Problem 38 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes \(\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is (-1,2). The foci of the hyperbola are at (-1, 2±\(\sqrt{85}\)). The equations for asymptotes are \(y = 2 ± \frac{6}{7}(x+1)\).

Step 1: Find Center, semi-major axis(a), semi-minor axis(b)

The equation of the hyperbola is in the form \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) or \(\frac{(y-k)^2}{a^2} - \(\frac{(x-h)^2}{b^2}=1\). Here, (h,k) is the center, 'a' is semi-major axis and 'b' is semi-minor axis. From the given equation, the center (h,k) of the hyperbola is (-1,2), 'a' (semi-major axis) is \(\sqrt{36}\) or 6, 'b' (semi-minor axis) is \(\sqrt{49}\) or 7.

Step 2: Calculate c (distance from center to foci)

The distance 'c' is given by the equation \(c=\sqrt{a^2+b^2}\). Substituting a=6 and b=7 in the equation, we get c=\(\sqrt{6^2+7^2}\) or c=\(\sqrt{85}\).

Step 3: Find foci

The foci of the hyperbola are at (h, k±c) since this is a vertical hyperbola. Substituting h=-1, k=2 and c=\(\sqrt{85}\), the foci are at (-1, 2±\(\sqrt{85}\)).

Step 4: Formulate the equations of the asymptotes

The equations for asymptotes of a vertical hyperbola are given by \(k ± \frac{a}{b}(x-h)\) . Substituting the values of h=-1, k=2, a=6 and b=7, we get the equations of asymptotes as \(y = 2 ± \frac{6}{7}(x+1)\).

Key Concepts

Center of a Hyperbola

The center of a hyperbola is a pivotal concept when it comes to understanding this type of conic section. Think of it as the heart of the hyperbola from which all other features are defined. In the equation \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2}=1\) or \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\), the \(h\) and \(k\) represent the coordinates of the center. For our example hyperbola \(\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1\), the center is at the point (-1, 2).

Understanding where the center lies is essential because it helps us graph the hyperbola accurately. The center also marks the midpoint between the vertices and the foci, and it is at the intersection of the asymptotes, which are invisible lines that the hyperbola approaches but never touches. Conceptualizing the center can greatly simplify the process of graphing and analyzing a hyperbola.

Vertices of a Hyperbola

The vertices of a hyperbola are another key feature and are closely linked to the center. The vertices are the points of intersection between the hyperbola and its major axis. For a vertical hyperbola like ours, the vertices can be found by adding and subtracting the \(a\) value from the \(k\) coordinate of the center along the vertical axis. In this case, since our \(a\) value is the square root of 36, which is 6, we add and subtract this from the \(k\) value of the center (2).

The vertices of the given hyperbola are therefore at the points (-1, 2-6) and (-1, 2+6), or more simply, at (-1, -4) and (-1, 8). These vertices are instrumental when sketching the hyperbola because they provide concrete points through which the curve will pass. They offer a visual boundary for the shape of the hyperbola and are crucial for plotting its precise curvature on a graph.

Asymptotes of a Hyperbola

The asymptotes of a hyperbola act as the guiding lines for its shape, dictating the direction in which the curves of the hyperbola will extend infinitely. Essentially, they are the 'crosshairs' that will never be touched by the hyperbola, yet they frame it. For our hyperbola equation \(\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1\), the equations of the asymptotes are based on the slopes found using the \(a\) and \(b\) values from the hyperbola's equation. The general form for the asymptotes of a vertical hyperbola is \(y=k±\frac{a}{b}(x-h)\).

By plugging in the values from our example, \(h=-1\), \(k=2\), \(a=6\), and \(b=7\), we derive the equations for the asymptotes to be \(y=2±\frac{6}{7}(x+1)\). These lines will slant across the graph and will help shape the hyperbola, providing a clear outline for its open arms to curl along endlessly. Recognizing how to calculate and draw the asymptotes aids students in visualizing the overall structure of the hyperbola and its behavior as it stretches towards infinity.