Problem 55 - A Graphical Approach to Precalculus with Limits

Question

CHECKING ANALYTIC SKILLS Graph each hyberbola by hand. Give the domain and range. Give the center in Exercises \(55-61 .\) Do not use a calculator. $$\frac{(x-1)^{2}}{9}-\frac{(y+3)^{2}}{25}=1$$

Step-by-Step Solution

Answer

Center at (1, -3). Domain: \((-\infty, \infty)\), Range: \((-\infty, \infty)\).

Step 1: Identify the Standard Form of Hyperbola

The given hyperbola equation is \(\frac{(x-1)^{2}}{9}-\frac{(y+3)^{2}}{25}=1\). This equation is in the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), indicating a hyperbola centered at \((h, k)\) with a horizontal transverse axis.

Step 2: Find the Center of Hyperbola

From the equation \(\frac{(x-1)^{2}}{9}-\frac{(y+3)^{2}}{25}=1\), we can see that \(h = 1\) and \(k = -3\). Thus, the center of the hyperbola is \((1, -3)\).

Step 3: Determine a and b Values

In the equation, \(a^2 = 9\) and \(b^2 = 25\). Therefore, \(a = 3\) and \(b = 5\). These values represent the distances from the center to the vertices and co-vertices along the x and y axes, respectively.

Step 4: Plot the Vertices and Co-Vertices

For the vertices along the x-axis (since the hyperbola opens horizontally), add and subtract \(a\) from the x-coordinate of the center. The vertices are at \((1 \pm 3, -3)\), or \((4, -3)\) and \((-2, -3)\). The co-vertices, along the y-axis, are at \((1, -3 \pm 5)\), or \((1, 2)\) and \((1, -8)\).

Step 5: Sketch the Asymptotes

The equations for the asymptotes of a hyperbola in this form are \(y = k \pm \frac{b}{a}(x-h)\). Substituting the values, the asymptotes are \(y = -3 \pm \frac{5}{3}(x-1)\). Calculate these lines to guide your sketch of the hyperbola.

Step 6: Graph the Hyperbola

Draw the hyperbola by sketching branches opening horizontally from each vertex, approaching but never touching the asymptotes. The branches go through the vertices \((4, -3)\) and \((-2, -3)\) and curve towards the asymptotes.

Step 7: Determine Domain and Range

The asymptotes imply the hyperbola stretches infinitely. The domain, concerning x-axis freedom, is all real numbers \((-\infty, \infty)\). The range, concerning y-coordinate freedom, is also all real numbers \((-\infty, \infty)\), as there are no vertical boundaries.

Key Concepts

Standard Form

The standard form of a hyperbola is crucial for understanding its fundamental characteristics, including its orientation, center, and other elements like vertices. In mathematical terms, a hyperbola can be represented by an equation of the form:

• If it opens horizontally: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
• If it opens vertically: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]

The equation given in the exercise \(\frac{(x-1)^2}{9} - \frac{(y+3)^2}{25} = 1\) follows the horizontal opening form. Here, the point \((h, k)\) is \((1, -3)\), revealing the hyperbola's center. This form not only tells us about the center but also signifies how the hyperbola is aligned.
Horizontal alignment means the transverse axis is along the x-axis, aiding in determining the positions of vertices and asymptotes.

Vertices and Co-Vertices

Vertices and co-vertices define the main axes of a hyperbola. They are derived from the center by adjusting the values of \(a\) and \(b\), which are the semi-major and semi-minor axis lengths respectively. For the equation given, we have:

• \(a^2 = 9\) leading to \(a = 3\)
• \(b^2 = 25\) leading to \(b = 5\)

For a hyperbola like ours, that opens horizontally, you find the vertices by moving \(a\) units left and right from the center along the x-axis.
From \((1, -3)\), adding and subtracting \(a\) gives us vertices at \((4, -3)\) and \((-2, -3)\).
Meanwhile, the co-vertices are determined by moving \(b\) units up and down from the center along the y-axis, resulting in co-vertices at \((1, 2)\) and \((1, -8)\).
These points help define the bounding rectangle, framing the hyperbola's branches.

Asymptotes

Asymptotes in a hyperbola are diagonal lines that the curve approaches but never actually touches. They provide a crucial guide for sketching the hyperbola. The asymptotes are defined using the relationship between the values \(a\), \(b\), and the center \((h, k)\).
In our exercise, these asymptotes are determined by:

• Using the formula \[ y = k \pm \frac{b}{a}(x - h) \]
• Substituting our known values gives the asymptote equations as\[ y = -3 \pm \frac{5}{3}(x-1) \]

These linear equations yield two lines that intersect at the center and guide the hyperbola's shape, offering boundaries to the curve’s extent.
Thus, one line has a positive slope, and the other, a negative slope, creating symmetric passageways that the hyperbola follows.