Hyperbolas - MATH 160, Precalculusbanach.millersville.edu/~bob/math160/hyperbolas/main.pdf ·...

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.


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


Anatomy of Hyperbola

Branch Branch

Focus FocusVertex Vertex

Center


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)


Center, Foci, Vertices

Hh ,k LHh-c ,k L Hh+c ,k L

Hh-a ,k L Hh+a ,k L


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.


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


y2

9− (x − 2)2

27= 1.


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


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)


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 .


Graph

- 10 - 5 5 10x

- 10

- 5

5

10

y


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


(y − 3)2

9− (x − 3)2

9= 1.


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


(y − 3)2

9− (x − 3)2

9= 1.


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).



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).



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.


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)


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)


Examples


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)


Examples


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)


Examples



4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0

(hyperbola)

9x2 + 4y2 − 90y + 8x + 28 = 0

(ellipse)


Examples



4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0 (hyperbola)

9x2 + 4y2 − 90y + 8x + 28 = 0

(ellipse)


Examples



4x2 + 25y2 + 250y + 16x + 520 = 0 (ellipse)4y2 − 2x2 − 2y − 8x + 5 = 0 (hyperbola)

9x2 + 4y2 − 90y + 8x + 28 = 0 (ellipse)


Homework

Read Section 6.4.Exercises: 1, 5, 9, 13, . . . , 65, 69


Hyperbolas - MATH 160, Precalculusbanach.millersville.edu/~bob/math160/hyperbolas/main.pdf ·...

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