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Hyperbola

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Construct a rectangle by moving 4 units up and down from the vertices. ... z 1000-1000 z = 594. 2z = 594 or z = 297. Hyperbola Find an Equation. V ... – PowerPoint PPT presentation

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Title: Hyperbola


1
Hyperbola
  • Conic Sections

2
Hyperbola
  • The plane can intersect two nappes of the cone
    resulting in a hyperbola.

3
Hyperbola - Definition
A hyperbola is the set of all points in a plane
such that the difference in the distances from
two points (foci) is constant.
d1 d2 is a constant value.
4
Finding An Equation
  • Hyperbola

5
Hyperbola - Definition
What is the constant value for the difference in
the distance from the two foci? Let the two foci
be (c, 0) and (-c, 0). The vertices are (a, 0)
and (-a, 0).
d1 d2 is the constant.
If the length of d2 is subtracted from the left
side of d1, what is the length which remains?
d1 d2 2a
6
Hyperbola - Equation
Find the equation by setting the difference in
the distance from the two foci equal to 2a.
d1 d2 2a
7
Hyperbola - Equation
Simplify
Remove the absolute value by using or -.
Get one square root by itself and square both
sides.
8
Hyperbola - Equation
Subtract y2 and square the binomials.
Solve for the square root and square both sides.
9
Hyperbola - Equation
Square the binomials and simplify.
Get xs and ys together on one side.
10
Hyperbola - Equation
Factor.
Divide both sides by a2(c2 a2)
11
Hyperbola - Equation
Let b2 c2 a2
where c2 a2 b2
If the graph is shifted over h units and up k
units, the equation of the hyperbola is
12
Hyperbola - Equation
where c2 a2 b2
RecognitionHow do you tell a hyperbola from an
ellipse?
AnswerA hyperbola has a minus (-) between the
terms while an ellipse has a plus ().
13
Graph - Example 1
  • Hyperbola

14
Hyperbola - Graph
Graph
Center
(-1, -2)
The hyperbola opens in the x direction because
x is positive.
Transverse Axis
y -2
15
Hyperbola - Graph
Graph
Vertices
(2, -2) (-4, -2)
Construct a rectangle by moving 4 units up and
down from the vertices.
Construct the diagonals of the rectangle.
16
Hyperbola - Graph
Graph
Draw the hyperbola touching the vertices and
approaching the asymptotes.
Where are the foci?
17
Hyperbola - Graph
Graph
The foci are 5 units from the center on the
transverse axis.
Foci (-6, -2) (4, -2)
18
Hyperbola - Graph
Graph
Find the equation of the asymptote lines.
4
3
Use point-slope formy y1 m(x x1) since
the center is on both lines.
-4
Slope
Asymptote Equations
19
Graph - Example 2
  • Hyperbola

20
Hyperbola - Graph
Sketch the graph without a grapher
RecognitionHow do you determine the type of
conic section?
AnswerThe squared terms have opposite signs.
Write the equation in hyperbolic form.
21
Hyperbola - Graph
Sketch the graph without a grapher
22
Hyperbola - Graph
Sketch the graph without a grapher
Center
(-1, 2)
Transverse Axis Direction
Up/Down
Equation
x-1
Vertices
Up/Down from the center or
23
Hyperbola - Graph
Sketch the graph without a grapher
Plot the rectangular points and draw the
asymptotes.
Sketch the hyperbola.
24
Hyperbola - Graph
Sketch the graph without a grapher
Plot the foci.
Foci
25
Hyperbola - Graph
Sketch the graph without a grapher
Equation of the asymptotes
26
Finding an EquationA problem for CSI!
  • Hyperbola

27
Hyperbola Find an Equation
The sound of a gunshot was recorded at one
microphone 0.54 seconds before being recorded at
a second microphone. If the two microphones are
2,000 ft apart. Provide a model for the possible
locations of the gunshot. (The speed of sound is
1100 ft/sec.)
The time between the shots can be used to
calculate the difference in the distance from the
two microphones.
1100 ft/sec 0.54 sec 594 ft. The constant
difference in distance from the microphones is
594 ft.
Since the difference is constant, the equation
must be a hyperbola. The points on the hyperbola
are possible positions for the gunshot.
28
Hyperbola Find an Equation
Two microphones are stationed 2,000 ft apart.
The difference in distance between the
microphones is 594 ft.
Let the center be at (0,0). The foci must be
2,000 ft apart.
V
The vertices are a possible position for the
gunshot. The difference in the distance must be
594 feet between the vertices.
Let the vertices be at (z, 0). Assuming zgt0,
then(z-(-1000)) (1000-z) 594z1000-1000z
5942z 594 or z 297.
29
Hyperbola Find an Equation
V(297, 0)
V(-2970, 0)
V
V
Oops! We could have remembered the constant
difference in distance is 2a! 2a 594, a 297.
Start finding the model of the hyperbola.
2972 88209
The distance from the center to the foci (c) is
1000 ft. Find b.
30
Hyperbola Find an Equation
V(294, 0)
V(294, 0)
V
V
The model is
31
Hyperbola Find an Equation
The gunshot was calculated to be at some point
along the hyperbola.
32
Conic Section Recogition
33
Recognizing a Conic Section
Parabola -
One squared term. Solve for the term which is
not squared. Complete the square on the squared
term.
Ellipse -
Two squared terms. Both terms are the same
sign.
Circle -
Two squared terms with the same coefficient.
Hyperbola -
Two squared terms with opposite signs.
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