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HyperbolasMATH 160, Precalculus
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan Hyperbolas
Objectives
In this lesson we will learn to:write the equations of hyperbolas in standard form,find the asymptotes of hyperbolas,graph hyperbolas,use the properties of hyperbolas to solve real-worldproblems,classify conics from their general equations.
J. Robert Buchanan Hyperbolas
Definition of Hyperbola
DefinitionA hyperbola is the set of all points (x , y) in a plane, thedifference of whose distances from two distinct fixed points(foci) is a positive constant.
d1d2
Focus Focus
H x , y L
J. Robert Buchanan Hyperbolas
Standard Equation
Standard Equation of a HyperbolaThe standard for of the equation of a hyperbola with center(h, k) is
(x − h)2
a2 − (y − k)2
b2 = 1 (axis horizontal)
(y − k)2
a2 − (x − h)2
b2 = 1. (axis vertical)
The vertices are a units from the center. The foci are c unitsfrom the center (where c2 = a2 + b2). If the center of thehyperbola is at the origin, the equation becomes one of
x2
a2 −y2
b2 = 1 (axis horizontal)
y2
a2 −x2
b2 = 1. (axis vertical)
J. Robert Buchanan Hyperbolas
Example
Find the standard form of the equation of a hyperbola withvertices at (2,±3) and foci at (2,±6).
The center of the hyperbola will be at the midpoint of the linewhose endpoints are (2,−3) and (2, 3).
(h, k) = (2, 0)
Since the vertices are a units away from the center, a = 3.Since the foci are c units away from the center, c = 6.
c2 = 36 = a2 + b2 = 9 + b2 =⇒ b2 = 27
Thus the equation of the hyperbola is
y2
9− (x − 2)2
27= 1.
J. Robert Buchanan Hyperbolas
Example
Find the standard form of the equation of a hyperbola withvertices at (2,±3) and foci at (2,±6).The center of the hyperbola will be at the midpoint of the linewhose endpoints are (2,−3) and (2, 3).
(h, k) = (2, 0)
Since the vertices are a units away from the center, a = 3.Since the foci are c units away from the center, c = 6.
c2 = 36 = a2 + b2 = 9 + b2 =⇒ b2 = 27
Thus the equation of the hyperbola is
y2
9− (x − 2)2
27= 1.
J. Robert Buchanan Hyperbolas
Asymptotes (1 of 2)
A hyperbola has two asymptotes which intersect at the centerof the hyperbola.
Hh ,k LHh-c ,k L Hh+c ,k L
Hh-a ,k L Hh+a ,k L
J. Robert Buchanan Hyperbolas
Asymptotes (2 of 2)
Asymptotes of a HyperbolaThe equations of the asymptotes of a hyperbola are
y = k ± ba
(x − h) (axis horizontal)
y = k ± ab
(x − h). (axis vertical)
J. Robert Buchanan Hyperbolas
Example
Sketch the hyperbola whose equation isx2
36− y2
4= 1.
The axis is horizontal, a = 6, b = 2, the center is at the origin,the vertices are at (±6, 0), and the asymptotes are
y = ±13
x .
J. Robert Buchanan Hyperbolas
Example
Sketch the hyperbola whose equation isx2
36− y2
4= 1.
The axis is horizontal, a = 6, b = 2, the center is at the origin,the vertices are at (±6, 0), and the asymptotes are
y = ±13
x .
J. Robert Buchanan Hyperbolas
Example
Find the standard form of the equation of a hyperbola whosevertices are (3, 0) and (3, 6) and whose asymptotes arey = 6− x and y = x .
The center of the hyperbola will be at the midpoint of the linewhose endpoints are (3, 0) and (3, 6).
(h, k) = (3, 3)
Since the vertices are a units away from the center, a = 3. Theaxis of the hyperbola is vertical, therefore
±1 = ±ab
= ±3b
=⇒ b = 3
Thus the equation of the hyperbola is
(y − 3)2
9− (x − 3)2
9= 1.
J. Robert Buchanan Hyperbolas
Example
Find the standard form of the equation of a hyperbola whosevertices are (3, 0) and (3, 6) and whose asymptotes arey = 6− x and y = x .The center of the hyperbola will be at the midpoint of the linewhose endpoints are (3, 0) and (3, 6).
(h, k) = (3, 3)
Since the vertices are a units away from the center, a = 3. Theaxis of the hyperbola is vertical, therefore
±1 = ±ab
= ±3b
=⇒ b = 3
Thus the equation of the hyperbola is
(y − 3)2
9− (x − 3)2
9= 1.
J. Robert Buchanan Hyperbolas
Application (1 of 3)
You and a friend live 4 miles apart (on the same “east-west”street) and are talking on the telephone. You hear a clap ofthunder from lightning in a storm, and 18 seconds later yourfriend hears the thunder. Find an equation that gives thepossible places where the lightning could have occurred(assume distances are measured in feet and sound travels at1100 feet per second).
J. Robert Buchanan Hyperbolas
Application (2 of 3)
The difference in distances between you and the lightning andyour friend and the lightning is
(18)(1100) = 19, 800 feet.
The locus of all points 19, 800 feet closer to you than yourfriend will be one branch of the hyperbola
x2
a2 −y2
b2 = 1.
Assume the center of the hyperbola is at the origin. You andyour friend are at the foci located at
(±(2)(5280), 0) = (±10560, 0) = (±c, 0).
J. Robert Buchanan Hyperbolas
Application (3 of 3)
Since the 4-mile (21,120-foot) distance between you and yourfriend can be thought of as
19, 800 + 2(c − a)
then19, 800 + 2(10, 560− a) = 21, 120
which implies a = 9900 and a2 = 98, 010, 000 and thus
b2 = c2 − a2 = (10560)2 − (9900)2 = 13, 503, 600.
The equation of the hyperbola is
x2
98, 010, 000− y2
13, 503, 600= 1.
J. Robert Buchanan Hyperbolas
General Equations of Conics
Classifying a Conic from its General Equation
The graph of
Ax2 + Cy2 + Dx + Ey + F = 0
is one of the following.Circle: A = C
Parabola: AC = 0 (either A = 0 or C = 0 but not both)Ellipse: AC > 0 (A and C have the same sign)
Hyperbola: AC < 0 (A and C have opposite signs)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0
(circle)
y2 − 6y − 4x + 20 = 0
(parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0
(ellipse)
4y2 − 2x2 − 2y − 8x + 5 = 0
(hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0
(ellipse)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0 (circle)y2 − 6y − 4x + 20 = 0
(parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0
(ellipse)
4y2 − 2x2 − 2y − 8x + 5 = 0
(hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0
(ellipse)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0 (circle)y2 − 6y − 4x + 20 = 0 (parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0
(ellipse)
4y2 − 2x2 − 2y − 8x + 5 = 0
(hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0
(ellipse)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0 (circle)y2 − 6y − 4x + 20 = 0 (parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0
(hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0
(ellipse)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0 (circle)y2 − 6y − 4x + 20 = 0 (parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0 (hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0
(ellipse)
J. Robert Buchanan Hyperbolas
Examples
Classify the graphs of the following equations as circles,ellipses, parabolas, or hyperbolas.
x2 + y2 − 4x − 6y − 32 = 0 (circle)y2 − 6y − 4x + 20 = 0 (parabola)
4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0 (hyperbola)
9x2 + 4y2 − 90y + 8x + 28 = 0 (ellipse)
J. Robert Buchanan Hyperbolas
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