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Chapter 11: Problem 9

Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$ \frac{(x+1)^{2}}{9}-\frac{(y-3)^{2}}{16}=1 $$

Short Answer

Expert verified
Center: (-1, 3); Vertices: (2, 3), (-4, 3); Foci: (4, 3), (-6, 3); Asymptotes: y = 3 卤 4/3(x + 1).

Step by step solution

01

Standard Form Identification

The given equation is \( \frac{(x+1)^{2}}{9}-\frac{(y-3)^{2}}{16}=1 \). This is the standard form of a horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), where the center is \((h, k)\), \(a^2 = 9\), and \(b^2 = 16\).
02

Find the Center

From the standard form, \( (h, k) = (-1, 3) \). This is derived from \((x+1)^2\) and \((y-3)^2\).
03

Calculate Vertices

For a horizontal hyperbola, the vertices are located at \((h \pm a, k)\). Here, \(a = \sqrt{9} = 3\). Therefore, the vertices are \((-1 \pm 3, 3)\), or \((2, 3)\) and \((-4, 3)\).
04

Identify Foci

Foci are calculated using \(c^2 = a^2 + b^2\). So, \(c^2 = 9 + 16 = 25 \), hence \(c = 5\). The foci are \((h \pm c, k)\), which are \((-1 \pm 5, 3)\), or \((4, 3)\) and \((-6, 3)\).
05

Determine Asymptotes

Asymptotes for a horizontal hyperbola are given by \(y = k \pm \frac{b}{a}(x - h)\). Substituting \(a = 3\), \(b = 4\), \(h = -1\), and \(k = 3\), the asymptotes are \(y = 3 \pm \frac{4}{3}(x + 1)\).
06

Sketch the Graph

Using the center \((-1, 3)\), plot the vertices \((2, 3)\) and \((-4, 3)\), and the foci \((4, 3)\) and \((-6, 3)\). Draw the rectangle bounded by \(a\) and \(b\) to guide the hyperbola's direction, and sketch the asymptotes \(y = 3 \pm \frac{4}{3}(x + 1)\). Finally, draw the hyperbola opening left and right, centered at \((-1, 3)\).

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Key Concepts

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

Horizontal Hyperbola
A horizontal hyperbola is a specific type of hyperbola where the transverse axis runs horizontally. This simply means that the opening of the hyperbola is left and right from its center.
For a hyperbola to be horizontal, its standard equation must have the form:
  • \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
Here,
  • \((h, k)\) represents the center of the hyperbola. In our exercise, these values are \( (-1, 3) \).
  • The positive term in the equation (\( (x-h)^2 \)) indicates that this is a horizontal hyperbola.
  • \( a \) and \( b \) are the distances related to the shape and orientation of the hyperbola.
The orientation determines the layout on a graph, and in this case, the hyperbola opens left and right.
Vertices
The vertices of a hyperbola are critical points that determine its shape and help in plotting the graph.
For a horizontal hyperbola, the vertices are located at a fixed distance from the center along the horizontal axis.
This distance is given by the value \( a \), where \( a = \sqrt{9} = 3 \) in our case.
Using the center \((h, k) = (-1, 3)\), the vertices can be calculated as follows:
  • \((h + a, k) = (-1 + 3, 3) = (2, 3)\)
  • \((h - a, k) = (-1 - 3, 3) = (-4, 3)\)
Thus, the vertices \((2, 3)\) and \((-4, 3)\) are along the horizontal line \( y = 3 \), showing the direction in which the hyperbola opens.
Foci
Foci are two important points on the major axis of a hyperbola that help in accurately defining its shape. They lie further out from the center than the vertices.
To locate the foci, we use the relationship \( c^2 = a^2 + b^2 \). From the problem:
  • \( a^2 = 9 \) and \( b^2 = 16 \)
  • Thus, \( c^2 = 9 + 16 = 25 \) => \( c = 5 \)
Given the center \((h, k) = (-1, 3)\), the foci are found by placing \( c \) units to the left and right of the center:
  • \((h + c, k) = (-1 + 5, 3) = (4, 3)\)
  • \((h - c, k) = (-1 - 5, 3) = (-6, 3)\)
These foci \((4, 3)\) and \((-6, 3)\) are crucial for graphing because they help in identifying the true shape and orientation of the hyperbola on the graph.
Asymptotes
Asymptotes are lines that the branches of the hyperbola approach but never touch. They guide the shape of the hyperbola and give a sense of its openness.
For a horizontal hyperbola, asymptotes are given by the formula:
  • \( y = k \pm \frac{b}{a}(x - h) \)
Using our problem values:
  • \( k = 3 \), \( h = -1 \), \( b = 4 \), \( a = 3 \)
  • The equations become:
  • \( y = 3 \pm \frac{4}{3}(x + 1) \)
Which translates to two asymptotes:
  • \( y = 3 + \frac{4}{3}(x + 1) \)
  • \( y = 3 - \frac{4}{3}(x + 1) \)
These lines are essential when sketching the hyperbola, ensuring it doesn't touch the asymptotes but approaches them as it extends outward.

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