Problem 27 - College Algebra

Question

Find the equations of the asymptotes for each hyperbola. \(\frac{(x-3)^{2}}{5^{2}}-\frac{(y+4)^{2}}{2^{2}}=1\)

Step-by-Step Solution

Answer

The asymptote equations are \( y = \frac{2}{5}x - \frac{26}{5} \) and \( y = -\frac{2}{5}x - \frac{14}{5} \).

Step 1: Identify the standard form

The given hyperbola is in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Here, \(h = 3\), \(k = -4\), \(a = 5\), and \(b = 2\). This is the standard form of a hyperbola with the transverse axis parallel to the x-axis.

Step 2: Use the formula for asymptotes

For hyperbolas with the transverse axis parallel to the x-axis, the asymptotes are given by the equations \( y - k = \pm \frac{b}{a} (x - h) \).

Step 3: Substitute known values

Plug in the values for \(h\), \(k\), \(a\), and \(b\) into the formula for asymptotes. Here, \(h = 3\), \(k = -4\), \(a = 5\), \(b = 2\).

Step 4: Calculate the asymptote equations

Using the formula \( y + 4 = \pm \frac{2}{5} (x - 3) \), we expand to find the two asymptote equations: \( y + 4 = \frac{2}{5}(x - 3) \) and \( y + 4 = -\frac{2}{5}(x - 3) \).

Step 5: Simplify the equations

Simplify each asymptote equation to its standard linear form. The first equation becomes \( y = \frac{2}{5}x - \frac{6}{5} - 4 \) which simplifies to \( y = \frac{2}{5}x - \frac{26}{5} \). The second equation becomes \( y = -\frac{2}{5}x + \frac{6}{5} - 4 \) which simplifies to \( y = -\frac{2}{5}x - \frac{14}{5} \).

Key Concepts

Hyperbola Standard Form

Understanding hyperbolas starts with recognizing their unique structure, called the hyperbola's standard form. The standard form of a hyperbola with a horizontal transverse axis is given by \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]. This is different from the form of an ellipse, as it involves subtraction. Here,

• \( h \) and \( k \) define the center of the hyperbola, found at the point \( (h, k) \).
• \( a \) represents the distance from the center to each vertex on the transverse axis.
• \( b \) indicates the distance from the center to the endpoints of the conjugate axis.

This standard form helps in identifying the nature of the hyperbola and assists in finding crucial elements like asymptotes and axes."},{

Asymptote Equations

Asymptotes are lines that a hyperbola approaches but never actually touches. They provide a boundary that defines the overall shape. To find the asymptote equations for a hyperbola in standard form, you can use the formula: \[ y - k = \pm \frac{b}{a} (x - h) \]. This enables us to generate two linear equations, which correspond to the hyperbola's asymptotes. By using the parameters

• \( h = 3 \), \( k = -4 \)
• \( a = 5 \), and \( b = 2 \)

from our exercise, we substitute into the equation to obtain:

• \( y + 4 = \frac{2}{5}(x - 3) \)
• \( y + 4 = -\frac{2}{5}(x - 3) \)

These equations describe the asymptotes of the given hyperbola, which are essential for sketching and understanding the geometric properties.

Transverse Axis

A crucial concept when examining hyperbolas is the transverse axis, which serves as the primary axis of symmetry. It is the line segment that passes through the center and vertices of the hyperbola. For hyperbolas with a horizontal transverse axis, such

• as in our exercise, this segment aligns with the x-axis.

The transverse axis length is dependent on the value of \( a \), as it's exactly twice this distance, identified by \( 2a \). In our example, \( a = 5 \), making the transverse axis length \( 2 \times 5 = 10 \). The transverse axis is critical because it determines how 'wide' the hyperbola is, significantly influencing its overall shape and orientation.

Standard Linear Form

Linear equations for asymptotes are often presented in what's known as the standard linear form, \[ y = mx + c \]. This form allows one to easily identify the slope \( m \), which represents the steepness, and \( c \), the y-intercept. Starting from the asymptote equations

• \( y + 4 = \frac{2}{5}(x - 3) \)
• \( y + 4 = -\frac{2}{5}(x - 3) \)

we simplify these to the easier-to-read form:

• \( y = \frac{2}{5}x - \frac{26}{5} \)
• \( y = -\frac{2}{5}x - \frac{14}{5} \)

By rewriting the equations this way, it becomes straightforward to graph them or use them in further calculations. Recognizing and converting into the standard linear form ensures clarity in mathematical communication and application.