Problem 32 - Intermediate Algebra

Question

Graph each hyperbola. See Example 3. $$ \frac{(y-1)^{2}}{9}-\frac{x^{2}}{9}=1 $$

Step-by-Step Solution

Answer

The hyperbola is centered at (0, 1) with vertices at (0, 4) and (0, -2); asymptotes are y = x + 1 and y = -x + 1.

Step 1: Identify the Standard Form

The given equation of the hyperbola is \( \frac{(y-1)^2}{9} - \frac{x^2}{9} = 1 \). This matches the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) for a hyperbola that opens vertically.

Step 2: Determine the Center

From the standard equation \( \frac{(y-1)^2}{9} - \frac{x^2}{9} = 1 \), we identify the center as \((h, k)\), which in this case is \((0, 1)\).

Step 3: Find the Vertices

The vertices of the hyperbola are determined by \( (h, k \pm a) \). Given \( a^2 = 9 \), we have \( a = 3 \). Therefore, the vertices are at \((0, 1+3)\) and \((0, 1-3)\), which gives us points \((0, 4)\) and \((0, -2)\).

Step 4: Find the Co-vertices

The co-vertices, though not directly required for the graph, help in understanding the asymptotes. They are determined by \((h \pm b, k)\). With \( b^2 = 9 \), \( b = 3 \). Thus, the co-vertices are at \((3, 1)\) and \((-3, 1)\).

Step 5: Find the Asymptotes

The asymptotes of the hyperbola can be found using \( y - k = \pm \frac{a}{b} (x - h) \). Substituting \( a = 3 \), \( b = 3 \), \( h = 0 \), and \( k = 1 \), we get the equations \( y - 1 = \pm (x) \). Hence, the asymptotes are \( y = x + 1 \) and \( y = -x + 1 \).

Step 6: Graph the Hyperbola

To graph the hyperbola, plot the center \((0,1)\), vertices \((0,4)\) and \((0,-2)\), and draw the asymptotes \(y = x + 1\) and \(y = -x + 1\). The hyperbola branches appear to be extending vertically along the \(y\)-axis, approaching but never touching the asymptotes.

Key Concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is essential for graphing it correctly. The typical equation for a hyperbola that opens directly or vertically is given by:

• For a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
• For a horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)

In these equations, \((h, k)\) represent the center of the hyperbola. The variable \(a\) indicates the distance from the center to each vertex along the axis of symmetry, and \(b\) represents the distance to the co-vertices. Once you identify the standard form, as in the exercise equation \( \frac{(y-1)^2}{9} - \frac{x^2}{9} = 1 \), it becomes straightforward to determine the orientation and important features of the hyperbola, such as its center, vertices, and asymptotes.

Vertices of a Hyperbola

The vertices of a hyperbola are crucial points that help shape the graph. They lie along the line that passes through the center and are \(a\) units away from the center along the axis of symmetry.
In the equation \( \frac{(y-1)^2}{9} - \frac{x^2}{9} = 1 \), \(a^2 = 9\) which means \(a = 3\).
For a vertically oriented hyperbola like this one, the vertices are placed at \((h, k \pm a)\).

• The calculation for the vertices in this example would be: \((0, 1 + 3)\) and \((0, 1 - 3)\)
• This results in the points \((0, 4)\) and \((0, -2)\)

The vertices represent where the two branches of the hyperbola begin to curve away from the axes, and they are symmetrically placed on opposite sides of the center.

Asymptotes of a Hyperbola

Asymptotes are the guidelines the branches of a hyperbola approach but never actually touch. They provide valuable orientation for sketching the hyperbola.
To find asymptotes for the hyperbola, we use the formula:

• \(y - k = \pm \frac{a}{b}(x - h)\)

In our given equation, \(a = 3\) and \(b = 3\), which simplifies the asymptote formulas to:

• \(y - 1 = \pm (x)\)

Thus, the asymptotes are the lines:

• \(y = x + 1\)
• \(y = -x + 1\)

Plotting these lines on the graph provides a clear framework for how the branches of the hyperbola extend towards infinity, ensuring that the branches snugly curve around these guidelines.

Center of a Hyperbola

The center of a hyperbola is like its heart, acting as the central hub from which all other key features are drawn. In the standard form of a hyperbola equation, it's specified by the coordinates \((h, k)\).
For the equation \( \frac{(y-1)^2}{9} - \frac{x^2}{9} = 1 \), the center is clearly indicated as \((0, 1)\).

• This position serves as the point of symmetry for the hyperbola.
• From the center, vertices and co-vertices are equidistant, ensuring balance in the hyperbola’s structure.

When graphing, marking the center correctly is vital as it acts as the reference point for positioning all other elements like vertices and asymptotes. Always check the signs within the equation to accurately determine its location.