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Chapter 10: Problem 54

Write an equation of the hyperbola that satisfies each set of conditions. vertices \((6,-6)\) and \((0,-6),\) foci \((3 \pm \sqrt{13},-6)\)

Short Answer

Expert verified
The equation of the hyperbola is \(\frac{(x-3)^2}{9} - \frac{(y+6)^2}{4} = 1.\)

Step by step solution

01

Determine the Orientation

The vertices are given as (6,-6) and (0,-6), which means the hyperbola's vertices lie along the x-axis because the y-coordinate is the same for both. Therefore, the hyperbola opens horizontally. The standard equation of a hyperbola that opens horizontally is \[(x-h)^2/a^2 - (y-k)^2/b^2 = 1.\]
02

Find the Center

The center of the hyperbola is the midpoint of the line segment joining the vertices. The midpoint formula \(\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\) gives us:\[ h = \frac{6+0}{2} = 3, \, k = \frac{-6+-6}{2} = -6.\]Thus, the center is \((3, -6)\).
03

Calculate \(a^2\)

The distance between the vertices is equal to \(2a\). The distance between the vertices \((6, -6)\) and \((0, -6)\) is \(6 - 0 = 6.\) Thus, \(2a = 6,\) so \(a = 3,\) and therefore, \(a^2 = 9.\)
04

Calculate \(c^2\) Using the Foci

The distance from the center to each focus is \(c\). The foci are given as \((3 \pm \sqrt{13}, -6)\), so \(c = \sqrt{13}\) and \(c^2 = 13.\)
05

Find \(b^2\) Using \(a^2\) and \(c^2\)

For a hyperbola, the relationship \(c^2 = a^2 + b^2\) must hold. Substituting the known values, we get:\[ c^2 = a^2 + b^2.\]\[ 13 = 9 + b^2.\]Solving for \(b^2\), we find \[ b^2 = 4.\]
06

Write the Equation of the Hyperbola

Substitute \(h\), \(k\), \(a^2\), and \(b^2\) into the standard hyperbola equation:\[ \frac{(x-3)^2}{9} - \frac{(y+6)^2}{4} = 1.\]

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Key Concepts

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

Vertices of a Hyperbola
In a hyperbola, the vertices are key points where each branch of the hyperbola is closest to its center. They define how "wide" a hyperbola opens along its axis. Given vertices \(6,-6\) and \(0,-6\), it's clear that they lie horizontally. This indicates the hyperbola opens horizontally, which means the standard form of the equation is typically: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.\]
  • The vertices are equally spaced from the center in opposite directions along the transverse axis.
  • The distance between the vertices, which is \(2a\), helps determine the "stretch" of the hyperbola along the transverse axis.
  • In this problem, since the distance between the vertices is \(6\), we find that \(a = 3\).
In summary, the vertices play an essential role in determining the hyperbola's width and direction.
Foci of a Hyperbola
Every hyperbola has two foci (plural of focus), which are points used to define the hyperbola's shape and properties. Foci of a hyperbola are always located along the major axis. For this specific hyperbola, the foci are \(3 \pm \sqrt{13}, -6\).
  • The distance from the center to each focus is denoted as \(c\).
  • Using the given foci, it's determined that \(c = \sqrt{13}\).
  • For hyperbolas, the relationship is \(c^2 = a^2 + b^2\).
The foci help us understand the "eccentricity" and overall behavior of the hyperbola. Unlike ellipses, the foci of hyperbolas are positioned outside the vertices, which influences the curve's sharp opening angle.
Center of a Hyperbola
The center of a hyperbola is the midpoint between its two vertices. This point serves as a reference for the symmetry of the hyperbola. The center provides a pivot point around which the branches curve away. Given vertices \(6,-6\) and \(0,-6\), the center can be calculated using the midpoint formula:
  • Since the center lies on both the transverse and conjugate axes, it's crucial for locating the hyperbola in the coordinate plane.
  • The coordinates of the center are used directly in the equation of the hyperbola as \(h\) and \(k\).
  • In our equation, they affect how we shift the standard form along both the x and y axes.
In essence, the center provides the anchor point for the hyperbola's orientation and location within the coordinate system.

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Most popular questions from this chapter

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ x^{2}+y^{2}+4 x+2 y-6=0 $$

Each equation is of the form \(A x^{2}+B x y+C y^{2}+D x+\) \(E y+F=0 .\) Identify the values of \(A, B,\) and \(C\). $$ -3 x^{2}+x y+2 y^{2}+4 x-7 y=0 $$

Find an equation for a hyperbola centered at the origin with a horizontal transverse axis of length 8 units and a conjugate axis of length 6 units.

OPEN ENDED Find and graph a counterexample to the following statement. If the equation of a hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\) then \(a^{2} \geq b^{2} .\)

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ 6 y^{2}=2 x^{2}+12 $$
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