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Chapter 9: Problem 35

Determine the equation in standard form of the hyperbola that satisfies the given conditions. Vertices at (3,0),(-3,0)\(;\) foci at (4,0),(-4,0)

Short Answer

Expert verified
The equation of the hyperbola is \( x^2/9 - y^2/7 = 1 \)

Step by step solution

01

Find the center of the hyperbola

The center of the hyperbola is the midpoint of both vertices and foci. Since the vertices are at (3,0) and (-3,0), the midpoint or center is at ((3-3)/2 , (0+0)/2) = (0,0).
02

Identify the value of 'a'

'a' is the distance from the center to a vertex. Since one of the vertices is at (3,0) and the center is at (0,0), 'a' is the distance between these two points which is 3.
03

Identify the value of 'c'

'c' is the distance from the center to a focus. Since the focus is at (4,0) and the center is at (0,0), 'c' is 4.
04

Calculate value of 'b'

We calculate 'b' using the equation \( c^2 = a^2 + b^2 \), so \( b^2 = c^2 - a^2 \). Plugging in the values of 'a' and 'c' we found, \( b^2 = 4^2 - 3^2 = 7 \). Thus, 'b' is \( \sqrt{7} \).
05

Write the equation of the hyperbola

The standard equation of a hyperbola with its center at (0,0) and vertices along the x-axis is \( x^2/a^2 - y^2/b^2 = 1 \). Substituting the values we found for 'a' and 'b', we get the equation \( x^2/3^2 - y^2/(\sqrt{7})^2 = 1 \). After simplification, we arrive at \( x^2/9 - y^2/7 = 1 \).

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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 an essential concept to understand. Hyperbolas have two branches and can open either horizontally or vertically.
If a hyperbola opens horizontally, its standard equation looks like this:
  • \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
If it opens vertically, it will be the opposite:
  • \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

Here, \(a\) and \(b\) are real numbers that help describe the hyperbola鈥檚 shape. \(a^2\) is always under the positive term, and \(b^2\) is under the negative term.
This is derived from the formula \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to a focus. The knowledge of whether the hyperbola is horizontal or vertical depends on the given vertices or foci information.
Vertices
Vertices are specific points on the hyperbola. They are where the graph changes direction. Usually, the distance from the center to a vertex is represented by \(a\).
In the exercise example, the given vertices are \((3,0)\) and \((-3,0)\). These points tell us several things:
  • The hyperbola opens horizontally because the vertices align on the x-axis.
  • The center of the hyperbola is halfway between the vertices, which is calculated as \((0,0)\).
  • The value of \(a\) is 3, which is simply the distance from the center to any of the vertices.

Understanding the position and significance of the vertices helps in shaping the hyperbola and measuring its openness or width.
Foci
Foci are the fixed points located inside each branch of the hyperbola. They are crucial because they determine the shape and equation of the hyperbola.
In our example, the foci are at \((4,0)\) and \((-4,0)\). Their locations tell us important information:
  • The distance from the center to each focus is \(c = 4\), which directly relates to the eccentricity and open nature of the hyperbola.
  • Since the foci are further along the x-axis than the vertices, it confirms the hyperbola opens horizontally.
  • The relationship \(c^2 = a^2 + b^2\) allows us to use the foci to calculate \(b\) once we know \(a\).

Foci give us constant characteristics of the hyperbola, shaping the graph鈥檚 curvature and its direction of opening.
Center of Hyperbola
The center of a hyperbola is the point equidistant from its vertices and foci. It serves as the reference point around which the hyperbola forms.
In the original problem, the center is easily determined to be \((0,0)\), being the midpoint between the sets of vertices and also between the foci. This point is crucial for plotting the hyperbola and is always denoted by
  • \((h,k)\)
Since the hyperbola鈥檚 center here is the origin, it significantly simplifies calculations and positioning for other features like vertices and foci.
The location of the center assists with easily applying the formula for the hyperbola鈥檚 equation and understanding how the hyperbola will look on a graph.

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Most popular questions from this chapter

A video game developer gives a parametric representation of the motion of one of the game's characters, at time \(t\), as $$ x=f(t) \quad \text { and } \quad y=g(t) $$ where the table of values for \(f\) and \(g\) are as given. $$\begin{array}{cccccc} t & 0 & 2 & 4 & 6 & 8 \\ \hline x=f(t) & 0 & 2 & 2 & 0 & 0 \end{array}$$ $$\begin{array}{ccccccc} t & 0 & 2 & 4 & 6 & 8 \\ \hline y=g(t) & 0 & 0 & 2 & 2 & 0 \end{array}$$ Sketch the motion of the game character in the \(x y\) plane, indicating the direction of increasing \(t .\) Assume that the path between successive points is a straight line.

This set of exercises will draw on the ideas presented in this section and your general math background. There are hyperbolas other than the types studied in this section. For example, some hyperbolas satisfy an equation of the form \(x y=c,\) where \(c\) is a nonzero constant. In which quadrant(s) of the coordinate plane does the hyperbola with equation \(x y=10\) lie? the hyperbola with equation \(x y=-10 ?\)

Identify and graph the conic section given by each of the equations. $$r=\frac{6}{6+3 \sin \theta}$$

Sara kicks a soccer ball from the ground with an initial velocity of 120 feet per second at an angle of \(30^{\circ}\) to the horizontal. (a) Find the parametric equations that give the position of the ball as a function of time. (b) When is the ball at its maximum height, to the nearest hundredth of a second? What is its maximum height, to the nearest tenth of a foot? (c) How far did the ball travel? Round your answer to the nearest foot.

Determine the equation in standard form of the hyperbola that satisfies the given conditions. Vertices at (4,6),(-4,6)\(;\) slope of one asymptote is -2
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