Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 113 pptx

E.2 Defining Conics Algebraically 1101
F
1
and F
2
are called the foci of the ellipse.
To sketch an ellipse, you can
tack one end of a string to each
focus and, with the point of
a pen, hold the string taut. by
moving around the foci you'll
trace out an ellipse.
d
1
+ d
2
is constant
d
1
F
1
F
2
d
2
Pen
Figure E.5 Ellipse
The farther apart the two foci, the more elongated the ellipse. The closer they are to
one another, the more circular the ellipse looks. If the two foci merge to one point, then
the figure is a circle.
A hyperbola

A hyperbola is the set of points (x, y) the difference of whose distances from two
distinct fixed points F
1
and F
2
is constant.
F
1
and F
2
are called the foci of the hyperbola.
|d
2
– d
1
| is constant
d
1
d
1
d
2
d
2
F
1
F
2
Figure E.6
E.2 DEFINING CONICS ALGEBRAICALLY

Next we relate geometric and algebraic representations of conics. First the conclusions
are presented. Then a road map is given. The actual trips from geometric to algebraic
respresentations areleft asproblems atthe endof thesection. Signposts are provided. To find
these problems, look under the appropriate conic section; they lead the section of problems
for the particular conic.
1102 APPENDIX E Conic Sections
Relating the Geometric and Algebraic Representations of Conics
If a circle has center (0, 0) and the distance from the center is denoted by r, then the
circle is given by the equation
x
2
+ y
2
= r
2
.
y
y
x
x
r
Circle: center (0, 0)
radius r
Figure E.7
If a parabola has focus (0, c) and directrix y =−c,then it will have a vertex at (0, 0)
and be given by the equation
y =
1
4c
x

2
.
y
x
c
Focus
directrix
Parabola:
focus (0, c)
directrix y = – c
y = – c
Figure E.8
If an ellipse has foci at (c,0)and (−c,0)and the sum of the distances from any point
on the ellipse to the foci is denoted by 2a, then the ellipse will have x-intercepts of
(a,0)and (−a,0)and be given by the equation
x
2
a
2
+
y
2
b
2
= 1,
where b
2
= a
2
− c

2
. (See Figure E.9(a).)
If an ellipse has foci at (0, c) and (0, −c) and the sum of the distances from any point
on the ellipse to the foci is denoted by 2a, then the ellipse will have y-intercepts of
(0, a) and (0, −a) and be given by the equation
y
2
a
2
+
x
2
b
2
= 1,
where b
2
= a
2
− c
2
. (See Figure E.9(b).)
E.2 Defining Conics Algebraically 1103
y
y
xx
a
a
c
c

–c
–c
–a
–a
–b
–b
b
b
Ellipse with foci (c, 0), (–c, 0)
(a)
Ellipse with foci (0, c), (0, –c)
(b)
Figure E.9
If a hyperbola has foci at (c,0)and (−c,0)and the magnitude of the difference of
the distances from any point on the hyperbola to the foci is denoted by 2a, then the
hyperbola will have x-intercepts of (a,0)and (−a,0)and be given by the equation
x
2
a
2
−
y
2
b
2
= 1,
where b
2
= c
2

− a
2
. (See Figure E.10(a).)
If a hyperbola has foci at (0, c) and (0, −c) and the magnitude of the difference of
the distances from any point on the hyperbola to the foci is denoted by 2a, then the
hyperbola will have y-intercepts of (0, a) and (0, −a) and be given by the equation
y
2
a
2
−
x
2
b
2
= 1,
where b
2
= c
2
− a
2
. (See Figure E.10(b).)
y
y
xx
a
a
b
b

c
c
–a
–a
–b
–b
–c
–c
y = x
–b
a
y = x
– a
b
y = x
b
a
y = x
a
b
Hyperbola with foci (c, 0) (– c, 0)
(a)
Hyperbola with foci (0, c) (0, –c)
(b)
Figure E.10
1104 APPENDIX E Conic Sections
Roadmaps
Circle: Let (x, y) be a point on the circle with center (0, 0). Use the distance formula
(or the Pythagorean Theorem) to conclude that if the distance from (x, y) to (0, 0) is
r, then x

2
+ y
2
= r
2
.
Parabola: Let (x, y) be a point on the parabola with focus F at (0, c) and directrix L
at y =−c.Using the distance formula, we know that the distance from (x, y) to F is

x
2
+ (y − c)
2
and the distance from (x, y) to L is

(y + c)
2
. Equate the distances,
square both sides, and simplify.
Ellipse: Let (x, y) be a point on the ellipse with foci at (c,0)and (−c,0).Denote the
sum of the distances from any point on the ellipse to the foci by 2a. (This distance is
some positive number, and denoting it by 2a is convenient.) Using the distance formula
we have that

(x + c)
2
+ y
2
+


(x − c)
2
+ y
2
= 2a.
Isolate one radical and square both sides to eliminate it. Then isolate the other radical
and square both sides to eliminate the second radical. Simplifying and denoting the
quantity a
2
− c
2
by b
2
gives the desired result.
Hyperbola: Let (x, y)be a point on the hyperbola with foci at (0, c) and (0, −c). Denote
the magnitude (absolute value) of the differences of the distances from any point on
the hyperbola to the foci by 2a. (Denoting this difference by 2a is convenient.) Using
the distance formula we obtain

(x + c)
2
+ y
2
−

(x − c)
2
+ y
2
=±2a.

Isolate one radical and square both sides to eliminate it. Then isolate the other radical
and square both sides to eliminate the second radical. Simplifying and denoting the
quantity c
2
− a
2
by b
2
gives the desired result.
Conics Given as Ax
2
+ Bxy + Cy
2
+ Dx + Ey + F = 0
Consider the equation
Ax
2
+ Cy
2
= P ,
where A, C, and P are constants. This is
i. the equation of a circle if A = C, and the signs of A, C, and P agree;
ii. the equation of an ellipse if A, C, and P have the same signs;
iii. the equation of a hyperbola if A and C have opposite signs (to find out which way the
hyperbola opens, look for the x- and y-intercepts—you’ll find only one pair);
iv. the equation of a parabola if either A = 0orC=0.
Consider the graph of 4(x − 2)
2
+ 9(y − 3)
2

= 36. This is the ellipse 4x
2
+ 9y
2
= 36
shifted 2 units right and 3 units up. This example can be generalized.
The graph of A(x − h)
2
+ C(y − k)
2
= P is the graph of the conic section Ax
2
+
Cy
2
= P shifted right h units and up k units. Multiplying out gives an equation of the form
Ax
2
+ Cy
2
+ Dx + Ey + F = 0,
E.2 Defining Conics Algebraically 1105
where A, C, D, E, and F are constants.
Conversely, an equation of the form Ax
2
+ Cy
2
+ Dx + Ey + F = 0 can, by com-
pleting the square twice, be put in the form A(x − h)
2

+ C(y −k)
2
=P . By looking at the
signs of A and C as described above, one can determine the nature of the conic section.
◆
EXAMPLE E.1 Put the conic section
2x
2
− y
2
+ 4x + 6y − 8 =0
into the form A(x − h)
2
+ C(y −k)
2
=P and determine what it looks like.
SOLUTION
2x
2
− y
2
+ 4x + 6y − 8 =0
2x
2
+ 4x − y
2
+ 6y = 8
2(x
2
+ 2x) −(y

2
− 6y) =8
2

(x + 1)
2
− 1

−

(y − 3)
2
− 9

= 8
2(x + 1)
2
− 2 − (y − 3)
2
+ 9 = 8
2(x + 1)
2
− (y − 3)
2
= 1
Therefore, this conic is the hyperbola 2x
2
− y
2
= 1 (a hyperbola with x-intercepts at ±

1
√
2
)
shifted left 1 unit and up 3 units.
◆
In addition to shifting the conic Ax
2
+ Cy
2
= P horizontally and vertically, we can
rotate the conic. The resulting algebraic equation is of the form
Ax
2
+ Bxy +Cy
2
+Dx +Ey + F = 0,
where A, B, C, D, E, and F are constants.
It can be shown
1
that this equation gives
a parabola when B
2
− 4AC = 0,
an ellipse when B
2
− 4AC < 0 (a circle if, in addition, A = C),
a hyperbola when B
2
− 4AC > 0.

Notice that the sign of the expression B
2
− 4AC discriminates between the three types of
conics. This is not coincidental
In addition to having fascinating geometric characteristics, conic sections have many
practical applications.
1
This can be shown most elegantly using linear algebra; therefore the more accessible but less appealing argument is not given
here. Both arguments involve introducing a new, rotated coordinate system.
1106 APPENDIX E Conic Sections
E.3 THE PRACTICAL IMPORTANCE OF CONIC SECTIONS
Reflection Properties
The reflective properties of parabolas, hyperbolas, and ellipses are most useful when
considering surfaces formed by rotating the two-dimensional figures around the line through
the two foci in the latter two cases and through the focus and perpendicular to the directrix
in the former.
Reflecting properties of parabolas: A light source placed at the focus will be reflected
parallel to the axis of symmetry of the parabola. Conversely, sound or light waves
coming in parallel to the axis of symmetry of the parabola are reflected through the
focus, thereby concentrating them there.
This property makes parabolas a useful shape in the design of headlights and search-
lights, reflecting mirrors in telescopes, satellite dishes, radio antennas, and micro-
phones designed to pick up a conversation far away.
F
Reflective property
of a parabola
Parabolic headlight with
the bulb at the focus
Figure E.11
Reflecting property of ellipses: Sound or light emanating from one focus and reflecting

off the ellipse will pass through the other focus.
F
1
F
2
Reflection property
of an ellipse
Figure E.12
There are what are known as “whispering galleries” under the elliptic dome in
the Capitol building of the United States as well as in the Mormon Tabernacle in Salt
Lake City, Utah. A person standing at one focus and whispering to his neighbor can
be heard quite clearly by an individual located at the other focus. As you can imagine,
this can be a potential problem for those unaware of the reflection properties of the
ellipse.
In 1980 an ingenious medical treatment for kidney stones, called lithotripsy (from
the Greek word for “stone breaking”), was introduced. The treatment is based on the
reflective properties of the ellipse. Intense sound waves generated outside the body are
focused on the kidney stone, bombarding and thereby destroying it without invasive
E.3 The Practical Importance of Conic Sections 1107
surgery. Over 80% of patients with kidney stones can be treated with lithotripsy; the
recovery time is substantially less and the mortality rate much lower than with the
traditional surgery .
2
Reflecting property of hyperbolas: Light traveling along a line through one focus of a
hyperbola will be reflected off the surface of the hyperbola along the line through the
point of reflection and the other focus.
F
1
F
2

Reflection property
of a hyperbola
Figure E.13
The reflective property of hyperbolas is used in the construction of camera and telescope
lenses as well as radio navigation systems, such as the long-range navigation system
(LORAN).
Conic Sections and Astronomy
The path of any projectile under the force of gravity is a parabolic arc. Around 1600
Kepler found that planets have elliptical orbits around the sun. This discovery involved
painstaking calculations using measurements collected by Tycho Brahe in the late 1500s
over a period of more than two decades. Kepler’s conclusions (and the accuracy of Tycho
Brahe’s measurements) are rather astounding, given that the foci of most of the planets
are very close together, making their paths very nearly circular.
3
Newton later put Kepler’s
observations on solid theoretical grounds.
Comets can travel in elliptical orbits (like Halley’s comet) or in parabolic orbits. The
moon has an elliptical orbit with the earth at one focus.
PROBLEMS FOR APPENDIX E
Parabolas
1. Begin with the geometric characterization of a parabola. Suppose that the focus F is at
(0, c) and the directrix L is at y =−c.You will show that the set of points equidistant
from F and L is the parabola y =
1
4c
x
2
.
(a) Use the fact that the axis of symmetry is perpendicular to the directrix and passes
through the focus to deduce that the vertex of the parabola is at (0, 0).

2
The story of the development of this treatment is fascinating. It came out of scientists looking at aircraft. You can get more in-
formation on the scientific aspects at and on the mathematics of it at :
edu:80/m261vis/litho.
3
For more information on Kepler, consult a history of mathematics book, such as A History of Mathematics An Introduction
by Victor Katz, Addison Wesley 1998, pp. 409–416.
1108 APPENDIX E Conic Sections
5
2
y
x
(2.5, 2)
(b) Let (x, y) be a point on the parabola described above. Equating the distance
between the point (x, y) and F with the distance between the point (x, y) and
L gives the equation

x
2
+ (y − c)
2
=

(y + c)
2
.
Explain.
(c) Square both sides of the equation in part (b) to solve for y.
2. In Problem 1 you showed that a parabola with focus F at (0, c) and the directrix L at
y =−ccan be written in the form

y =
1
4c
x
2
.
(a) Find the equation of the parabola with vertex (0, 0) and focus (0, 1). Sketch its
graph. In the sketch show the location of the focus and directrix.
(b) Find the equation of the parabola with vertex (0, 0) and directrix y = 2. Sketch its
graph. In the sketch show the location of the focus and directrix.
3. In Problem 1 you showed that a parabola with focus F at (0, c) and the directrix L at
y =−ccan be written in the form
y =
1
4c
x
2
.
Find the focus and directrix of the parabola y = 2x
2
.
4. If the focus and directrix of a parabola are moved farther apart, does the parabola get
narrower or does it get wider? Explain.
5. Find the equation of a parabola with its vertex at the origin and its focus at (0, 6).
6. Find the equation of a parabola with vertex at (0, 0) and directrix y = 5.
7. A headlight is made of reflective material in the shape of a parabolic dish. In order to
take advantage of the fact that light emanating from the focus will be reflected by the
parabolic bowl in the direction of the axis of symmetry, where should the light source
be placed if the dish is 5 inches in diameter and 2 inches in depth?
(Hint: Look at a cross section of the reflector and introduce a coordinate system.

Let the y-axis lie along the axis of symmetry of the reflector and let the origin lie at the
vertex of the reflector.)
8. The light filament in the bulb of a floodlight is 1.5 inches from the vertex of the
paraboloid reflector. (The cross sections of the reflector taken through the vertex are all
identical parabolas.)
(a) Find the equation of the parabolic cross section.
(b) If the reflector dish is 10 inches in diameter, how deep is it?
E.3 The Practical Importance of Conic Sections 1109
Ellipses
9. Let (x, y) be a point on the ellipse with foci at (0, c) and (0, −c). You will arrive at the
formula
x
2
a
2
+
y
2
b
2
= 1,
where a
2
− C
2
= b
2
.
(a) Denote the sum of the distances from any point (x, y) on the ellipse to the foci
by 2a. (This distance is some positive number; denoting it by 2a is convenient.)

Using the distance formula we know that

(x + c)
2
+ y
2
+

(x − c)
2
+ y
2
= 2a.
Isolate

(x + c)
2
+ y
2
and square both sides of the equation to eliminate it.
After multiplying out you should be able to simplify the result to
a

(x + c)
2
+ y
2
= a
2
+ cx.

(b) Square both sides of the equation a

(x + c)
2
+ y
2
= a
2
+ cx to eliminate the
radical. Show that the result can be expressed in the form
(a
2
− c
2
)x
2
+ a
2
y
2
= a
2
(a
2
− c
2
). (E.1)
(c) The distance between the two foci is 2c. This must be less than the sum of the
distance between (x, y) and the foci. Thus 2a>2c,ora>c. Consequently,
(a

2
− c
2
)>0 and we can divide both sides of Equation E.1 by a
2
(a
2
− c
2
) to
get
x
2
a
2
+
y
2
a
2
− c
2
= 1.
Denote a
2
− c
2
by b
2
to get the desired result.

10. (a) In Problem 9 you showed that the ellipse with foci at (0, c) and (0, −c) is given
algebraically by
x
2
a
2
+
y
2
b
2
= 1.
Explain from an algebraic point of view how you know that b<a.
(b) What is the geometric significance of the constants a and b?
Hyperbolas
11. Let (x, y) be a point on the hyperbola with foci at (0, c) and (0, −c). Denote the
magnitude (absolute value) of the differences of the distances from any point on the
hyperbola to the foci by 2a.
(a) Using the distance formula show that

(x + c)
2
+ y
2
−

(x − c)
2
+ y
2

=±2a.
1110 APPENDIX E Conic Sections
(b) Carry on as in Problem 9. Isolate one radical and square both sides to eliminate it.
Then isolate the other radical and square both sides to eliminate the second radical.
Show that you obtain
(c
2
− a
2
)x
2
− a
2
y
2
= a
2
(c
2
− a
2
).
(c) Argue that c
2
− a
2
> 0. Once this is done, we can set b
2
= c
2

− a
2
. Simplify to
obtain
x
2
a
2
−
y
2
b
2
= 1.