HYPERBOLAS

Introduction
A hyperbola is one of the conic sections that most students have not encoun-
tered formally before, unlike circles and parabolas. Its graph consists of two
unbounded branches which extend in opposite directions. It is a misconception
that each branch is a parabola. This is not true, as parabolas and hyperbolas
have very di↵erent features. An application of hyperbolas in basic location and
navigation schemes are presented in an example and some exercises.

1.4.1. Definition and Equation of a Hyperbola

Consider the points F1(—5, 0) and F2(5, 0) as shown in Figure 1.21. What is the
absolute value of the di↵erence of the distances of A(3.75, —3) from F1 a.nd fromΣ
F2? How about the absolute value of the di↵erence of the distances of B —5,16
from F1 and from F2? 3

|AF1 — AF2| = |9.25 — 3.25| = 6

|BF
— BF | = . 16 — 34 . = 6
1 2 .
3 3.
There are other points P such that| PF 1 —PF 2 = 6. The collection of all such
points forms a shape called a hyperbola, which
| consists of two disjoint branches.
For points P on the left branch, PF 2 — PF 1 = 6; for those on the right branch,
PF 1 — PF 2 = 6.

Figure 1.21 Figure 1.22

 Let F1 and F2 be two distinct points. The set of all points P , whose
distances from F1 and from F2 di↵er by a certain constant, is called a
hyperbola. The points F1 and F2 are called the foci of the hyperbola.

In Figure 1.22, given are two points on the x-axis, F1( c,

— 0) and F2(c, 0), the
foci, both c units away from their midpoint (0, 0). This midpoint is the center
 Figure 1.23 Figure 1.24

of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute
value of the di↵erence of the distances of P from F1 and F2, be 2a (the coefficient
2 will make computations simpler). Thus, |PF 1 — PF 2 | = 2a, and so
p p
(x + c)2 + y2 — (x — c)2 + y2 = 2a.
. .

Algebraic manipulations allow us to rewrite this into the much simpler

x2 y—2 p
a2 = 1, where b c2 — a2.
b2 =
p
When we let b = c2 — a2, we assumed c > a. To see why this is true, suppose
that P is closer to F2, so PF 1 — PF 2 = 2a. Refer to Figure 1.22. Suppose also
that P is not on the x-axis, so 4PF 1F 2 is formed. From the triangle inequality,
F1F2 + PF 2 > PF 1 . Thus, 2c> PF 1 — PF 2 = 2a, so c > a.
Now we present a derivation. For now, assume P is closer to F 2 so PF1 > PF2,
and
— PF 1 PF 2 = 2a.
PF 1 = 2a + PF 2
p p
(x + c) + y2 = 2a + (x — c)2 + y2
2

(x + c)2 + = 2a (x — c)2 +
⌘2 ⇣ +
2
⇣p y
2
py ⌘2
p
⇣ p cx — a2 = a (x⌘— 2
2 c) + y
2
2 2
— —
(c — a )x — a y = a (c — a2)
2 2 2 2 2 2 2
2 2(cx a ) = a (x c) p
bx —a y =ab
2 2 2 2 2 2
by letting b = c2 — a2 > 0
x2 y2
— 2 =1
a2 b
 We collect here the features of the graph of a hyperbola with standard equa-
tion
x2 y2
p — = 1.
Let c = a2 + b2. a2 b2
(1) center : origin (0, 0)
(2) foci : F1(—c, 0) and F2(c, 0)

• Each focus is c units away from the center.

• For any point on the hyperbola, the absolute value of the di↵erence of
its distances from the foci is 2a.

(3) vertices: V1(—a, 0) and V2(a, 0)

The vertices are points on the hyperbola, collinear with the center and
• foci.
• If y = 0, then x = ±a. Each vertex is a units away from the center.
• The segment V1V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = b x and y = — b x, the lines `1 and `2 in Figure 1.24
a a

• The asymptotes of the hyperbola are two lines passing through the
cen- ter which serve as a guide in graphing the hyperbola: each
branch of the hyperbola gets closer and closer to the asymptotes, in
the direction towards which the branch extends. (We need the
concept of limits from calculus to explain this.)
• An aid in determining the equations of the asymptotes: in the standard
2 y2
equation, replace 1 by 0, and in the resulting equation x — = 0, solve
for y. 2
a b 2
• To help us sketch the asymptotes, we point out that the asymptotes
`1 and `2 are the extended diagonals of the auxiliary rectangle drawn
in Figure 1.24. This rectangle has sides 2a and 2b with its diagonals
intersecting at the center C. Two sides are congruent and parallel to
the transverse axis V1V2. The other two sides are congruent and
parallel to the conjugate axis, the segment shown which is
perpendicular to the transverse axis at the center, and has length 2b.
Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola
with equation
x2 — y2 = 1.
9 7
Sketch the graph, and include these points and lines, the transverse and
conjugate axes, and the auxiliary rectangle.
Solution. Wpith a2 = 9 apnd b2 = 7, we have
a = 3, b = 7, and c = a2 + b2 = 4.
foci: F1(—4, 0) and F2(4, 0)
vertices: V1(—3, 0) and V2(3, 0)
p
asymptotes: y = 73 x and y = — p73 x
The graph is shown at the right. The cponju-
gate axis drawn has its endpoints b = 7 ⇡
2.7 units above and below the center. 2
Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are F1(
5, 0) and F2(5, 0), such that for any point on it, the absolute value of the
di↵—erence of its distances from the foci is 6. See Figure 1.21.
p
Solution. We have 2a = 6 and c = 5, so a = 3 and b = c2 — a2 = 4. The
x2 y2
hyperbola then has — = 1. 2
equation 9 16

Seatwork/Homework 1.4.1
1. Determine foci, vertices, and asymptotes of the hyperbola with equation
x2 y 2
— = 1.
16 20
Sketch the graph, and include these points and lines, along with the
auxiliary rectangle.
Answer: foci F1(—6, 0) and F2(6, 0), vertices V1(—4, 0) and V2(4, 0), asymp-
totes y = p5 x and y = — p5 x
2 2

p
2. Find the epquation in standard form of the hyperbola whose foci are F1 (—4 2, 0)
and F2(4 2, 0), such that for any point on it, the absolute value of the
x 2 — y2 = 1
di↵erence of its distances from the foci is 8. Answer:
16 16
1.4.2. More Properties of Hyperbolas

The hyperbolas we considered so far are “horizontal” and have the origin as
their centers. Some hyperbolas have their foci aligned vertically, and some have
centers not at the origin. Their standard equations and properties are given in
the box. The derivations are more involved, but are similar to the one above, and
so are not shown anymore.

Center Corresponding Hyperbola

(0, 0)

x2 y 2 y 2 x2
— 2 =1 — =1
a2 b a2 b2

(h, k)

(x — h)2 (y — k)2 (y — k)2 (x — h)2

— =1 — =1
a2 b2 a2 b2
transverse axis: horizontal transverse axis: vertical
conjugate axis: vertical conjugate axis: horizontal
 p
In all four cases above, we let c = a2 + b2. The foci 1F and F2 are c units
away from the center C. The vertices V1 and V2 are a units away from the center.
The transverse axis V1V2 has length 2a. The conjugate axis has length 2b and is
perpendicular to the transverse axis. The transverse and conjugate axes bisect
each other at their intersection point, C. Each branch of a hyperbola gets closer
and closer to the asymptotes, in the direction towards which the branch extends.
The equations of the asymptotes can be determined by replacing 1 in the
standard equation by 0. The asymptotes can be drawn as the extended diagonals
of the auxiliary rectangle determined by the transverse and conjugate axes.
Recall that, for any point on the hyperbola, the absolute value of the di↵erence of
its distances from the foci is 2a.
In the standard equation, aside from being positive, there are no other re-
strictions on a and b. In fact, a and b can even be equal. The orientation of the
hyperbola is determined by the variable appearing in the first term (the positive
term): the corresponding axis is where the two branches will open. For example,
if the variable in the first term is x, the hyperbola is “horizontal”: the transverse
axis is horizontal, and the branches open to the left and right in the direction of
the x-axis.
Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymp-
totes of the hyperbola with the given equation. Sketch the graph, and include
these points and lines, the transverse and conjugate axes, and the auxiliary rect-
angle.
(y + 2)2 (x — 7)2
(1) — =1
25 9
(2) 4x2 — 5y2 + 32x + 30y = 1

Solutipon. (1) Frpom a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =

a2 + b2 = 34 ⇡2 5.8. The hyperbola is vertical. To determine the asymp-
(x—7)2
totes, we write 25 —
(y+2)
9 = 0, which is equivalent to y +2 = ± 53 (x — 7).
We can then solve this for y.

center: C(7, —2)

p ⇡ (7, —7.8) and F2(7, —2+
foci: F1(7, —2 — 34) p34) ⇡ (7, 3.8)
vertices: V1(7, —7) and V2(7, 3)
asymptotes: y = 5 x — 41 and y = — 5 x + 29
3 3 3 3

The conjugate axis drawn has its endpoints b = 3 units to the left and right
of the center.
(2) We first change the given equation to standard form.

4(x2 + 8x) — 5(y2 — 6y) = 1

4(x2 + 8x + 16) — 5(y2 — 6y + 9) = 1+ 4(16) — 5(9)
4(x + 4)2 — 5(y — 3)2 = 20
(x + 4)2 (y — 3)2
— =1
5 4
p p
We have a = 5 ⇡ 2.2 and b = 2. Thus, c = a2 + b2 = 3. The hyperbola
2
is horizontal. To determine the asymptotes, we write (x+4) 5 — (y—
3)2 = 0
2
which is equivalent to y — 3 = ± p5 (x + 4), and solve for y. 4

center: C(—4, 3)
foci: F1( 7, 3) and F2( 1, 3)
— p p
vertices: V1(—4 — 5, 3) ⇡ (—6.2, 3) and V2(—4+ 5, 3) ⇡ (—1.8, 3)
asymptotes: y = p2 x + p8 + 3 and y = — p2 x — p8 + 3
5 5 5 5

The conjugate axis drawn has its endpoints b = 2 units above and below
the center.
Example 1.4.4. The foci of a hyperbola are ( 5, —
3) and
— (9, 3). For any point on the
hyperbola, the absolute value of the di↵erence of its—of its distances from the foci
is 10. Find the standard equation of the hyperbola.

Solution. The midpoint (2, 3)—of the foci is the center of the hyperbola. Each focus
is c = 7 units away from the center. From the given di↵erence, 2a = 10 so a = 5.
Also, b2 = c2 a2 = 24.—The hyperbola is horizontal (because the foci are horizontally
aligned), so the equation is

(x 2)2 (y + 3)2
— —
25 = 1. 2
24
Example 1.4.5p. A hyperbola has vertices (—4, —5) and (—4, 9), and one of its
foci is (—4, 2 — 65). Find its standard equation.

Solution. The midpoint (—4, 2) of the vertices is the center of the hyperbola,
which is vertical (because the vertices are vertically alignped). Each
vertex is
a = 7 units away from the center. The given focus is c = 65 units away from
the center. Thus, b2 = c2 — a2 = 16, and the standard equation is

(y 2)2 (x + 4)2
— —
49 = 1. 2
16
Seatwork/Homework 1.4.2
1. Give the coordinates of the center, foci, vertices, and asymptotes of the hy-
perbola with equation 9x2 4y2 90x 32y = 305. Sketch the graph, and include
— with—the auxiliary
these points and lines, along — rectangle.
— p p
Answer: center C(5, —4), foci F1(5, —4 — 23 13) and F2(5, —4+2 13), vertices
V (5, —10) and V (5, 2), asymptotes y = — x + 7 and y = 3 x — 23
1 2 2 2 2 2

2. A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is (—2, 9). Find
Answer: (x — 7) — (y — 9) = 1
2 2
its standard equation.
36 45

1.4.3. Situational Problems Involving Hyperbolas

We now give an example on an application of hyperbolas.

Example 1.4.6. An explosion is heard by two stations 1200 m apart, located at
F1(—600, 0) and F2(600, 0). If the explosion was heard in F1 two seconds before
it was heard in F2, identify the possible locations of the explosion. Use 340 m/s
as the speed of sound.

Solution. Using the given speed of sound, we deduce that the sound traveled
340(2) = 680 m farther in reaching F 2 than in reaching F1. This is then the
di↵erence of the distances of the explosion from the two stations. Thus, the
explosion is on a hyperbola with foci are F 1 and F2, on the branch closer to
F 1.
 We have c = 600 and 2a = 680, so a = 340 and b2 = c2 — a2 = 244400.
The2explosion could therefore be anywhere on the left branch of the hyperbola
x y2
115600 — 244400 = 1. 2

Seatwork/Homework 1.4.3
?1. Two stations, located at M ( 1.5, 0) and N (1.5, 0) (units are in km),
—
simulta- neously send sound signals to a ship, with the signal traveling at
the speed of
0.33 km/s. If the signal from N was received by the ship four seconds before
the signal it received from M , find the equation of the curve containing
the y
2
possible location of the ship. Answer: — = 1 (right branch)
x
2 0.4356
1.8144

Exercises 1.4

1. Give the coordinates of the center, foci, vertices, and the asymptotes of the
hyperbola with the given equation. Sketch the graph, and include these points
and lines.
2 2
(a) x — y = 1
36 64
2 2
(b) y — x = 1
25 16
(c) (x — 1)2 — y2 = 4
(y + 2)2 (x + 3)2
(d) — =1
15 10
(e) 3x2 — 2y2 — 42x — 16y = —67
(f) 25x2 — 39y2 + 150x + 390y = —225
Answer:

Item Center Vertices Foci

(a) (0, 0) (±6, 0) (±10, 0)
p
(b) (0, 0) (0, ±5) (0, ± 41)
p
(c) (1, 0) (—1, 0), (3, 0) (1 ± 2 2, 0)
p
(d) (—3, —2) (—3, —2 ± 15) (—3, —7), (—3, 3)
p
(e) (7, —4) (3, —4), (11, —4) (7 ± 2 10, —4)
(f) (—3, 5) (—3, 0), (—3, 10) (—3, —3), (—3, 13)

Item Asymptotes
(a) y=±4x 3

(b) y = ± 54 x
(c) y = x 1, y = x + 1
q— q
(d) y=± 3x±3 3 —2
q2 q2
(e) y = ± 2x ⌥ 7 2 3 —
3

(f)4 y = ± p5 x ± p15 + 5
39 39

(a) (b)
 (c) (d)

(e) (f)
2. Find the standard equation of the hyperbola which satisfies the given condi-
tions.
(a) foci ( 4, 3) and ( 4, 13), the absolute value of the di↵erence of the
— —
distances of any point from the foci is 14
—
Answer (y — 5) (x + 4)2
2
— =1
: 49 15
(x 3)2
(b) vertices (—2, 8) and (8, 8), a focus (12, 8) — (y 8)2
25
— —
=1
Answer: 56
(c) center (—6, 9), a vertex (—6, 15), conjugate axis of length 12
—
Answer: (y 9)2 (x + 6)2
— =1
25 36
(d) asymptotes y = 4 x + 1 and y = — 4 x + 41 , a vertex (—1, 7)
3 3 3 3
(x — 5)2 (y — 7)2
— =1
Answer 36 64
:
Solution. The asymptotes intersect at (5, 7). This is the center. The
distance of the given vertex from the center is a = 6. This vertex and
center are aligned horizontally, so the hyperbola has equation of the
form
(x—h)2 — (y —k)2
a 2 b2
= 1. The asymptotes consequently
have the form y — k =
 b b b 4
±a — (x h), and thus, have slopes a . From the given asymptotes, a=3.
Since ±a = 6, then b = 8. The standard equation is then
(x 5)2 (y 7)2
— — —
36 = 1.
64
(e) asymptotes y = 1 x + 5 and y = — 1 x + 7 , a focus (1, 12)
3 3 3 3
(y — 2)2 (x — 1)2
— =1
Answer 10 90
:
Solution. The asymptotes intersect at (1, 2). This is the center. The
distance of the given focus from the center is c = 10. This focus and
center are aligned vertically, so the hyperbola has equation of the
form
(y—k) 2
(x = 1. The asymptotes consequently have the form y — k =
—h)2 —
a2 b2 a a 1
± b (x—h), and thus, have slopes b . From the given asymptotes, b = 3 ,
so b =±3a.
c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2
Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is
(y 2)2 (x 1)2
— — —
10 = 1.
90
3. Two control towers are located at points Q( — 500, 0) and R(500, 0), on a
straight shore where the x-axis runs through (all distances are in meters).
At the same moment, both towers sent a radio signal to a ship out at sea, each
traveling at 300 m/µs. The ship received the signal from Q 3 µs
(microseconds) before the message from R.
(a) Find the equation of the curve containing the possible location of the
x2 y2
ship. Answer: — = 1 (left branch)
202500 47500
(b) Find the coordinates (rounded o↵ to two decimal places) of the ship if it
is 200 m from the shore (y = 200). Answer: (—610.76, 200)
Solution. Since the time delay between the two signals is 3 µs, then the di↵er-
ence between the distances traveled by the two signals is 300 3 = 900 m. The
ship is then on a hyperbola, consisting of points whose distances· from Q and R
(the foci) di↵er by 2a = 900. With a = 450 and c = 500 (the distance of each
focus from the center, the origin), we have b2 = c2 — a2 = 5002 — 450
2
2
= 47500.
Since a = 202500, the hyperbola then has equation 202500 —
2 y
= 1. Since
47500
x
2
the signal from Q was received first, the ship is closer to Q than R, so the
ship is on the x2 left 200
branch
2 of this hyperbola. Using y = 200, we then solve
202500 — 47500 = 1 for x< 0 (left branch), and we get x ⇡ —610.76.