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Chapter 7: Problem 8

Graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. $$ \frac{(x-4)^{2}}{8}-\frac{(y-2)^{2}}{18}=1 $$

Short Answer

Expert verified
Center: (4, 2); Vertices: (4卤2鈭�2, 2); Foci: (4卤鈭�26, 2); Asymptotes: y = 2 卤 3/2(x-4); Transverse axis: along x.

Step by step solution

01

Identify the Standard Form

The given equation of the hyperbola is \( \frac{(x-4)^{2}}{8}-\frac{(y-2)^{2}}{18}=1 \). This is in the standard form for a hyperbola: \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \), where \((h, k)\) is the center, \(a^{2} = 8\), and \(b^{2} = 18\).
02

Determine the Center

From the standard form \((x-h)^{2}\) and \((y-k)^{2}\), the center \((h, k)\) is \((4, 2)\).
03

Calculate the Vertices

The vertices are located \(a\) units from the center along the transverse axis. Here, \(a^2 = 8\) so \(a = \sqrt{8} = 2\sqrt{2}\). Since the hyperbola is horizontal (as it subtracts the \(y\) term), the vertices are at \((4 \pm 2\sqrt{2}, 2)\).
04

Find the Foci

Calculate \(c\) using \(c^{2} = a^{2} + b^{2}\). So, \(c^{2} = 8 + 18 = 26\) and \(c = \sqrt{26}\). The foci are at \((4 \pm \sqrt{26}, 2)\).
05

Determine Asymptotes

The asymptotes for a horizontal hyperbola are given by \(y = k \pm \frac{b}{a}(x-h)\). Here, \(k = 2\), \(h = 4\), \(a = 2\sqrt{2}\), and \(b = 3\sqrt{2}\). Thus, the asymptotes are \(y = 2 \pm \frac{3\sqrt{2}}{2\sqrt{2}}(x-4)\), simplifying to \(y = 2 \pm \frac{3}{2}(x-4)\).
06

Transverse and Conjugate Axes

The transverse axis is parallel to the \(x\)-axis and includes the points \((4 \pm 2\sqrt{2}, 2)\). The conjugate axis is parallel to the \(y\)-axis and is not directly visible from the equation; it includes points vertically centered at \((4, 2 \pm 3\sqrt{2})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Center of a Hyperbola
The center of a hyperbola is the midpoint around which the hyperbola opens. This point functions similarly to the center of a circle or an ellipse, but in the case of the hyperbola, it signifies where the branches start to diverge in the opposite directions. The general equation of a hyperbola in standard form is \( \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \). In this equation, \((h, k)\) represents the center. Let鈥檚 look at the exercise example:
  • The given equation is \( \frac{(x-4)^{2}}{8} - \frac{(y-2)^{2}}{18} = 1 \). This indicates that the center of the hyperbola is at \((4, 2)\).
  • The center is determined by the values inside the parentheses: \((x-h)\) and \((y-k)\).
Knowing the center is essential because it helps locate other important features like the vertices and foci. By finding these coordinates, we set the foundation to graph the hyperbola accurately.
Defining the Vertices of a Hyperbola
Vertices are the points on the hyperbola that lie closest to the center on its branches. They help in understanding how the hyperbola is oriented. For a horizontal hyperbola, vertices extend along the x-axis from the center. Here's how we find them:
  • Vertices are located \(a\) units away from the center along the transverse axis. In our example, \(a^2 = 8\), which gives \(a = \sqrt{8} = 2\sqrt{2}\).
  • Since the hyperbola is horizontal, the calculation leads to vertices at \((4 \pm 2\sqrt{2}, 2)\).
The vertices indicate the direction in which the hyperbola opens. This specific placement along the transverse axis highlights whether your hyperbola stretches horizontally or vertically.
Finding the Foci of a Hyperbola
The foci (singular: focus) are key points that further define the hyperbola's shape. They represent the points from which the distances to any point on the hyperbola differ by a constant amount. To find the foci:
  • We calculate \(c\) using the relationship \(c^{2} = a^{2} + b^{2}\). For our equation, \(c^{2} = 8 + 18 = 26\), so \(c = \sqrt{26}\).
  • In this horizontal hyperbola, the foci are positioned at \((4 \pm \sqrt{26}, 2)\).
The foci extend along the same axis as the vertices, which for our example is horizontal. These points are significant as they help determine the eccentricity and overall stretch of the hyperbola.
Understanding the Asymptotes of a Hyperbola
Asymptotes are lines that the hyperbola approaches but never actually touches. They give the hyperbola its unique open-ended shape. For hyperbolas, these lines are a crucial guide for sketching their branches.To find the asymptotes, we use the following formula for a horizontal hyperbola:
  • The equation: \(y = k \pm \frac{b}{a}(x-h)\).
  • From the given example, substituting the values: \(k = 2\), \(h = 4\), \(a = 2\sqrt{2}\), \(b = 3\sqrt{2}\).
  • This results in the asymptotes being \(y = 2 \pm \frac{3}{2}(x-4)\).
Asymptotes help in the proper sketching of the graph showing the trajectory and boundary lines that the hyperbola's branches follow. Knowing these aids in comprehending how the hyperbola behaves at infinity and gives a complete picture of its layout.

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