Download Ellipses - MATH 160, Precalculus

Ellipses
MATH 160, Precalculus
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan
Ellipses
Objectives
In this lesson we will learn to:
write equations of ellipses in standard form,
graph ellipses,
use the properties of ellipses to model and solve real-world
problems,
find the eccentricities of ellipses.
J. Robert Buchanan
Ellipses
The Ellipse
Definition
An ellipse is the set of all points (x, y ) in the plane, the sum of
whose distances from two distinct fixed points called foci is
constant.
Hx,yL
d1
d2
Focus
Focus
J. Robert Buchanan
Ellipses
Anatomy of an Ellipse
Center
Vertex
Focus
Vertex
Focus
Line through the foci intersects the ellipse at two points
called vertices.
Chord connecting the vertices is called the major axis.
Chord perpendicular to the major axis and passing through
the center is called the minor axis.
J. Robert Buchanan
Ellipses
Standard Equation for an Ellipse (1 of 2)
Standard Equation of an Ellipse
The standard form of the equation of an ellipse with center
(h, k ) and major and minor axes of lengths 2a and 2b
respectively where 0 < b < a, is
(x − h)2 (y − k )2
+
a2
b2
2
(x − h)
(y − k )2
+
b2
a2
= 1
(major axis is horizontal)
= 1
(major axis is vertical)
The foci lie on the major axis c =
center.
J. Robert Buchanan
√
a2 − b2 units from the
Ellipses
Standard Equation for an Ellipse (2 of 2)
If the center of the ellipse is at the origin (0, 0), the equations
take the form:
x2 y2
+ 2
a2
b
x2 y2
+ 2
b2
a
= 1
(major axis is horizontal)
= 1
(major axis is vertical)
J. Robert Buchanan
Ellipses
Example (1 of 3)
Find the standard form of the equation of an ellipse with center
at (2, −1), a vertex at (2, 1/2), and a minor axis of length 2.
J. Robert Buchanan
Ellipses
Example (1 of 3)
Find the standard form of the equation of an ellipse with center
at (2, −1), a vertex at (2, 1/2), and a minor axis of length 2.
Note that (h, k ) = (2, −1), 2b = 2 (which implies b = 1), and
the vertex 3/2 units above the center (which implies a = 3/2).
J. Robert Buchanan
Ellipses
Example (1 of 3)
Find the standard form of the equation of an ellipse with center
at (2, −1), a vertex at (2, 1/2), and a minor axis of length 2.
Note that (h, k ) = (2, −1), 2b = 2 (which implies b = 1), and
the vertex 3/2 units above the center (which implies a = 3/2).
Thus the equation of the ellipse is
(x − 2)2 (y + 1)2
+
12
(3/2)2
(x − 2)2 4(y + 1)2
+
1
9
J. Robert Buchanan
Ellipses
= 1
= 1.
Example (2 of 3)
Sketch the graph of the ellipse whose equation is
16x 2 + 25y 2 − 32x + 50y + 16 = 0
J. Robert Buchanan
Ellipses
Example (2 of 3)
Sketch the graph of the ellipse whose equation is
16x 2 + 25y 2 − 32x + 50y + 16 = 0
16(x 2 − 2x) + 25(y 2 + 2y ) = −16
16(x 2 − 2x + 1) + 25(y 2 + 2y + 1) = −16 + 16 + 25
16(x − 1)2 + 25(y + 1)2 = 25
(x − 1)2 (y + 1)2
+
= 1.
(5/4)2
12
J. Robert Buchanan
Ellipses
Example (2 of 3)
Sketch the graph of the ellipse whose equation is
16x 2 + 25y 2 − 32x + 50y + 16 = 0
16(x 2 − 2x) + 25(y 2 + 2y ) = −16
16(x 2 − 2x + 1) + 25(y 2 + 2y + 1) = −16 + 16 + 25
16(x − 1)2 + 25(y + 1)2 = 25
(x − 1)2 (y + 1)2
+
= 1.
(5/4)2
12
Center: (1, −1), Vertices: (−1/4, −1) and (9/4, −1), endpoints
of minor axis (1, 0) and (1, −2).
J. Robert Buchanan
Ellipses
Graph
0.0
y
-0.5
-1.0
-1.5
-2.0
0.0
0.5
1.0
x
J. Robert Buchanan
Ellipses
1.5
2.0
Example (3 of 3)
Find the coordinates of the center, foci, and vertices for the
ellipse whose equation is
36x 2 + 9y 2 + 48x − 36y − 72 = 0
J. Robert Buchanan
Ellipses
Example (3 of 3)
Find the coordinates of the center, foci, and vertices for the
ellipse whose equation is
36x 2 + 9y 2 + 48x − 36y − 72
4
2
36 x + x + 9(y 2 − 4y )
3
4
4
36 x 2 + x +
+ 9(y 2 − 4y + 4)
3
9
2 2
36 x +
+ 9(y − 2)2
3
2
x + 23
(y − 2)2
√
+ √
( 31/3)2 (2 31/3)2
J. Robert Buchanan
Ellipses
= 0
= 72
= 72 + 16 + 36
= 124
= 1.
Ellipse Features
Center: (−2/3, 2)
√
Vertices: (−2/3, 2 ± 2 31/3)
q
√
124
31
c:
31/3
9 − 9 =
√
Foci: (−2/3, 2 ± 93/3)
J. Robert Buchanan
Ellipses
Reflective Property of Ellipses
A ray emanating from one focus will always reflect off the
ellipse and pass through the other focus.
f
f
J. Robert Buchanan
Ellipses
Elliptical Orbits
The first artificial satellite to orbit Earth was Sputnik I (launched
by the former Soviet Union in 1957). Its highest point above the
Earth’s surface was 947 km, and its lowest point was 228 km.
The center of the Earth was at one focus of the elliptical orbit
and the radius of the Earth is 6378 km. Find the equation of the
orbit.
J. Robert Buchanan
Ellipses
Solution
Imagine one focus (the center of the Earth) is at (0, 0). The
major axis of the orbit is
2a = (6378 + 228) + (6378 + 947)
a = 6965.5 km.
If we think of the major axis as horizontal, the vertices of
the orbit are (−7325, 0) and (6606, 0).
The center of the elliptical orbit is the point
((6605 − 7325)/2, 0) = (−359.5, 0), thus c = 359.5.
√
Therefore b = 6965.52 − 359.52 = 6956.2 km.
Equation of the elliptical orbit:
(x + 359.5)2
y2
+
=1
(6965.5)2
(6956.2)2
J. Robert Buchanan
Ellipses
Graph of Sputnik Orbit
J. Robert Buchanan
Ellipses
Eccentricity
The measure of how “oval” an ellipse is, is called its eccentricity.
Definition
The eccentricity e of an ellipse is given by the ratio
e=
J. Robert Buchanan
c
.
a
Ellipses
Eccentricity
The measure of how “oval” an ellipse is, is called its eccentricity.
Definition
The eccentricity e of an ellipse is given by the ratio
e=
c
.
a
Remarks:
For every ellipse 0 < e < 1.
When the eccentricity is close to 0, the ellipse is more
circular than oval.
J. Robert Buchanan
Ellipses
Sputnik Revisited
Find the eccentricity of Sputnik’s orbit.
J. Robert Buchanan
Ellipses
Sputnik Revisited
Find the eccentricity of Sputnik’s orbit.
Since a = 6965.5 and c = 359.5 then
e=
359.5
= 0.0516115
6965.5
indicating a nearly circular orbit.
J. Robert Buchanan
Ellipses
Homework
Read Section 6.3.
Exercises: 1, 5, 9, 13, . . . , 57, 61
J. Robert Buchanan
Ellipses