Problem 16 - College Algebra

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

Center: (-1, 6), Vertices: (-1, 12) and (-1, 0), Foci: \((-1, 6 \pm 2\sqrt{13})\), Asymptotes: \(y = \frac{3}{2}(x+1) + 6\) and \(y = -\frac{3}{2}(x+1) + 6\).

Step 1: Identify the form of the equation

The given equation \( \frac{(y-6)^2}{36} - \frac{(x+1)^2}{16} = 1 \) is already in the standard form of a vertical hyperbola, which is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). This identifies it as a vertical hyperbola where the center is at \((h, k)\).

Step 2: Locate the center

From the equation \( \frac{(y-6)^2}{36} - \frac{(x+1)^2}{16} = 1 \), we can identify the center of the hyperbola as \((h, k) = (-1, 6)\).

Step 3: Determine the vertices

The vertices of a vertical hyperbola are located \(a\) units above and below the center. Given \(a^2 = 36\), we calculate \(a = \sqrt{36} = 6\). Thus, the vertices are at \((h, k \pm a) = (-1, 6 \pm 6)\), which are \((-1, 12)\) and \((-1, 0)\).

Step 4: Locate the foci

For hyperbolas, \(c^2 = a^2 + b^2\). Given \(a^2 = 36\) and \(b^2 = 16\), we find \(c^2 = 36 + 16 = 52\), so \(c = \sqrt{52} = 2\sqrt{13}\). Thus, the foci are at \((h, k \pm c) = \left(-1, 6 \pm 2\sqrt{13}\right)\).

Step 5: Identify the equations of asymptotes

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

Key Concepts

Standard Form of Hyperbola

A hyperbola is a fascinating geometric shape defined by its standard form equation. In general, a hyperbola can have two different orientations: horizontal and vertical. The standard form for a vertical hyperbola is given by:\[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]Here, \((h, k)\) represents the center of the hyperbola. The denominators, \({a^2}\) and \({b^2}\), define the distances from the center to the vertices and the asymptotes, respectively. For horizontal hyperbolas, the roles of \(x\) and \(y\) would be swapped.

• The term with \(y\) in the numerator indicates it's a vertical hyperbola.
• The hyperbola given in the problem, \( \frac{(y-6)^2}{36} - \frac{(x+1)^2}{16} = 1 \), is already in standard form.

The center of this hyperbola is located at \((-1, 6)\), as shown by the equations \(y-6\) and \(x+1\). Understanding this form helps to identify the various characteristics and confidently work through the subsequent steps.

Vertices of Hyperbola

Vertices are crucial in defining the shape of a hyperbola. For a vertical hyperbola, vertices lie along the vertical line passing through the center. The standard form provides a simple way to determine these key points.Recall that \(a\) represents the distance from the center to each vertex. In our example, the equation shows that \(a^2 = 36\). Thus, \(a = \sqrt{36} = 6\).

• The vertices are calculated as \( (h, k \pm a) = (-1, 6 \pm 6) \).
• Resulting coordinates are \((-1, 12)\) and \((-1, 0)\).

These vertices give the hyperbola its stretched shape, extending upwards and downwards from the center. Clearly seeing the position of these vertices helps solidify our understanding of the hyperbola's orientation and scale.

Foci of Hyperbola

Foci are deep within the structure of a hyperbola, adding layers to its geometric properties. Unlike the vertices, foci are slightly further away from the center. The rule here is: \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to each focus.In the equation, \(a^2 = 36\) and \(b^2 = 16\). Solving \(c^2 = 36 + 16 = 52\) gives us \(c = \sqrt{52} = 2\sqrt{13}\).

• Then, the foci are located at \( (h, k \pm c) = \left(-1, 6 \pm 2\sqrt{13}\right) \).
• This results in the foci points at the positions \((-1, 6 + 2\sqrt{13})\) and \((-1, 6 - 2\sqrt{13})\).

Understanding the placement of foci within the hyperbola is essential for appreciating its focal alignment, distinguishing it markedly from other conic sections like ellipses.

Equations of Asymptotes

Asymptotes are the imaginary lines that a hyperbola approaches but never touches. For vertical hyperbolas, these lines pass diagonally, crossing the center and extending infinitely.The standard formula for the equations of asymptotes of a vertical hyperbola is:\[y - k = \pm \frac{a}{b}(x-h)\]Substituting the known values from our problem: \(a = 6\), \(b = 4\), \(h = -1\), and \(k = 6\):

• The equation becomes \(y - 6 = \pm \frac{6}{4}(x + 1)\).
• Simplified, it splits into two lines: y = \frac{3}{2}(x + 1) + 6\ and y = -\frac{3}{2}(x + 1) + 6\.

These two lines guide the hyperbola’s open arms, showing how the graph will stretch towards these lines without ever reaching them. Asymptotes help illustrate the hyperbola's expansive nature, reinforcing its role as a constantly broadening shape.