Problem 11 - Precalculus: Functions and Graphs

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$\frac{(y+2)^{2}}{9}-\frac{(x+2)^{2}}{4}=1$$

Step-by-Step Solution

Answer

Vertices: \((-2, 1)\), \((-2, -5)\); Foci: \((-2, -2 \pm \sqrt{13})\); Asymptotes: \(y + 2 = \frac{3}{2}(x + 2)\), \(y + 2 = -\frac{3}{2}(x + 2)\).

Step 1: Identify the Hyperbola Form and Values

The given equation is \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\). This is in the standard form for a vertical hyperbola, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where the center is \((h, k)\), \(a^2 = 9\), and \(b^2 = 4\). Thus, \(a = 3\), \(b = 2\), and the center \((h, k)\) is \((-2, -2)\).

Step 2: Determine the Vertices

For a vertical hyperbola centered at \((-2, -2)\), the vertices are located \(a\) units up and down from the center. Therefore, the vertices are \((-2, -2+3) = (-2, 1)\) and \((-2, -2-3) = (-2, -5)\).

Step 3: Calculate the Foci

The distance from the center to each focus is given by \(c\) where \(c^2 = a^2 + b^2\). Thus, \(c^2 = 9 + 4 = 13\), so \(c = \sqrt{13}\). The foci are \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\).

Step 4: Find the Equations of the Asymptotes

The asymptotes for a vertical hyperbola \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) are given by the equations \((y-k) = \pm\frac{a}{b}(x-h)\). Thus, substituting the values, the equations are \[y + 2 = \pm\frac{3}{2}(x + 2)\].

Step 5: Sketch the Hyperbola and Key Features

Plot the center \((-2, -2)\), the vertices \((-2, 1)\) and \((-2, -5)\), and the foci \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\). Draw the asymptotes following the equations \(y + 2 = \frac{3}{2}(x + 2)\) and \(y + 2 = -\frac{3}{2}(x + 2)\) as dashed lines. Sketch the hyperbola opening vertically with respect to the center, approaching but never touching the asymptotes.

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are crucial points that help in understanding its shape and orientation. For the hyperbola given by \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\), the vertices can be easily determined once you know the center and the orientation.
Given that the center of the hyperbola is \((-2, -2)\) and it is oriented vertically,

• The distance from the center to each vertex is the value \(a\), which is 3 in this case.
• This means the vertices are located 3 units above and below the center along the y-axis.
  The two vertices are thus at \((-2, 1)\) and \((-2, -5)\).

Foci of a Hyperbola

The foci of a hyperbola are fixed points inside each branch of the hyperbola. They play a critical role in defining the shape and the curves of the hyperbola.
To find the foci for our hyperbola

• We compute the distance \(c\) using the relationship \(c^2 = a^2 + b^2\).
• Given that \(a = 3\) and \(b = 2\), we calculate \(c^2 = 9 + 4 = 13\) thus, \(c = \sqrt{13}\).
• The foci, located \(c\) units up and down from the center along the y-axis, are at \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\).

Asymptotes of a Hyperbola

The asymptotes are lines that the hyperbola approaches but never touches. They define the direction in which the branches of the hyperbola extend.
For a vertical hyperbola like ours, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the equations of the asymptotes are:

• \((y-k) = \pm \frac{a}{b}(x-h)\), where \((h, k)\) is the center.
• Substituting our values, we get:
  \(y + 2 = \pm \frac{3}{2}(x + 2)\).
  These are the equations of the asymptotes for this hyperbola.

Equation of a Hyperbola

The equation of a hyperbola helps to specify its geometry and orientation. It consists of a standard form that identifies key parameters.
Our hyperbola is given by the equation \(\frac{(y+2)^2}{9} - \frac{(x+2)^2}{4} = 1\), which is characteristic of a vertically oriented hyperbola:

• The center \((h, k)\) is at \((-2, -2)\).
• Parameters \(a^2 = 9\) and \(b^2 = 4\) give us \(a = 3\) and \(b = 2\) respectively, directing the distances to vertices and foci.Understanding how these parameters fit into the formula allows for easy determination of other features like vertices and asymptotes.

Graphing Hyperbolas

Graphing a hyperbola involves identifying and plotting several key components. Follow these steps to sketch your hyperbola:

• Begin by plotting the center \((-2, -2)\).
• Next, draw the vertices \((-2, 1)\) and \((-2, -5)\). These determine the general "spine" of the hyperbola.
• Locate the foci \((-2, -2 + \sqrt{13})\) and \((-2, -2 - \sqrt{13})\) within each curve branch.
• Draw the asymptotes given by \(y + 2 = \frac{3}{2}(x + 2)\) and \(y + 2 = -\frac{3}{2}(x + 2)\) as dashed lines. These lines give approximate guides for the hyperbola's approach but will never intersect the graph.
• Finally, sketch the two branches of the hyperbola, making sure they curve around but never touch the asymptotes.

This comprehensive visual process reinforces understanding and aids in accurate representation of hyperbolas.