Problem 16 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(\frac{(y-6)^{2}}{36}-\frac{(x+1)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The vertices are (-1, 12) and (-1, 0), foci are (-1, 6±2√13), and asymptotes are y=3/2x+15/2 and y=-3/2x+9/2.

Step 1: Identify the Standard Form of Hyperbola

The given equation \( \frac{(y-6)^{2}}{36} - \frac{(x+1)^{2}}{16} = 1 \) is already in the standard form of a hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). This indicates that the hyperbola is vertical.

Step 2: Determine the Center

In the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center of the hyperbola is at the point \((h, k)\). Here, \(h = -1\) and \(k = 6\), so the center is \((-1, 6)\).

Step 3: Find the Vertices

For a vertical hyperbola, the vertices are located \(a\) units above and below the center. Here, \(a^2 = 36\) so \(a = 6\). The vertices are at \((-1, 6 \pm 6) = (-1, 12)\) and \((-1, 0)\).

Step 4: Determine the Foci

The foci are located \(c\) units from the center along the axis of symmetry. For a hyperbola, \(c^2 = a^2 + b^2\). Here, \(b^2 = 16\) so \(c^2 = 36 + 16 = 52\) and \(c = \sqrt{52} = 2\sqrt{13}\). The foci are at \((-1, 6 \pm 2\sqrt{13})\).

Step 5: Write Equations of Asymptotes

The equations of the asymptotes for a vertical hyperbola are given by \(y = k \pm \frac{a}{b}(x - h)\). Substituting \(a = 6, b = 4, h = -1,\) and \(k = 6\), the asymptotes are \(y = 6 \pm \frac{3}{2}(x + 1)\), which simplifies to \(y = \frac{3}{2}x + \frac{15}{2}\) and \(y = -\frac{3}{2}x + \frac{9}{2}\).

Key Concepts

Standard Form

The standard form of a hyperbola's equation is crucial because it provides insight into its overall shape and orientation. A hyperbola's equation often appears as:

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

In these equations, the variables \(h\) and \(k\) are the coordinates of the center of the hyperbola. The values \(a\) and \(b\) represent the distances from the center to the vertices and the co-vertices, respectively. The given equation, \( \frac{(y-6)^{2}}{36} - \frac{(x+1)^{2}}{16} = 1 \), is already in standard form which immediately tells us the hyperbola is vertical. The constants inform us that the center of the hyperbola is at \((-1, 6)\), setting the stage for finding the other key components of the hyperbola.

Vertices

Vertices are essential in defining and visualizing the shape of a hyperbola. For a vertical hyperbola, the vertices are positioned along the line that passes through the center and runs parallel to the \(y\)-axis. From the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), we derive the vertices using the formula:

• Vertices: \((h, k \pm a)\)

In our exercise, we have \(a^2 = 36\), giving us \(a = 6\). So, the vertices are at \((-1, 6 + 6) = (-1, 12)\) and \((-1, 6 - 6) = (-1, 0)\). These points indicate where the hyperbola intersects the line passing through the center and running vertically. Knowing the location of the vertices helps to identify how wide the branches of the hyperbola open.

Foci

Foci are points located further out from the center than the vertices, and they help define the shape and "stretch" of the hyperbola. The foci are found at a distance \(c\) from the center along the major axis. For a hyperbola, \(c\) is determined using the equation \(c^2 = a^2 + b^2\). In this problem:

• \(c^2 = 36 + 16 = 52\)
• \(c = \sqrt{52} = 2\sqrt{13}\)

Thus, the foci of our hyperbola are positioned at \((-1, 6 + 2\sqrt{13})\) and \((-1, 6 - 2\sqrt{13})\). Understanding the position of the foci is crucial because they relate to the eccentricity and directness of the branches in a hyperbola, offering deeper insights into its geometry.

Asymptotes

Asymptotes are lines that pass through the center of the hyperbola and illustrate how the branches approach these lines but never actually touch them. The asymptotes guide the hyperbola's branches as they extend towards infinity. For a vertical hyperbola like ours, the equations for the asymptotes are:

• \(y = k \pm \frac{a}{b}(x - h)\)

In substituting \(a = 6, b = 4, h = -1,\) and \(k = 6\), we derive:

• \(y = 6 + \frac{3}{2}(x + 1)\) and \(y = 6 - \frac{3}{2}(x + 1)\)
• Simplifying these to: \(y = \frac{3}{2}x + \frac{15}{2}\) and \(y = -\frac{3}{2}x + \frac{9}{2}\)

Understanding asymptotes aids in sketching the hyperbola and providing further insights into the behavior and nature of its branches as they extend outward. Asymptotes act like boundaries enclosing the shape, ensuring the exponential nature of the hyperbola is portrayed correctly.