Problem 36 - College Algebra

Question

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. $$ \frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1 $$

Step-by-Step Solution

Answer

Center: (2, -3); Vertices: (2±2√2, -3); Foci: (2±√35, -3).

Step 1: Identify the Center

The equation of the hyperbola is given as \( \frac{(x-2)^{2}}{8}-\frac{(y+3)^{2}}{27}=1 \). This indicates that the center of the hyperbola is at \((h, k)=(2, -3)\).

Step 2: Determine the Vertices

For a hyperbola of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the vertices are located \(a\) units from the center along the x-axis. Here, \(a^2 = 8\), so \(a = \sqrt{8} = 2\sqrt{2}\). Thus, the vertices are \((2+2\sqrt{2}, -3)\) and \((2-2\sqrt{2}, -3)\).

Step 3: Find the Foci

The distance to the foci, \(c\), is found using the relation \(c^2 = a^2 + b^2\). Here \(b^2 = 27\), so \(c^2 = 8 + 27 = 35\) and \(c = \sqrt{35}\). Thus, the foci are located at \((2+\sqrt{35}, -3)\) and \((2-\sqrt{35}, -3)\).

Step 4: Sketch the Graph

Draw the hyperbola centered at \((2, -3)\). Mark the vertices at \((2+2\sqrt{2}, -3)\) and \((2-2\sqrt{2}, -3)\). Then, plot the foci at \((2+\sqrt{35}, -3)\) and \((2-\sqrt{35}, -3)\). The hyperbola opens horizontally since the \(x\) term comes first.

Key Concepts

Equations of Hyperbolas

The equation of a hyperbola has a standard form which is crucial to its identification and graphing. In the equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), each symbol represents a significant component. The variables \(h\) and \(k\) denote the center of the hyperbola, while \(a\) and \(b\) are related to the distances from the center to the vertices and axes of symmetry. The positive term, \((x-h)^2/a^2\), indicates the direction in which the hyperbola opens. If the term with \(x\) is positive, it opens along the x-axis. Conversely, if the \(y\) term is positive, it opens along the y-axis. Understanding this form helps determine the hyperbola's orientation and critical points such as vertices and foci.

Graphing Hyperbolas

Graphing hyperbolas involves a few key steps. First, identify the center of the hyperbola from the equation. In the exercise, the center is at \((2, -3)\). This center acts as the base point for plotting other components. Next, calculate the distances \(a\) and \(b\), which are derived from \(a^2\) and \(b^2\) in the equation. Using these, determine the vertices and foci. The vertices fall on the transverse axis, which dictates the direction the hyperbola opens—horizontally or vertically. Finally, using these points as a guide, draw the two curves of the hyperbola, ensuring they approach the asymptotes, which are imaginary lines guiding the shape. This process creates a visual map of the equation.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These include circles, ellipses, parabolas, and hyperbolas—each with unique properties. Hyperbolas are distinct because they have two separate curves or branches, resulting from the intersecting plane being parallel to the cone's axis. Their equations reflect this duality with subtraction instead of addition, unlike ellipses. Understanding conic sections involves recognizing these curves and their defining features, such as their centers, axes, and symmetry properties. Hyperbolas, in particular, are characterized by their symmetry about both axes and their open-ended structures.

Vertices of Hyperbolas

Vertices are critical points on a hyperbola, marking where the curve is closest to the transverse axis. For the hyperbola \( \frac{(x-2)^2}{8} - \frac{(y+3)^2}{27} = 1 \), the vertices are calculated by determining \(a\), which is \(\sqrt{8} = 2\sqrt{2}\). These vertices are then positioned \(a\) units from the center \((2, -3)\) along the x-axis, at \((2+2\sqrt{2}, -3)\) and \((2-2\sqrt{2}, -3)\). They act as a foundation for the hyperbola's shape and provide direction for its opening. Precisely plotting these points is essential for accurately sketching the hyperbola's graph.