8.5 Graph and Write
Equations of Hyperbolas
p.518
What are the parts of a hyperbola?
What are the standard form equations of a
hyperbola?
How do you know which way it opens?
Given a & b, how do you find the value of c?
How do you graph a hyperbola?
Why does drawing a box make graphing easier?
How do you write the equation from a graph?
Hyperbolas
• Like an ellipse but instead of the sum of
distances it is the difference
• A hyperbola is the set of all points P such that
the differences from P to two fixed points,
called foci, is constant
• The line thru the foci intersects the hyperbola
@ two points (the vertices)
• The line segment joining the vertices is the
transverse axis, and it’s midpoint is the
center of the hyperbola.
• Has 2 branches and 2 asymptotes
• The asymptotes contain the diagonals of a
rectangle centered at the hyperbolas center
Asymptotes
Vertex (-a,0)
(0,b)
Vertex (a,0)
Focus
Focus
(0,-b)
Vertical transverse axis
2
2
y
x
 2 1
2
a
b
Standard Form of Hyperbola w/
center @ origin
Transverse
Equation
Asymptotes Vertices
Axis
x2 y2
 2  1 Horizontal
2
a
b
2
y=+/- (b/a)x
(+/-a,0)
y=+/- (a/b)x
(0,+/-a)
2
y
x
 2  1 Vertical
2
a
b
Foci lie on transverse axis, c units from the center c2 = a2+b2
Graph 4x2 – 9y2 = 36
• Write in standard form (divide through by 36)
• a=3 b=2 – because x2 term is ‘+’ transverse
axis is horizontal & vertices are (-3,0) & (3,0)
• Draw a rectangle centered at the origin.
• Draw asymptotes.
• Draw hyperbola.
Graph 25y2 – 4x2 = 100. Identify the vertices, foci, and
asymptotes of the hyperbola.
SOLUTION
STEP 1 Rewrite the equation in standard form.
25y2 – 4x2 = 100
25y2
100 –
y2
4 –
STEP 2
Write original equation.
4x2 = 100
Divide each side by 100.
100
100
y2
Simplify.
=
1
25
Identify the vertices, foci, and asymptotes. Note
that a2 = 4 and b2 = 25, so a = 2 and b = 5. The
y2 - term is positive, so the transverse axis is
vertical and the vertices are at (0, +2). Find the foci.
c2 = a2 – b2 = 22 – 52 = 29.
so c = 29. The foci are at ( 0, + 29. ) (0, + 5.4).
The asymptotes are y = + a x or y = + 2 x
5
b
STEP 3
Draw the hyperbola. First draw
a rectangle centered at the
origin that is 2a = 4 units high
and 2b = 10 units wide. The
asymptotes pass through
opposite corners of the
rectangle. Then, draw the
hyperbola passing through the
vertices and approaching the
asymptotes.
Graph the equation. Identify the vertices, foci, and
asymptotes of the hyperbola.
2
y2
x
1.
16 – 49 = 1
SOLUTION
2
y2
x
STEP 1 The equation is in standard form.
16 – 49 = 1
STEP 2
Identify the vertices, foci, and asymptotes. Note that
a2 = 16 and b2 = 49, so a = 4 and b = 7. The x2 term is positive, so the transverse axis is horizontal
and the vertices are at (+4, 0). Find the foci.
c2 = a2 + b2 = 42 + 72 = 65.
so c = + 65.
The foci are at ( + 65 , 0)
STEP 3
Draw the hyperbola. First draw a rectangle centered at
the origin that is 2a = 8 units high and 2b = 14 units
wide. The asymptotes pass through opposite corners
of the rectangle. Then, draw the hyperbola passing
through the vertices and approaching the asymptotes.
Write the equation of a hyperbola
with foci (0,-3) & (0,3) and vertices
(0,-2) & (0,2).
• Vertical because foci & vertices lie on the y-axis
• Center @ origin because focus & vertices are
equidistant from the origin
• Since c=3 & a=2, c2 = b2 + a2
2
2
2
•
9=b +4
•
5 = b2
•
+/-√5 = b
y
x
 1
4
5
Write an equation of the hyperbola with foci at
(– 4, 0) and (4, 0) and vertices at (– 3, 0) and (3, 0).
SOLUTION
The foci and vertices lie
on the x-axis equidistant
from the origin, so the
transverse axis is
horizontal and the center
is the origin. The foci are
each 4 units from the
center, so c = 4. The
vertices are each 3 units
from the center, so a = 3.
Because c2 = a2 + b2, you have b2 = c2 – a2. Find b2.
b2 = c2 – a2 = 42 – 32 = 7
Because the transverse axis is horizontal, the
standard form of the equation is as follows:
y2
x2
32 – 7
y2
x2
9 – 7
= 1
= 1
Substitute 3 for a and 7 for b2.
Simplify
Write an equation of the hyperbola with the given foci
and vertices.
4.
Foci: (– 3, 0), (3, 0)
Vertices: (– 1, 0), (1, 0)
SOLUTION The foci and vertices lie on the x-axis equidistant
from the origin, so the transverse axis is horizontal
and the center is the origin. The foci are each 3
units from the center, so c = 3. The vertices are
each 1 units from the center, so a = 1.
Because c2 = a2 + b2, you have b2 = c2 – a2. Find b2.
b2 = c2 – a2 = 32 – 12 = 8
Because the transverse axis is horizontal, the
standard form of the equation is as follows:
y2
x2
12 – 8
2
y
x2 – 8
= 1
= 1
Substitute 1 for a and 8 for b2.
Simplify
5.
Foci: (0, – 10), (0, 10)
Vertices: (0, – 6), (0, 6)
The foci and vertices lie on the y-axis equidistant
from the origin, so the transverse axis is vertical
and the center is the origin. The foci are each 10
units from the center, so c = 10. The vertices are
each 6 units from the center, so a = 6.
Because c2 = a2 + b2, you have b2 = c2 – a2. Find b2.
b2 = c2 – a2 = 102 – 62 = 64
SOLUTION
Because the transverse axis is horizontal, the
standard form of the equation is as follows:
x2
y2
362 – 64 = 1
x2
y2
= 1
36 – 64
Substitute 6 for a and 64 for b2.
Simplify
Photography
You can take panoramic photographs
using a hyperbolic mirror. Light rays
heading toward the focus behind the
mirror are reflected to a camera
positioned at the other focus as
shown. After a photograph is taken,
computers can “unwrap” the
distorted image into a 360° view.
•
Write an equation for the cross section of the mirror.
•
The mirror is 6 centimeters wide. How tall is it?
SOLUTION
From the diagram, a = 2.81 and c = 3.66.
STEP 1
To write an equation, find b2. b2 = c2 – a2 = 3.662 – 2.812 5.50
Because the transverse axis is vertical, the standard
form of the equation for the cross section of the mirror
is as follows: y2
y2
x2
x2
–
= 1 or 7.90 – 5.50 = 1
2
2.81
5.50
STEP 2
Find the y-coordinate at the mirror’s bottom edge.
Because the mirror is 6 centimeters wide, substitute
x = 3 into the equation and solve.
y2
32
–
7.90
5.50
y2
=
1
20.83
Substitute 3 for x.
Solve for y2.
Solve for y.
y
– 4.56
So, the mirror has a height of – 2.81 – (– 4.56) = 1.75 cm.
• What are the parts of a hyperbola?
Vertices, foci, center, transverse axis & asymptotes
• What are the standard form equations of a hyperbola?
x2 y2
 2 1
2
a
b
y 2 x2
 2 1
2
a
b
• How do you know which way it opens?
Transverse axis is always over a
• Given a & b, how do you find the value of c?
c2 = a 2 + b2
• How do you graph a hyperbola?
Plot a and b, draw a box with diagonals. Draw the
hyperbola following the diagonals through the vertices.
• Why does drawing a box make graphing
easier?
The diagonals of the box are the asymptotes of
the hyperbola.
• How do you write the equation from a graph?
Identify the transverse axis, find the value of a
and b (may have to use c2 = a2 + b2) and
substitute into the equation.
8.5 Assignment, day 1
Page 521, 3-13 odd,
19-25 odd
10.5 Hyperbolas, day 2
What are the standard form equations of a
hyperbola if the center has been translated?
How do you graph a translated hyperbola?
How do you write the equation of a translated
hyperbola?
Translated Hyperbolas
In the following equations the point (h,k) is
the center of the hyperbola.
Horizontal axis
Vertical axis
x  h 
2
a2

y  k

b2
y  k 
2
2
1
a
2

x  h

2
b
Remember c2 = a2 + b2
2
1
Graphing the Equation of a Translated Hyperbola
Graph (y +
1) 2
(x + 1) 2
–
= 1.
4
(–1, 0)
SOLUTION
The y 2-term is positive, so the
transverse axis is vertical. Since
a 2 = 1 and b 2 = 4, you know that
a = 1 and b = 2.
(–1, –1)
(–1, –2)
Plot the center at (h, k) = (–1, –1). Plot the vertices 1 unit above and
below the center at (–1, 0) and (–1, –2).
Draw a rectangle that is centered at (–1, –1) and is 2a = 2 units high
and 2b = 4 units wide.
Graphing the Equation of a Translated Hyperbola
Graph (y +
1) 2
(x + 1) 2
–
= 1.
4
(–1, 0)
SOLUTION
The y 2-term is positive, so the
transverse axis is vertical. Since
a 2 = 1 and b 2 = 4, you know that
a = 1 and b = 2.
(–1, –1)
(–1, –2)
Draw the asymptotes through the corners of the rectangle.
Draw the hyperbola so that it passes through the vertices and
approaches the asymptotes.
2
2
(y
–
3)
(x
+
1)
Graph
–
= 1
4
9
SOLUTION
STEP 1 Compare the given equation to the
standard forms of equations of
hyperbolas. The equation’s form
tells you that the graph is a
hyperbola with a vertical transverse
axis. The center is at (h, k) = (– 1, 3).
Because a2 = 4 and b2 = 9, you
know that a = 2 and b = 3.
STEP 2
Plot the center, vertices, and foci. The vertices lie a = 2 units
above and below the center, at (21, 5) and (21, 1). Because c2
= a2 + b2 = 13, the foci lie c = 13
3.6 units above and
below the center, at (– 1, 6.6) and (– 1, – 0.6).
STEP 3
Draw the hyperbola. Draw a rectangle centered at (21, 3)
that is 2a = 4 units high and 2b = 6 units wide. Draw the
asymptotes through the opposite corners of the
rectangle. Then draw the hyperbola passing through the
vertices and approaching the asymptotes.
3. Graph
2
(y
–
4)
(x + 3)2 –
= 1
9
SOLUTION
STEP 1 Compare the given equation to the standard
forms of equations of hyperbolas. The
equation’s form tells you that the graph is a
hyperbola with a vertical transverse axis. The
center is at (h, k) = (– 3, 4). Because a2 = 1 and
b2 = 4, you know that a = 1 and b = 2.
STEP 2 Plot the center, vertices, and foci. The vertices lie
a = 1 units above and below the center, at (–2, 4)
and (–4, 4). Because c2 = a2 + b2 = 5, the foci lie c
= 5 units above and below the center, at (– 3 + 5
,4 ) and (– 3, – 5 , 4).
STEP 3
Draw the hyperbola. Draw a rectangle centered at
( –3, 4) that is 2a = 2 units high and 2b = 4 units
wide. Draw the asymptotes through the opposite
corners of the rectangle.
Write the equation of the hyperbola
in standard form.
16y2 −36x2 + 9 = 0
16y2 −36x2 = −9
1
16
1
16
2
1
36 x 36


0
9 1
2 36
2
16 y
9
2
y
x


1
9  9 
   
 16   36 
2
2
x
y

1
1  9 
   
 4   16 
Write an equation for the
hyperbola.
Vertices at (5, −4)
and (5,4) and foci
at (5,−6) and (5,6).
Draw a quick graph.
Equation will be:
2
2
 y  k   x  h   1
2
2
a
b
y  k 
2
2

x  h

(5,6)
2
2
1
a
b
Center (0,5) (h,k)
a = 4, c = 6
c2 = a2 + b2
62 = 4 2 + b 2
36 = 16 + b2
2
20 = b2
y
16
(5,4)
Center?
 y  k 2  x  h 2
a2
b2
1
(5,−4)
(5,−6)

x  5

2
20
1
• What are the standard form equations of a
hyperbola if the center has been translated?
( x  h) ( y  k )

1
2
2
a
b
2
2
( y  k ) ( x  h)

1
2
2
a
b
2
2
8.5 Assignment, day 2
Page 521, 4-10 even, 18-24 even
Page 531, 5, 6, 11, 19-20