Hyberbola
Conic Sections
Hyperbola

The plane can intersect
two nappes of the cone
resulting in a
hyperbola.
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.
Finding An Equation
Hyperbola
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
Hyperbola - Equation
Find the equation by setting the difference in the
distance from the two foci equal to 2a.
| d1 – d2 | = 2a
d1  ( x  c)  y
2
2
d 2  ( x  c) 2  y 2
Hyperbola - Equation
Simplify:
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a
Remove the absolute value by using + or -.
( x  c) 2  y 2  ( x  c) 2  y 2  2a
Get one square root by itself and square both sides.
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a

( x  c)  y
2
2
 
2
( x  c )  y  2a
2
2

2
( x  c ) 2  y 2  ( x  c ) 2  y 2  4 a ( x  c ) 2  y 2  4a 2
Hyperbola - Equation
( x  c ) 2  y 2  ( x  c ) 2  y 2  4 a ( x  c ) 2  y 2  4a 2
Subtract y2 and square the binomials.
x 2  2 xc  c 2  x 2  2 xc  c 2  4a ( x  c) 2  y 2  4a 2
Solve for the square root and square both sides.
4 xc  4a 2  4a ( x  c) 2  y 2
xc  a 2   a ( x  c)2  y 2
 xc  a 
2 2

  a ( x  c)  y
2
2

2
Hyperbola - Equation
 xc  a 
2 2

  a ( x  c)  y
2
2

2
x 2c 2  2 xca 2  a 4  a 2  ( x  c )2  y 2 
Square the binomials and simplify.
x 2 c 2  2 xca 2  a 4  a 2  x 2  2 xc  c 2  y 2 
x 2c 2  2 xca 2  a 4  a 2 x 2  2 xca 2  a 2c 2  a 2 y 2
x 2c 2  a 4  a 2 x 2  a 2c 2  a 2 y 2
Get x’s and y’s together on one side.
x 2 c 2  a 2 x 2  a 2 y 2  a 2c 2  a 4
Hyperbola - Equation
x 2 c 2  a 2 x 2  a 2 y 2  a 2c 2  a 4
Factor.
x2  c2  a2   a2 y 2  a2 c2  a2 
Divide both sides by a2(c2 – a2)
x2  c2  a2 
2
2
a2  c2  a2 
a y
 2 2
 2 2
2
2
2
2
a  c  a  a c  a  a c  a2 
x2
y2
 2
1
2
2
a c  a 
Hyperbola - Equation
x2
y2
 2
1
2
2
a c  a 
Let b2 = c2 – a2
2
2
x
y
 2 1
2
a b
where c2 = a2 + b2
If the graph is shifted over h units and up k
units, the equation of the hyperbola is:
Hyperbola - Equation
 x  h
a2
2
y k


b2
2
1
where c2 = a2 + b2
Recognition:
How do you tell a hyperbola from an
ellipse?
Answer:
A hyperbola has a minus (-) between the
terms while an ellipse has a plus (+).
Example #1
Hyperbola
Find the center, vertices, foci, and
asymptotes of the hyperbola. Then
graph.
4 x  9 y  36
2
2
Example #2
Hyperbola
Find the standard form of the
equation for the hyperbola. Then
graph and find center, vertices, foci
and asymptotes
10 y  5x  40 y  10 x  15  0
2
2
Finding an Equation
Hyperbola
Hyperbola – Find an Equation
Find the equation of a hyperbola with foci at (2, 6) and
(2, -4). The transverse axis length is 6.
Finding an Equation
Hyperbola
Hyperbola – Find an Equation
One focus is at
 2, 1  58 
Finding an Equation
A problem for CSI!
Hyperbola
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.
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 z>0, then
(z-(-1000)) – (1000-z) = 594
z+1000-1000+z = 594
2z = 594 or z = 297.
Hyperbola – Find an Equation
V(-2970, 0)
V(297, 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.
2
2
x
y
 2 1
88209 b
2972 = 88209
The distance from the center to the foci (c) is 1000 ft. Find b.
Hyperbola – Find an Equation
V(294, 0)
V(294, 0)
V
V
c2  a 2  b2
10002  297 2  b 2
b 2  911791
The model is:
2
x
88209

2
y
911791
1
Hyperbola – Find an Equation
The gunshot was calculated to be at some point along
the hyperbola.
Conic Section
Recogition
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”.
Assignment:
(Please do your assignment on graph paper!!)

Finish Wksheet #26, 27, 34, 39, 48, 49,
and p. 803 #68
(Yesterday’s Assign):Wksheet #8-11**, 24,
25, 39
**Graph and find center, vertices, foci, and
asymptotes.

Story Problems
1. Find the area of the shaded region
assuming the quadrilateral inside the circle
is a square. The equation of the circle is
x2 + y2 = 36.
36p - 72 square units
Story Problem:
2. A mirror is shaped like a paraboloid of
revolution and will be used to concentrate
the rays of the sun at its focus, creating a
heat source. If the mirror is 20 feet across
at its opening and is 6 feet deep, where
will the heat source be concentrated?
4.17 ft along the base, on the axis of symmetry
Story Problem
3. An arch for a bridge over a highway is in
the form of half an ellipse. The top of the
arch is 20 feet above the ground level.
The highway has four lanes, each 12 feet
wide; a center safety strip 8 feet wide; and
two side strips, each four feet wide. What
should the span of the bridge be if the
height 28 feet from the center is to be 13
feet?
73.69 feet
Story Problem:
Suppose two people standing 1 mile (5280 feet)
apart both see a flash of lightning. After a
period of time, the first person standing at point
A hears the thunder. Two seconds later, the
second person standing at point B hears the
thunder. If the person standing at point B is due
west of the person standing at point A and if the
lightning strike is known to occur due north of
the person standing at point A, where did the
lightning strike? (Note: speed of sound is 1100
ft/sec)
5236 feet North of point A