Problem 27 - Intermediate Algebra

Question

Graph each hyperbola. See Example 3. $$ \frac{(y+1)^{2}}{1}-\frac{(x-2)^{2}}{4}=1 $$

Step-by-Step Solution

Answer

The hyperbola is centered at (2, -1) with vertices at (2, 0) and (2, -2), and asymptotes given by y = 1/2x - 2 and y = -1/2x.

Step 1: Identify the Standard Form of Hyperbola

The given equation \( \frac{(y+1)^2}{1} - \frac{(x-2)^2}{4} = 1 \) resembles the standard form of a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) where \( (h, k) \) is the center. Compare the equation to the standard form to identify \( h, k, a^2, \) and \( b^2 \).

Step 2: Determine the Center

From the standard form, we identify \( h = 2 \) and \( k = -1 \). Therefore, the center of the hyperbola is at point \( (2, -1) \).

Step 3: Identify \( a \) and \( b \)

From the given equation, \( a^2 = 1 \) hence \( a = 1 \), and \( b^2 = 4 \) hence \( b = 2 \). This means the hyperbola opens vertically with \( a \) representing vertical distances and \( b \) representing horizontal distances.

Step 4: Locate the Vertices and Co-vertices

The vertices are \( a \) units up and down from the center \( (2, -1) \). Therefore, the vertices are \( (2, -1+1) = (2, 0) \) and \( (2, -1-1) = (2, -2) \). The co-vertices are \( b \) units horizontally from the center: \( (2+2, -1) = (4, -1) \) and \( (2-2, -1) = (0, -1) \).

Step 5: Find the Asymptotes

The equation of the asymptotes of a vertical hyperbola is \( y = k \pm \frac{a}{b}(x-h) \). Substitute \( a = 1 \), \( b = 2 \), \( h = 2 \), and \( k = -1 \) to get the equations: \( y = -1 \pm \frac{1}{2}(x-2) \). This simplifies to the two lines: \( y = -1 + \frac{1}{2}(x-2) \) and \( y = -1 - \frac{1}{2}(x-2) \), which further simplify to \( y = \frac{1}{2}x - 2 \) and \( y = -\frac{1}{2}x \).

Step 6: Sketch the Hyperbola

First, plot the center \( (2, -1) \). Plot the vertices \( (2, 0) \) and \( (2, -2) \), and co-vertices \( (0, -1) \) and \( (4, -1) \). Draw the asymptotes as dashed lines through the center with slopes \( \pm \frac{1}{2} \). Sketch the hyperbola opening vertically, approaching the asymptotes but never touching them.

Key Concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is crucial for graphing and analyzing these unique curves. The standard form for a vertical hyperbola is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). This formula helps us identify several key components:

• \( h \) and \( k \) are the coordinates of the center of the hyperbola.
• \( a^2 \) is always associated with the term containing \((y-k)^2\) or \((x-h)^2\), depending on the orientation of the hyperbola.
• \( b^2 \) is linked with the other term.

In our exercise, the given equation is \( \frac{(y+1)^{2}}{1}-\frac{(x-2)^{2}}{4}=1 \), matching the standard form of a vertical hyperbola. Placing it into the standard structure, you can easily read off these essential quantities and prepare to graph.

Center of a Hyperbola

The center of a hyperbola serves as the central reference point from which everything else in the hyperbola is gauged. It is at the intersection of the transverse and conjugate axes. For our specific equation, \( \frac{(y+1)^2}{1} - \frac{(x-2)^2}{4} = 1 \), the center is derived from \( (h, k) \) in the standard form.
By comparing, we discern that \( h = 2 \) and \( k = -1 \). Therefore, the center of this hyperbola is located at the point \( (2, -1) \).
Identifying the center is vital because it helps to determine other features such as vertices, co-vertices, and the path of nearby asymptotes.

Vertices and Co-vertices

Vertices and co-vertices are the special points on a hyperbola that guide its shape and orientation.

• **Vertices:** Located \( a \) units away from the center along the transverse axis.
  In a vertical hyperbola, they sit above and below the center. From \( (2, -1) \), moving 1 unit gives us vertices at \( (2, 0) \) and \( (2, -2) \).
• **Co-vertices:** Situated \( b \) units from the center along the conjugate axis.
  For our equation, moving 2 units horizontally from the center yields co-vertices at \( (0, -1) \) and \( (4, -1) \).

These points help in shaping the rectangular box used to sketch the asymptotes and guide the general curvature of the hyperbola.

Asymptotes of a Hyperbola

Asymptotes in a hyperbola are lines that the curve approaches but never touches. They provide a framework within which the branches of the hyperbola sit. For a vertical hyperbola, the asymptotes are given by the equations:\[ y = k \pm \frac{a}{b}(x-h) \]Substituting \( a = 1 \), \( b = 2 \), \( h = 2 \), and \( k = -1 \) from the exercise, we derive the asymptotes:

• \( y = -1 + \frac{1}{2}(x-2) \)
• \( y = -1 - \frac{1}{2}(x-2) \)

Simplifying these gives us the linear equations: \( y = \frac{1}{2}x - 2 \) and \( y = -\frac{1}{2}x \).
Plotting these lines as dashed lines on the graph helps in shaping the trajectory of the hyperbola's branches correctly. They guide the sketch, ensuring it correctly approaches these boundaries without crossing them.