center is straightforward to find:

• The center \( (h, k) \) is derived from the terms \( (x-h) \) and \( (y-k) \).
• Here, \( h = -5 \) and \( k = 3 \).

Thus, for this hyperbola, the center is at the point \((-5, 3)\). The center acts as a reference for dividing and balancing the hyperbola's structure and positioning its asymptotes.

Vertically-Oriented Hyperbola

A vertically-oriented hyperbola, as the name suggests, opens upwards and downwards along the y-axis.
The classification into vertically versus horizontally oriented depends on the position of \( a^2 \) and \( b^2 \) in the equation. For a vertically-oriented hyperbola:

• The equation takes the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).
• Here, the \( y \) term appears first and dictates the hyperbola’s vertical nature.

This orientation affects how the graph of the hyperbola will look and how we determine other properties like vertices, foci, and asymptotes.
Understanding these opens the door to easily sketching and analyzing the behavior of a vertically-oriented hyperbola.