8.6 The Hyperbola
Some ships navigate using a radio
navigation system called LORAN,
which is an acronym for LOng RAnge
Navigation. A ship receives radio
signals from pairs of transmitting
stations that send signals at the same
time. The LORAN equipment detects
the difference in the arrival times of
the signals and uses the locus
definition of the hyperbola to
determine the ship’s location.
To determine the equations of
hyperbolas, the absolute values of
numbers are used. The absolute value
of a real number is its distance from
zero on a real number line. For a real
number represented by x, the absolute
value is written |x |, which means the
positive value, or magnitude, of x.
The diagram shows that the absolute value
of −3 is 3 and the absolute value of 3 is 3.
| –3| = 3
–4
–3
–2
–1
3 units
| 3| = 3
0
1
2
3
4
We b C o n n e c t i o n
www.school.mcgrawhill.ca/resources/
To learn more about LORAN, visit the above web
site. Go to Math Resources, then to
MATHEMATICS 11, to find out where to
go next. Write a brief report about the
origins of LORAN.
3 units
A hyperbola is the set or locus of points P in the plane such that the
absolute value of the difference of the distances from P to two fixed points
F1 and F2 is a constant.
F1P − F2P = k
The two fixed points, F1 and F2, are called the foci of the hyperbola. The
line segments F1P and F2P are called the focal radii of the hyperbola.
8.6 The Hyperbola • MHR 637
I NVESTIGATE & I NQUIRE
You will need two clear plastic rulers, a sheet of paper, and a pencil.
Step 1 Draw a 10-cm line segment near the centre
of the piece of paper. Label the two endpoints F1
and F2.
Step 2 Choose a length, k centimetres, which is less
than the length F1F2. For this investigation, use
k = 4 cm.
F1
F2
Step 3 Choose a pair of lengths, a centimetres and b centimetres, such that the
absolute value of their difference equals 4, that is, |a − b | = 4. For example, choose
a = 9 cm and b = 5 cm, or a = 5 cm and b = 9 cm, since |9 − 5|= 4 and |5 − 9| = 4.
Step 4 Use both rulers to mark
two points that are 9 cm from F1
and 5 cm from F2. Then, use both
rulers to mark two
points that are 5 cm
9 cm
from F1 and 9 cm
from F2.
F1
5 cm
F2
9 cm
5 cm
5 cm
9 cm
F1
5 cm
F2
9 cm
Step 5 Repeat steps 3 and 4 using different values of a and b, such that |a − b | = 4,
until you have marked enough points to define two curves.
Step 6 Draw a smooth curve through each set of points. The two curves form a
hyperbola.
1.
How many axes of symmetry does the hyperbola have?
In relation to F1 and F2, where is the point of intersection of the axes of
symmetry?
2.
3. In Step 4, you marked four points for the chosen values of a and b. Are there
any values of a and b for which only two points can be marked? If so, what are
the values of a and b? Describe the locations of the points on the hyperbola.
The vertices of this hyperbola are located on F1F2. How is the distance between
the vertices of the two curves related to the value of k ? Explain.
4.
638 MHR • Chapter 8
EXAMPLE 1 Finding the Equation of a Hyperbola From its Locus Definition
Use the locus definition of the hyperbola to find an equation of the hyperbola
with foci F1(−5, 0) and F2(5, 0), and with the constant difference between the
focal radii equal to 8.
SOLUTION
Let P(x, y) be any point on the hyperbola.
The locus definition of the hyperbola can be stated algebraically as
|F1P − F2P| = 8.
Use the formula for the length of a line segment, l = (x
− x
)2 + (y,
− y1)2 to
2
1
2
rewrite F1P and F2P.
l = (x
− x
)2 + (y
− y1)2
2
1
2
2
F1P = (x
5))
− (−y
+ (
− 0)2
= (x
+ 5)2
+ y2
2
F2P = (x
− 5)
+ (y −
0)2
= (x
− 5)2
+ y2
Without loss in generality, we assume that F1P > F2P and substitute.
Substitute: (x
+ 5)2
+ y2 − (x
− 5)2
+ y2 = 8
Isolate a radical:
(x
+ 5)2
+ y2 = 8 + (x
− 5)2
+ y2
Square both sides:
(x + 5)2 + y2 = 64 + 16(x
− 5)2
+ y2 + (x − 5)2 + y2
x2 + 10x + 25 + y2 = 64 + 16(x
− 5)2
+ y2 + x2 − 10x + 25 + y2
Isolate the radical:
20x − 64 = 16(x
− 5)2
+ y2
Divide both sides by 4:
5x − 16 = 4(x
− 5)2
+ y2
Square both sides:
25x2 − 160x + 256 = 16((x − 5)2 + y2)
Simplify:
25x2 − 160x + 256 = 16(x2 − 10x + 25 + y2)
25x2 − 160x + 256 = 16x2 − 160x + 400 +16y2
9x2 − 16y2 = 144
2
x2 − y = 1
Divide both sides by 144:
16 9
2
x2 − y = 1.
An equation of the hyperbola is 16 9
8.6 The Hyperbola • MHR 639
The hyperbola in Example 1 can be
modelled graphically, as shown.
The hyperbola has two axes of
symmetry. The points (−4, 0) and (4, 0)
lie on one axis of symmetry. The line
segment joining these two points is
called the transverse axis. In this case,
the length of the transverse axis is 8.
The endpoints of the transverse axis
are the vertices of the hyperbola.
y
3
y = – –x
4
4
3
y = –x
4
(0, 3)
2
F1(–5, 0)
–8
–6
(–4, 0)
–4
–2
(4, 0)
0
2
–2
–4
(0, –3)
–6
The points (0, −3) and (0, 3) lie on
the other axis of symmetry. The line
segment joining these two points is called the conjugate axis.
In this case, the length of the conjugate axis is 6. The endpoints of the
conjugate axis are called the co-vertices of the hyperbola.
The point of intersection of the transverse axis and the conjugate axis is
called the centre of the hyperbola. In this case, the centre is the origin,
(0, 0).
The lines x = −4 and x = 4 form a rectangle with the lines y = −3 and y = 3.
The graph of the hyperbola lies between the diagonals of this rectangle.
The diagonals of this rectangle are called the asymptotes of the hyperbola.
These are the lines that the hyperbola approaches for large values of x and y.
The equation of the hyperbola from Example 1 can be written as
2
x2 − y = 1
2 2
4
3
In this form of the equation, notice that 4 is half the length of the transverse
axis, or half the difference between the focal radii, and 3 is half the length of
the conjugate axis. Notice also that the equations of the asymptotes are
3
3
y = x and y = – x.
4
4
The coordinates of the foci are (−5, 0) and (5, 0).
2
2
2
Notice that 4 + 3 = 5 .
640 MHR • Chapter 8
4
F2(5, 0)
6
8
x
The standard form of the equation of a hyperbola
centred at the origin, with the transverse axis
along the x-axis and the conjugate axis along the
y-axis, is
2
x2 − y = 1
2 2
a
b
• The length of transverse axis along the x-axis is
2a.
• The length of conjugate axis along the y-axis is
2b.
• The vertices are V1(−a, 0) and V2(a, 0).
• The co-vertices are (0, −b) and (0, b).
• The foci are F1(−c, 0) and F2(c, 0), which are on
the transverse axis.
• The equations of the asymptotes are
y = b x and y = – b x.
a
a
2
2
2
•a +b =c
The standard form of the equation of a
hyperbola centred at the origin, with the
transverse axis along the y -axis and the
conjugate axis along the x-axis, is
y2 x2
2 − 2 = 1
a
b
• The length of transverse axis along the
y-axis is 2a.
• The length of conjugate axis along the
x-axis is 2b.
• The vertices are V1(0, −a) and V2(0, a).
• The co-vertices are (−b, 0) and (b, 0).
• The foci are F1(0, −c) and F2(0, c),
which are on the transverse axis.
• The equations of the asymptotes are
y = ax and y = – a x.
b
b
2
2
2
•a +b =c
y
y
(0, b)
axis
F2 (0, c)
(0, –b)
V 2 (0, a)
F2 (c, 0)
x
axis
axis
V 2 (a, 0)
(–b, 0) conjugate
axis
(b, 0)
x
0
transverse
transverse
V 1 (–a, 0) 0
conjugate
F1 (–c, 0)
V1 (0, –a)
F 1 (0, –c)
8.6 The Hyperbola • MHR 641
EXAMPLE 2 Sketching the Graph of a Hyperbola With Centre (0, 0)
2
y
x2 = 1. Label the foci and the asymptotes.
Sketch the graph of the hyperbola − 36 4
SOLUTION
y2 x2
Since the equation is in the form 2 − = 1, the centre is (0, 0) and the
2
a
b
transverse axis is along the y-axis.
2
Since a = 36, a = 6, and the vertices are V1(0, −6) and V2(0, 6).
Since b2 = 4, b = 2, and the co-vertices are (−2, 0) and (2, 0).
The length of the transverse axis is 2a = 2(6)
= 12
The length of the conjugate axis is 2b = 2(2)
=4
The foci are F1(0, −c) and F2(0, c).
To find c, use a2 + b2 = c2, where a = 6 and b = 2.
c2 = a2 + b2
= 62 + 22
= 36 + 4
= 40
c = 40
= 210
The coordinates of the foci are (0, −210
) and (0, 210
), or
approximately (0, −6.32) and (0, 6.32).
a x and y = – a x.
The equations of the asymptotes are y = b
b
Substituting for a and b gives
6 x, or y = 3x, and y = – 6x , or y = −3x.
y=
2
2
To sketch the graph of the hyperbola, plot the vertices and the
co-vertices. Use the lines y = 6, y = −6, x = 2, and x = −2 to
construct a rectangle. Sketch the asymptotes by extending the
diagonals of the rectangle.
Since the transverse axis is along the y-axis, the hyperbola must
open up and down. To sketch a branch of the hyperbola, start at
a vertex and approach the asymptotes. Sketch the other branch
in the same way. Label the foci and the asymptotes.
642 MHR • Chapter 8
y
8
y = –3x
y = 3x
F2 (0, 2 10)
6
V 2 (0, 6)
4
2
(–2, 0)
–4
–2
(2, 0)
0
2
4
–2
–4
V1 (0 1 , –6)
–6 F (0, –2 10)
1
–8
–10
x
A hyperbola may not be centred at the origin. As in the case of the circle
and the ellipse, a hyperbola can have a centre (h, k). The translation rules
that apply to the circle and the ellipse also apply to the hyperbola.
The standard form of the equation of a
hyperbola centred at (h, k), with the
transverse axis parallel to the x-axis and the
conjugate axis parallel to the y-axis, is
2
(x – h)2 − ( y − k) = 1
a2
b2
• The length of the transverse axis parallel to
the x-axis is 2a.
• The length of the conjugate axis parallel to
the y-axis is 2b.
• The vertices are V1(h − a, k) and V2(h + a, k).
• The co-vertices are (h, k − b) and (h, k + b).
• The foci are F1(h − c, k) and F2(h + c, k),
which are on the transverse axis.
• a2 + b2 = c2
y
y
(h, k + b)
transverse
(h, k) axis
V1(h – a, k)
0
h–c h–a
k–b
V2(h + a, k)
F2(h, k + c)
k+a
V2(h, k + a)
axis
k
conjugate
F1(h – c, k)
k+c
F2(h + c, k)
k
h h+a h+c
conjugate axis
(h + b, k)
(h, k)
(h – b, k)
transverse
axis
k+b
The standard form of the equation of a
hyperbola centred at (h, k), with the
transverse axis parallel to the y-axis, and
conjugate axis parallel to the x-axis, is
(y − k)2 (x − h)2
−
=1
b2
a2
• The length of the transverse axis parallel to
the y-axis is 2a.
• The length of the conjugate axis parallel to
the x-axis is 2b.
• The vertices are V1(h, k − a) and V2(h, k + a).
• The co-vertices are (h − b, k) and (h + b, k).
• The foci are F1(h, k − c) and F2(h, k + c),
which are on the transverse axis.
• a2 + b2 = c2
x
k–a
V1(h, k – a)
(h, k – b)
0
k–c
h–b
h
h+c
x
F1(h, k – c)
8.6 The Hyperbola • MHR 643
EXAMPLE 3 Sketching the Graph of a Hyperbola With Centre (h, k)
2
(x − 2)2 − ( y + 1) = 1. Label the foci.
Sketch the graph of the hyperbola 36
16
SOLUTION
The centre is C(h, k) = (2, −1).
2
(x − h)2 − ( y − k) = 1, the transverse axis
Since the equation is in the form a2
b2
is parallel to the x-axis.
a2 = 36, so a = 6
b2 = 16, so b = 4
The length of the transverse axis is 2a, or 12.
The length of the conjugate axis is 2b, or 8.
The vertices are V1(h − a, k) and V2(h + a, k).
Substitute the values of h, k, and a.
The vertices are V1(2 − 6, −1) and V2(2 + 6, −1), or V1(−4, −1) and
V2(8, −1).
The co-vertices are (h, k − b) and (h, k + b).
Substitute the values of h, k, and b.
The co-vertices are (2, −1 − 4) and (2, −1 + 4), or (2, −5) and (2, 3).
The foci are F1(h − c, k) and F2(h + c, k).
To find c, use a2 + b2 = c2, with a = 6 and b = 4.
c2 = 62 + 42
= 36 + 16
= 52
c = 52
= 213
The coordinates of the foci are (2 − 213
, −1) and (2 + 213
, −1), or
approximately (−5.21, −1) and (9.21, −1).
To sketch the graph of the hyperbola, plot the vertices and the co-vertices.
Then, construct the rectangle that goes through all four of these points.
Sketch the asymptotes by extending the diagonals of the rectangle.
644 MHR • Chapter 8
Since the transverse axis is
parallel to the x-axis, the
hyperbola must open left and
right. To sketch a branch of the
hyperbola, start at a vertex and
approach the asymptotes. Sketch
the other branch in the same
way. Label the foci.
y
4
(2, 3)
2
–8
–6
–4
–2
0
2
4
F1(2 – 2 13, –1) V1 (–4, –1)–2
6
8
10
V2(8, –1)
x
12
F2(2 + 2 13, –1)
–4
–6
(2, –5)
EXAMPLE 4 Writing an Equation of a Hyperbola
Write an equation for the hyperbola with vertices (2, −2) and (2, 8)
and co-vertices (0, 3) and (4, 3). Sketch the hyperbola.
SOLUTION
The vertices (2, −2) and (2, 8) lie on the transverse axis, so the
transverse axis is parallel to the y-axis.
y
8
The centre of the hyperbola is located midway between the vertices,
and is also located midway between the co-vertices.
Using the midpoint formula with the two vertices (2, −2) and (2, 8)
gives
(2, 8)
6
4
(4, 3)
(0, 3)
2
x1 + x2 y1 + y2
2 + 2 , −2 + 8
, = 2
2
2
2
0
= (2, 3)
–2
Since the transverse axis is parallel to the y-axis, and the centre is not
at the origin, the standard form of the equation of the ellipse is
(y − k)2 (x − h)2
−
= 1.
a2
b2
Since the centre is (2, 3), h = 2 and k = 3.
2
4
6
x
(2, –2)
The length of the transverse axis, which is parallel to the y-axis, is 10.
Since 2a = 10, a = 5.
The length of the conjugate axis, which is parallel to the x-axis, is 4.
Since 2b = 4, b = 2.
8.6 The Hyperbola • MHR 645
2
(y − 3)
(x − 2) = 1.
The equation of the hyperbola is − 25
4
2
The foci are F1(h, k − c) and F2(h, k + c).
To find c, use a2 + b2 = c2, with a = 5 and b = 2.
c2 = a2 + b2
= 52 + 22
= 25 + 4
= 29
c = 29
The foci are located at (h, k − c) and (h, k + c). Thus, the coordinates of
the foci are (2, 3 − 29
) and (2, 3 + 29
), or approximately (2, −2.39)
and (2, 8.39).
y
To sketch the graph of the hyperbola, plot the vertices and the
co-vertices. Then, construct the rectangle that goes through all
four of these points. Sketch the asymptotes by extending the
diagonals of the rectangle.
Since the transverse axis is parallel to the y-axis, the hyperbola
must open up and down. To sketch a branch of the hyperbola,
start at a vertex and approach the asymptotes. Sketch the other
branch in the same way. Label the foci.
10
F2(2, 3 + 29)
8
V2(2, 8)
6
4
(0, 3)
(4, 3)
2
–2
0
–2
–4
–6
Note that hyperbolas can be graphed using a graphing calculator. As with
circles and ellipses, the equations of hyperbolas must first be solved for y.
2
y
2
x2 = 1, results in y = ± 2 9.
For example, solving − + x
3
4
9
Enter both of the resulting equations in the Y= editor.
2
2 + x2 and Y2 = – 2 9
Y1 = 9
+ x
3
3
646 MHR • Chapter 8
2
4
6
V1(2, –2)
F1(2, 3 – 29)
x
Adjust the window variables, if necessary, and use the Zsquare instruction.
Key
Concepts
• A hyperbola is the set, or locus, of points P in the plane such that the
absolute value of the difference of the distances from P to two fixed points
F1 and F2 is a constant.
|F1P − F2P| = k
• The standard form of the equation of a hyperbola, with centre at the
2
y2
y2 x2
x
origin, is either 2 − 2 = 1 (transverse axis along the x-axis) or 2 − 2 = 1
a
b
a
b
(transverse axis along the y-axis).
• The standard form of the equation of a hyperbola, centred at (h, k), is
2
(x − h)2 − (y − k) = 1 (transverse axis parallel to the x-axis)
either a2
b2
2
(y − k)
(x − h)2 = 1 (transverse axis parallel to the y -axis).
or −
a2
b2
Communicate
Yo u r
Understanding
1. Describe how you would use the locus definition of the hyperbola to find
an equation of the hyperbola with foci F1(0, −5) and F2(0, 5), and with the
constant difference between the focal radii equal to 8.
2
(x − 2)2 ( y + 4)
2. Describe how you would sketch the graph of − = 1.
25
9
3. Describe how you would write an equation in standard form for the
hyperbola with vertices (3, −4) and (3, 6) and co-vertices (1, 1) and (5, 1).
4. How would you know if
a) an equation in standard form models an ellipse or a hyperbola?
b) a graph models an ellipse or a hyperbola?
5. Is a hyperbola a function? Explain.
8.6 The Hyperbola • MHR 647
Practise
A
1. Use the locus definition of the hyperbola
to find an equation for each of the following
hyperbolas, centred at the origin.
a) foci at (−5, 0) and (5, 0), with the
difference between the focal radii 6
b) foci at (0, −5) and (0, 5), with the
difference between the focal radii 6
c)
d)
y
y
6
8
4
6
2
4
–4
2
2. For each hyperbola, determine
i) the coordinates of the centre
ii) the coordinates of the vertices and
–2
0
2
4
–2
–2
0
2
x
–4
–2
co-vertices
iii) the lengths of the transverse and conjugate
axes
iv) an equation in standard form
v) the coordinates of the foci
vi) the domain and range
–6
–4
–6
–8
a)
y
2
–8
–6
–4
–2
0
2
4
6
8
–2
b)
y
2
–4
–2
0
–2
–4
648 MHR • Chapter 8
x
Determine an equation in standard form
for each of the following hyperbolas.
a) vertices (0, −4) and (0, 4), co-vertices
(−5, 0) and (5, 0)
b) vertices (−3, 0) and (3, 0), foci (−5, 0)
and (5, 0)
c) foci (−5, 0) and (5, 0), with constant
difference between focal radii 4
d) foci (0, −3) and (0, 3), with constant
difference between focal radii 5
4.
4
–6
Sketch the graph of each hyperbola. Label
the coordinates of the centre, the vertices, the
co-vertices, and the foci, and the equations of
the asymptotes.
2
2
2
y
y
x2
x
a) − = 1
b) − = 1
25 36
16 49
2
2
2
2
c) 25x − 4y = 100
d) 4y − 9x = 36
3.
2
4
6
x
x
5. For each hyperbola, determine
i) the coordinates of the centre
ii) the coordinates of the vertices and
d)
6
co-vertices
iii) the lengths of the transverse and conjugate
axes
iv) an equation in standard form
v) the coordinates of the foci
a)
y
4
2
0
y
–2
6
–4
4
–6
2
4
6
x
2
–2
0
2
4
6. Sketch the graph of each hyperbola. Label
the coordinates of the centre, the vertices, the
co-vertices, and the foci.
2
( y + 1)2
(x − 3)
a) − = 1
16
9
2
(y + 2)
(x − 1)2
b) − = 1
81
25
2
2
c) 16x − 9(y − 2) = 144
2
2
d) 4(y − 2) −9(x − 5) = 36
6x
–2
–4
y
b)
6
4
2
–8
–6
–4
–2
0
2
4
–2
–4
c)
y
–4
–2
0
–2
2
4
Determine an equation in standard form
for each of the following hyperbolas.
a) vertices (−2, −5) and (6, −5), foci
(−4, −5) and (8, −5)
b) centre (0, 3), one vertex (0, 5), and one
focus (0, 6)
c) centre (−3, 1), one focus (−5, 1), and
length of conjugate axis 4
d) centre (−2, 2), one focus (−2, 5), and
length of conjugate axis 6
7.
6
x
x
8.6 The Hyperbola • MHR 649
Apply, Solve, Communicate
B
A ship is monitoring the movement of a pod of
whales with its radar. The radar screen can be modelled as a coordinate grid
with the ship at the centre (0, 0). The pod appears to be moving along a
curve such that the absolute value of the difference of its distances from
(2, 7) and (2, −3) is always 6. Write an equation in standard form to
describe the path of the pod.
8. Marine biology
Some comets travel on hyperbolic paths and never
return. Suppose the hyperbolic path of a comet is modelled on a grid
with a scale of 1 unit = 3 000 000 km. The vertex of the path is at
(2, 0), and the focus is at (6, 0). Write an equation in standard form to
model the path of the comet.
9. Motion in space
10. Conjugate hyperbolas a) Two
hyperbolas are centred at (1, 2). One
has a transverse axis parallel to the x-axis, and the other has a transverse axis
parallel to the y -axis. They share the same pair of asymptotes. The equation
2
2
(y − 1)
(x − 2) = 1. Find the lengths of the
of one hyperbola is − 25
9
conjugate and transverse axes, and the equation in standard form, of the
other hyperbola.
b) The two hyperbolas in part a) are known as conjugate hyperbolas. Write
equations in standard form for another pair of conjugate hyperbolas with
their common centre not at the origin. Explain your reasoning.
Two park rangers were stationed at separate locations on
the same side of a lake. The rangers were 6 km apart. Both rangers saw a
lightning bolt strike the ground on the other side of the lake and heard the
clap of thunder.
a) If one ranger heard the clap of thunder 4 s before the other, write an
equation that describes all the possible locations of the thunder clap. Place
the two rangers on the y-axis, with the midpoint between the rangers at the
origin. The speed of sound is approximately 0.35 km/s.
b) Sketch the possible locations where the lightning bolt hit the ground.
Include the ranger stations in the sketch.
11. Application
650 MHR • Chapter 8
Use a graphing calculator to graph each of the following.
2
2
( y + 1)2
( y − 1)
(x + 2)
(x + 3)
a) − = 1
b) − = 1
9
4
4
16
12. Technology
2
The LORAN system uses the locus definition of the
hyperbola. The system measures the difference in the arrival times at a ship
or aircraft of radio signals from two stations. The difference in the arrival
times is converted to a difference in the distances of the ship or aircraft from
the two stations. Two pairs of stations sending radio signals define two
hyperbolas. The exact location of the ship or aircraft is a point of
intersection of the two hyperbolas.
a) On a grid with a scale in kilometres, plot and label the foci F1(−6, 0) and
F2(6, 0) of a hyperbola. The foci represent two stations, F1 and F2, that send
radio signals to a ship positioned at point P. Find an equation of the
hyperbola, if the difference in the arrival times of the radio signals at P is
converted to the difference in distances |F1P − F2P| = 8.
b) On the same grid as in part a), plot and label the foci F3(0, −5) and
F4(0, 5) of another hyperbola. The foci represent another two stations, F3
and F4, that send radio signals to the same ship at point P. Find an equation
of this hyperbola, if |F3P − F4P| = 4.
c) Sketch the two hyperbolas and estimate the coordinates of the points of
intersection. How many points of intersection are there?
d) Use a graphing calculator to find the coordinates of the points of
intersection of the two hyperbolas, to the nearest tenth. If the second
quadrant of the grid is the only quadrant that contains a body of water,
what are the coordinates of the ship?
13. Navigation
Use a graphing calculator to graph the family of
2
y2
x
hyperbolas − 2 = 1 for b = 1, 2, and 3.
25 b
1 1 1
b) Now, graph for b = , , .
2 3 4
c) How are the graphs alike? How are they different?
d) What happens to the hyperbola as b gets closer to 0?
14. Communication a)
C
15. Inquiry/Problem Solving
The eccentricity, e, of a hyperbola is
defined as e = c . Since c > a, it follows that e > 1.
a
a) Describe the general shape of a hyperbola whose eccentricity is close to 1.
b) Describe the shape if e is very large.
8.6 The Hyperbola • MHR 651
16. Equilateral hyperbolas
The following are examples of equilateral
hyperbolas.
2
2
2
2
2
2
2
2
i) x − y = 4
ii) x − y = 9
iii) y − x = 16
iv) y − x = 25
a) What do the equations have in common?
b) Graph each hyperbola. State the equations of its asymptotes.
c) What do the graphs have in common that is different from the other
hyperbolas you have graphed?
Consider a hyperbola with its transverse axis parallel to
the x-axis, foci at (−c, 0) and (c, 0), and vertices at (−a, 0) and (a, 0).
a) Use the formula for the length of a line segment, l = (x
− x
)2 + (y,
− y1)2
2
1
2
to show that, for a point P(x, y) on the hyperbola,
2
+ y2 − (x
+ c)2
+ y2 = 2a and (x
+ c)2
+ y2 − (x
− c)2
+ y2 = 2a.
(x
− c) 2
2
y
x
b) From the equations in part a), derive − 2 = 1.
2
2
c −b
b
2
2
y
x
c) Derive 2 − 2 = 1, the standard form for the hyperbola centred at
a
b
(0, 0) with transverse axis parallel to the x-axis.
17. Standard form
When one variable increases and another variable
decreases, such that their product is a constant, the relationship can be
represented by xy = k, k ≠ 0. The relationship can be stated verbally as y varies
inversely as x. The graph of the inverse variation is one branch of a hyperbola,
for x, y > 0. Complete the following using a graphing calculator.
a) Graph xy = 2, xy = 5, and xy =10. What are the similarities and differences
for xy = k?
b) Graph xy = −2, xy = −5, and xy = −10. Compare these graphs with the
graphs in part a).
c) Explain why the graphs in parts a) and b) do not go through the origin.
d) Can you graph xy = 0? Explain.
18. Inverse variation
A C H I E V E M E N T Check
Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
Find an equation for the locus of points such that the product of the slopes of the lines
from a point on the locus to the points (6, 0) and (−6, 0) is 4.
652 MHR • Chapter 8