Download Section 10.3 – Ellipses

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Section 10.3 – Ellipses
Objective a:
Definition of an ellipse & ellipses centered at (0, 0).
A graph that is an oval shape is called an ellipse. Just like we did in the last
section, we can define an ellipse as a set of points.
Definition: An ellipse is the set of all points P in a plane such that the sum
of the distance from one fixed point F1 to the point P and the distance from a
second fixed point F2 to the point P is constant. The two fixed points F1 and
F2 are called the foci (plural of focus) of the ellipse. An ellipse has
two axes of symmetry,
minor axis
the major axis and minor
P
axis. The major axis is the
"longer" axis and contains
d(F1, P)
the two foci. The midpoint
d(F2, P)
between the two foci is
the center ϑ of the ellipse
V1
F1
ϑ
F2
V2
and the points the ellipse
major
that intersect the major
axis
axis are the vertices V1 &
V2 (plural of vertex) of the
ellipse. The minor axis is
d(F1, P) + d(F2, P) = constant
perpendicular to the major axis and intersects the major axis at the center of
the ellipse. In the illustration above, the major axis runs parallel to the
x-axis. We could have just as easily drawn an ellipse with the major axis
running parallel to y-axis and all of the same definitions would still apply.
If we let a be the distance between
the center of the ellipse and one of
the vertices and let c be the distance
b
between the center of the ellipse and
a
one of the foci. Then the distance
V1
F1
ϑ
F2
V2
between V1 and F1 will be a – c and
the distance between F2 and V2 will
c
be a – c as well. Also, if we let b be
the distance between the center of
the ellipse and where the ellipse
intersects the minor axis, then there are two properties we can establish.
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1)
2)
d(F1, P) + d(F2, P) = 2a
b2 = a2 – c 2
Proof:
1) Recall that
d(F1, P) + d(F2, P) = constant
for any point P on the ellipse.
c
c
a–c
If we let P = V2, then d(F1, P)
= d(F1, ϑ) + d(ϑ
ϑ, F2) + d(F2, V2)
F1
ϑ
F2 a – c P = V2
=c+c+a–c
Also, d(F2, P) = a – c
Thus, d(F1, P) + d(F2, P)
= c + c + a – c + a – c = 2a
But, d(F1, P) + d(F2, P) = constant for any point on the ellipse, so
d(F1, P) + d(F2, P) = 2a
2) Now that we have established
P
the d(F1, P) + d(F2, P) = 2a, let P
be one of the points on the ellipse
a
b
that intersects the minor axis.
Then the line segment from F1 to P,
c
the line segment from P to F2, and
F1
ϑ
F2
the line segment from F2 to F1 form
an isosceles triangle. This means
that d(F1, P) = d(F2, P) = a. Also, the
line segment Pϑ
ϑ on the minor axis
bisects the isosceles triangle forming two right triangles with the lengths of
the legs of b and c and the length of the hypotenuse of a. By the
Pythagorean Theorem, a2 = b2 + c2 which implies b2 = a2 – c2
Now, we are ready to derive the
P: (x, y)
formula for the ellipse. For our
derivation, we will use an ellipse
centered at (0, 0) and with the
major axis parallel to the x-axis.
The derivation for the formula for
F1: (– c, 0)
the ellipse with the major axis
parallel to the y-axis works the
same way, just with the roles of x
and y reverse so we just need to
switch x and y in the answer to get the other form.
F2: (c, 0)
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Let F1: (– c, 0) and F2: (c, 0) be the foci and P: (x, y) be any point on the
ellipse. From property 1, then
(plug in the points)
d[F1, P] + d[F2, P] = 2a
d[(– c, 0), (x, y)] + d[(c, 0), (x, y)] = 2a
(x −(−c))2 +(y −0)2 +
(x + c)2 + y 2 +
(x −c)2 +(y −0)2 = 2a
(x −c)2 + y 2 = 2a
(x + c)2 + y 2 = 2a –
(
2
(x + c) + y
2
)
2
(apply the distance formula)
(add –
(x −c)2 + y 2
(
2
= 2a –
(simplify)
(x −c)2 + y 2 to both sides)
(square both sides)
(x −c) + y
2
)
2
(expand and simplify)
(x + c)2 + y2 = (2a)2 – 2(2a) (x −c)2 + y 2 + (x – c)2 + y2
(expand)
x2 + 2cx + c2 + y2 = 4a2 – 4a (x −c)2 + y 2 + x2 – 2cx + c2 + y2
(add – x2 + 2cx – c2 – y2 to both sides)
4cx = 4a2 – 4a (x −c)2 + y 2 (subtract 4a2 from both sides)
4cx – 4a2 = – 4a (x −c)2 + y 2
(divide both sides by – 4)
a2 – cx = a (x −c)2 + y 2
(square both sides)
(a
2
)
– cx
2
(
2
= a (x −c) + y
2
)
2
(expand and simplify)
a4 – 2a2cx + c2x2 = a2((x – c)2 + y2)
(expand and simplify)
a4 – 2a2cx + c2x2 = a2(x2 – 2cx + c2 + y2)
(distribute)
a4 – 2a2cx + c2x2 = a2x2 – 2a2cx + a2c2 + a2y2
(add 2a2cx – c2x2 – a2c2 to both sides)
a4 – a2c2 = a2x2 – c2x2 + a2y2 (factor a2 from a4 – a2c2 & x2 from a2x2 – c2x2)
a2(a2 – c2) = (a2 – c2)x2 + a2y2
a2b2 = b2x2 + a2y2
1=
x2
a2
+
y2
b2
or
x2
a2
(replace a2 – c2 by b2)
(divided both sides by a2b2)
+
y2
b2
=1
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If the major axis is along to the y-axis, then the equation is:
x2
b2
+
y2
a2
=1
Equation of an Ellipse Centered at the Origin
1)
The Ellipse with center (0, 0)
2)
The Ellipse with center (0, 0)
and equation
x2
a2
+
y2
b2
=1
and equation
has vertices of (± a, 0), foci of
(± c, 0), and endpoints on the
minor axis of (0, ± b) where
b2 = a2 – c2, a > b > 0, & c ≥ 0.
Major axis is along the x-axis.
x2
b2
+
y2
a2
=1
has vertices of (0, ± a), foci of
(0, ± c), and endpoints on the
minor axis of (± b, 0) where
b2 = a2 – c2, a > b > 0, & c ≥ 0.
Major axis is along the y-axis.
a
b
c
–a
–c
0
c
a
–b
0
b
–c
–b
–a
Find the equation of the ellipse. Then sketch the graph:
Ex. 1 Ellipse centered at the origin with a focus at (0, 4) and a vertex
at (0, – 6).
Solution:
Since the focus and vertex lie on the y-axis, then the major axis is
along y-axis, so we will be using the formula:
x2
b2
+
y2
a2
=1
The distance from the center (0, 0) to the given focus (0, 4) is 4 which
means c = 4 and the distance from the center to the given vertex
(0, – 6) is 6 which means a = 6 or a2 = 36 . Since b2 = a2 – c2, then
b2 = 62 – 42 = 36 – 16 = 20 which means b = 20 = 2 5
(Recall that a, b, and c are non-negative, so we can ignore the
negative solution b = – 2 5 )
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So, our equation is:
x2
20
+
y2
36
=1
To graph it, we will first plot the intercepts. The y-intercepts are the
vertices (0, ± 6) and the x-intercepts are (± b, 0) = (± 2 5 , 0). Now
draw the graph:
V2 = (0, 6)
(– 2
6
5
4
F2
3
2
1
5 , 0)
(2 5 , 0)
0
-1 0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1
-2
-3
F1
-4
-5
-6
V1 = (0, – 6)
Analyze the following equation:
Ex. 2 6x2 + 9y2 = 54
Solution:
To analyze the equation, we will find the center, the major axis, foci,
vertices, and then sketch the graph of the ellipse. First, we will need to
get the equation in the appropriate form:
6x2 + 9y2 = 54 (divide both sides by 54)
x2
9
+
y2
6
=1
Since the number under the x2 is larger than the number under the y2,
then the major axis runs along the x-axis and the ellipse is centered at
the origin. Also, a2 = 9 which means a = 3 and b2 = 6 which means
b = 6 . Plugging into b2 = a2 – c2, we can find c:
6 = 9 – c2 (solve for c2)
c2 = 3
c = 3 (c ≥ 0, so ignore the negative root)
Thus, the vertices are (± a, 0) = (± 3, 0), the foci are (± c, 0) = (± 3 , 0)
and the y-intercepts are (0, ± b) = (0, ±
graph:
6 ). Now, we can sketch the
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4
3
(0,
6)
2
1
F1
V1
F2
V2
0
-4
-3
-2
-1
-1
0
1
2
3
4
-2
-3
(0, –
6)
-4
Ex. 3
f(x) = ±
4−
x2
4
Solution:
First, we will need to get the equation in the appropriate form:
y = f(x) = ±
y2 = 4 –
x2
4
x2
16
x2
4
+ y2 = 4
+
y2
4
4−
x2
4
(square both sides)
(add
x2
4
to both sides)
(divide both sides by 4)
=1
Since the number under the x2 is larger than the number under the y2,
then the major axis runs along the x-axis and the ellipse is centered at
the origin. Also, a2 = 16 which means a = 4 and b2 = 4 which means
b = 2. Plugging into b2 = a2 – c2, we can find c:
4 = 16 – c2 (solve for c2)
c2 = 12
c = 12 = 2 3 (c ≥ 0, so ignore the negative root)
Thus, the vertices are (± a, 0) = (± 4, 0), the foci are (± c, 0)
= (± 2 3 , 0) and the y-intercepts are (0, ± b) = (0, ± 2). Now, we can
sketch the graph:
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4
3
(0, 2)
2
1
F1
V1
-4
-3
F2
0
-2
-1
-1
-2
0
1
2
3
V2
4
(0, – 2)
-3
-4
Note that if the equation had been f(x) =
4−
x2
4
, we would have only the
4−
upper half of the ellipse, whereas if the equation had been f(x) = –
x2
4
,
we would only have the lower half of the ellipse.
Graph of f(x) =
-4
-3
-2
-1
4−
x2
4
Graph of f(x) = –
4
4
3
3
2
2
1
1
0
0
-1
0
1
2
3
4 -4
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
0
4−
1
x2
4
2
3
4
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Ex. 4
x2 + y2 = 16
Solution:
First, we will need to get the equation in the appropriate form:
(divide both sides by 16)
x2 + y2 = 16
x2
16
+
y2
16
=1
Here, a2 and b2 have the same value and since b2 = a2 – c2, then c = 0.
Hence, both foci lie on the origin. The x-intercepts and y-intercepts are
all four units away from zero so the ellipse in this case is a circle with
6
radius of 4.
5
4
3
2
1
0
-1 0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1
-2
-3
-4
-5
-6
Objective b:
Ellipse with center (h, k).
We will now consider ellipses centered at (h, k). Such an ellipse is shifted h
units horizontal and k units vertical, so we will have (x – h) and (y – k) in
place of x and y respectively in the formulas for the ellipse. The vertices, foci,
and the endpoints on the minor axis will have h added to all the x-values and
k added to all the y-values.
Equation of an Ellipse Centered at (h, k)
1)
The Ellipse with center (h, k)
2)
and equation
(x − h)2
a2
+
(y − k)2
b2
=1
has vertices of (h ± a, k), foci of
(h ± c, k), and endpoints on the
minor axis of (h, k ± b) where
b2 = a2 – c2, a > b > 0, & c ≥ 0.
Major axis is parallel to the x-axis.
The Ellipse with center (h, k)
and equation
(x − h) 2
b2
+
(y − k)2
a2
=1
has vertices of (h, k ± a), foci of
(h, k ± c), and endpoints on the
minor axis of (h ± b, k) where
b2 = a2 – c2, a > b > 0, & c ≥ 0.
Major axis is parallel to the y-axis.
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(h, k + a)
(h, k + b)
(h, k + c)
(h – c, k)
(h, k)
(h – a, k)
(h + b, k)
(h – b, k)
(h + c, k)
(h + a, k)
(h, k)
(h, k – c)
(h, k – b)
(h, k – a)
Find the equation for the ellipse and then graph:
Ex. 5
The ellipse has foci at (5, – 3) and (– 1, – 3) and contains the
point (2, – 1).
Solution:
The center is midway between to two foci, thus the center is (h, k) =
(2, – 3) and c = 3. Since the foci lie on the line y = – 3, the major axis is
parallel to the x-axis, and thus the form of the equation we want to use
is
(x − h)2
a2
+
becomes:
(y − k)2
b2
(x − 2)2
a2
= 1. Plug in (h, k) = (2, – 3), and the equation
+
(y + 3) 2
b2
= 1. Since (2, – 1) lies two units above the
center on the minor axis, then b = 2. Thus, the points (h, k ± 2)
= (2, – 3 ± 2) = (2, – 1) and (2, – 5) are the endpoints of the minor axis.
Using b2 = a2 – c2, we can find a:
22 = a2 – 32
(solve for a2)
(take the principal square root since a > 0)
a2 = 13
a = 13
The vertices are (h ± a, k) = (2 ±
(5.61, – 3).
13 , – 3) ≈ (– 1.61, – 3) and
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2
1
0
-2
-1
-1
0
1
2
3
4
5
6
(2, – 1)
-2
(2 –
13 , – 3)
(2 +
-3
13 , – 3)
-4
(– 1, – 3)
-5
(2, – 5)
(5, – 3)
-6
Analyze the following equation:
Ex. 6
16x2 + 32x + 9y2 – 54y – 47 = 0
Solution:
First, we will need to get the equation in the appropriate form:
16x2 + 32x + 9y2 – 54y – 47 = 0
(add 47 to both sides)
2
2
(factor out 16 and 9 respectively)
16x + 32x + 9y – 54y = 47
2
2
16(x + 2x) + 9(y – 6y) = 47 (complete the square & add to both sides)
16(x2 + 2x + 1) + 9(y2 – 6y + 9) = 47+ 16(1) + 9(9) (simplify)
16(x2 + 2x + 1) + 9(y2 – 6y + 9) = 144
(rewrite as a perfect square)
2
2
16(x + 1) + 9(y – 3) = 144
(divide both sides by 144)
(x +1)2
9
+
(y −3) 2
16
=1
The center (h, k) = (– 1, 3). The major axis is parallel to the y-axis.
Since a = 4 and b = 3, we can use b2 = a2 – c2 to find c:
32 = 42 – c 2
(solve for c2)
c2 = 7
(take the principal square root since c > 0)
c= 7
The vertices are (h, k ± a) = (– 1, 3 ± 4) = (– 1, – 1) & (– 1, 7)
The foci are (h, k ± c) = (– 1, 3 ± 7 ) ≈ (– 1, 0.35) & (– 1, 5.65)
The endpoints of the minor axis are (h ± b, k) = (– 1 ± 3, 3)
= (– 4, 3) & (2, 3)
Now, we can sketch the graph:
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(– 1, 3 +
(– 1, 3 –
Objective c:
9
8
(– 1, 7)
7
7)
6
5
4
(– 4, 3)
(2, 3)
3
2
1
7)
0
-1 0 1 2 3 4 5 6
-6 -5 -4 -3 -2 -1
(– 1, – 1)-2
-3
Applications of Ellipses.
Ellipses have the phenomenon that if light or sound is emitted from one
focus, the waves get reflected off of the ellipse and sent to the other focus. A
lithotripter machine uses this principle to crush kidney stones using shock
waves. A beam is emitted from one focus of an elliptical tub and the patient
is placed in the tub in such a way that the kidney stone is at the other focus,
resulting in the kidney stone being crushed. This prevents harm to any other
part of the body. Another application of ellipses is the orbits of the planets.
The orbits of the planets are ellipses with the sun being at one focus of the
ellipses. A third application is in architecture in the construction of bridges
and tunnels; many times the shape used is a half of an ellipse or
semielliptical arch.
Solve the following:
Ex. 7 A one-way tunnel is designed in a shape of a semielliptical arch with a
height of 16.5 feet and 20 feet long at the base of the arch. If an oversize
truck is 12 feet wide, how high can it be to still pass through the tunnel?
Solution:
First, let's draw a diagram:
16.5 ft
12 ft
20 ft
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Since the base is 20 feet long, then the length from the center to the
edge is 10 feet which is less than 16.5 feet. Hence, the major axis is
parallel to the y-axis meaning that a = 16.5 ft and b = 10 ft. If we
impose our coordinate system such that the center is (0, 0), then the
equation of the ellipse is:
x2
100
+
y2
272.25
=1
Assuming the truck drives down the middle of the road, then the
distance from the center of the semiellipse to edge of the truck is 12/2
= 6 ft. Thus, x = 6 ft. Plug in the result into our equation and solve for y:
62
y2
+
100
272.25
2
64
y
= 100
272.25
=1
(subtract
36
100
from both sides)
(multiply both sides by 272.25)
y2 = 174.24
(use the square root property)
y = ± 13.2
(reject the negative solution since y is the height)
Thus, the truck needs to be less than 13.2 ft high.