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Chapter 7: Problem 17

Find the standard form of the equation of the hyperbola which has the given properties. Vertices (3,2),(13,2)\(;\) Endpoints of the Conjugate Axis (8,4),(8,0)

Short Answer

Expert verified
The standard form is \( \frac{(x-8)^2}{25} - \frac{(y-2)^2}{4} = 1 \).

Step by step solution

01

Identify the center of the hyperbola

The vertices of the hyperbola are given as (3,2) and (13,2). The center is the midpoint of the line segment joining the vertices. Use the midpoint formula: \( (x_1+x_2)/2, (y_1+y_2)/2 \). Substituting the values, \( (3+13)/2, (2+2)/2 = (8, 2) \). So, the center is (8, 2).
02

Determine the orientation and distance of the vertices

The vertices given, (3, 2) and (13, 2), lie on a horizontal line because their y-coordinates are the same. Thus, the hyperbola is horizontally oriented. The distance between the vertices is \(13 - 3 = 10\), so \(2a = 10\) and \(a = 5\).
03

Determine the distance of the endpoints of the conjugate axis

The endpoints of the conjugate axis, (8, 4) and (8, 0), give a vertical line because their x-coordinates are the same. The distance between them is \(4 - 0 = 4\), so \(2b = 4\) and \(b = 2\).
04

Write the standard form equation of the hyperbola

For a horizontally oriented hyperbola with center (h, k), vertices distance 2a, and conjugate axis distance 2b, the standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). With center (8, 2), \(a = 5\), and \(b = 2\), substituting into the equation we get: \[ \frac{(x-8)^2}{25} - \frac{(y-2)^2}{4} = 1 \].

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Key Concepts

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

Standard Form
The standard form of a hyperbola is a specific way to express its equation. It is crucial for identifying critical features, such as its center, vertices, and axes. For a hyperbola centered at (h, k) with a horizontal transverse axis, the equation is as follows:
\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]
In this equation,
  • \(h\) and \(k\) are the coordinates of the center.
  • \(a\) is half the distance between the vertices (on the transverse axis).
  • \(b\) is half the distance of the conjugate axis.
This form helps you visualize the structure of the hyperbola, making it easier to analyze and work with.
Vertices
Vertices of a hyperbola are key points where the curve intersects its transverse axis, which is the line segment connecting the vertices. They are symmetrically placed around the center. In our example, the vertices are at (3,2) and (13,2). The line's orientation tells us if the hyperbola opens horizontally or vertically.

Since the vertices share the same y-coordinate, it indicates a horizontal orientation. The vertices are
  • pivotal for determining the hyperbola's initial shape.
  • used to calculate \(a\), which is half the distance between them: \(2a = 10\) so \(a = 5\).
Conjugate Axis
The conjugate axis in a hyperbola is perpendicular to the transverse axis and crosses the transverse axis at the center. It does not intersect the hyperbola but is crucial for defining its box shape. For our equation, the endpoints of the conjugate axis are (8,4) and (8,0).

These points indicate that the conjugate axis is vertical because the x-coordinates are identical. Calculating the length of the conjugate axis's half provides \(b\):
  • The distance between (8,4) and (8,0) is \(4\), so \(2b = 4\), meaning \(b = 2\).
This information further confirms the horizontal orientation of the hyperbola.
Midpoint Formula
The midpoint formula is a reliable tool for locating the center of a hyperbola by determining the average of the coordinates of two endpoints. It is stated as follows:
\[(x_1+x_2)/2, (y_1+y_2)/2\]
Applying this formula to our vertices (3,2) and (13,2), we calculate:
  • The midpoint is \((3+13)/2, (2+2)/2 = (8, 2)\).
  • This point serves as the hyperbola's center, dividing both its transverse and conjugate axes symmetrically.
The center is crucial for setting up the hyperbola's standard form equation and confirming its geometric properties.

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Most popular questions from this chapter

The Earth's orbit around the sun is an ellipse with the sun at one focus and eccentricity \(e \approx 0.0167\). The length of the semimajor axis (that is, half of the major axis) is defined to be 1 astronomical unit (AU). The vertices of the elliptical orbit are given special names: 'aphelion' is the vertex farthest from the sun, and 'perihelion' is the vertex closest to the sun. Find the distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion.

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On page 510 in Section 7.3 , we discussed paraboloids of revolution when studying the design of satellite dishes and parabolic mirrors. In much the same way, 'natural draft' cooling towers are often shaped as hyperboloids of revolution. Each vertical cross section of these towers is a hyperbola. Suppose the a natural draft cooling tower has the cross section below. Suppose the tower is 450 feet wide at the base, 275 feet wide at the top, and 220 feet at its narrowest point (which occurs 330 feet above the ground.) Determine the height of the tower to the nearest foot.

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