### 10.4 - Ellipses

```10.4 - Ellipses
Ellipses - Warm Up
Solve each equation.
1. 27 = x2 + 11
2. x2 = 48
3. 84 = 120 – x2
Ellipses – Warm Up
Solutions
1. 27 = x2 + 11
16 = x2
x = ±
2. x2 = 48
3.
x =
±
x =
±4
16 = ±4
48
3
84 = 120 – x2
–36 = –x2
36 = x2
x =
±
36 = ±6
ELLIPSE TERMS
Vertex
EQUATION
Co-vertex
FORM
Focus
( x  h)2 ( y  k )2

1
a2
b2
CENTER
(h, k )
VERTICES
(h ± a , k)
a V=(h , k) c
CO-VERTICES (h, k ± b)
b
MAJOR AXIS
horizontal
Co-vertex
Vertex
MINOR AXIS
Major axis
Minor axis
c  a b
2
MAJOR length
2
2
2a
vertical
MINOR length
2b
FOCI
(h ± c , k)
ELLIPSE TERMS
Vertex
Minor axis
a
V=(h , k)
Co-vertex
b
Co-vertex
EQUATION
FORM
( x  h)2 ( y  k )2

1
b2
a2
CENTER
(h, k )
VERTICES
(h, k ± a)
c
CO-VERTICES (h ± b , k)
Focus
Vertex
MAJOR AXIS
MAJOR length
Major axis
MINOR AXIS
MINOR length
c  a b
2
2
2
FOCI
vertical
2a
horizontal
2b
(h , k ± c )
CONVERTING to STANDARD FORM
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


x² + 4y² + 4x – 24y + 24 = 0
Groups the x terms and y terms
x² + 4x + 4y² – 24y + 24 = 0
Complete the square
x² + 4x + 4(y² – 6y) + 24 = 0
x² + 4x + 4 + 4(y² – 6y + 9) = -24 + 4 +36
(x + 2)² + 4(y – 3)² = 16
Divide to put in standard form
(x + 2)²/16 + 4(y – 3)²/16 = 1
( x  2)2 ( y  3)2

1
16
4
Writing an Equation of an Ellipse
Write an equation in standard form of an ellipse that has a vertex at
(0, –4), a co-vertex at (3, 0), and is centered at the origin.
Since (0, –4) is a vertex of the ellipse, the other vertex is at (0, 4), and
the major axis is vertical.
Since (3, 0) is a co-vertex, the other co-vertex is at (–3, 0), and the
minor axis is horizontal.
So, a = 4, b = 3, a2 = 16, and b2 = 9.
x2
b2 +
y2
a2 = 1
x2
9 +
y2
16
= 1
An equation of the ellipse is
Standard form for an equation of an
ellipse with a vertical major axis.
Substitute 9 for b2 and 16 for a2.
x2
y2
9 + 16 = 1.
Graph and Label

b) Find coordinates of vertices,
covertices, foci
( x  3 ) 2 ( y  2) 2

1
9
4








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
Center = (-3,2)
Horizontal ellipse since the a² value is
under x terms
Since a = 3 and b = 2
Vertices are 3 points left and right
from center  (-3 ± 3, 2)
Covertices are 2 points up and down
 (-3, 2 ± 2)
Now to find focus points
Use c² = a² - b²
So c² = 9 – 4 = 5
c² = 5 and c = ±√5
Focus points are √5 left and right
from the center  F(-3 ±√5 , 2)
•
•
•
•
a) GRAPH
Plot Center (-3,2)
a = 3 (go left and right)
b = 2 (go up and down)
Graph and Label

b) Find coordinates of vertices,
covertices, foci
( x  3)2 ( y  1)2

1
3
16



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
Center = (3,-1)
Vertical ellipse since the a² value is
under y terms
Since a = 4 and b = 2
Vertices are 3 points up and down
from center  (3, -1 ± 2)
Covertices are 2 points left and right
 (3 ± 2, -1)
Now to find focus points
Use c² = a² - b²
So c² = 16 – 4 = 12
c² = 12 and c = ±2√3
Focus points are 2√3 up and down
from the center  F(3,-1 ±2√3)
•
•
•
•
a) GRAPH
Plot Center (3,1)
a = 4 (go up and down)
b = 2 (go left and right)
Write the equation of the ellipse given…
endpoints of major axis are at (-11,5) and (7,5)
endpoints of minor axis are at (-2,9) and (-2,1)
Draw a graph with given info
 Use given info to get measurements
◦ Major axis = 2a
◦ Major axis is 18 units,
◦ so a = 9
A = (-11,5)
◦ Minor axis = 2b
◦ Minor axis is 8 units,
◦ so b = 4
 Use standard form
( x  h)2 ( y  k )2

1
a2
b2
◦ Need values for h,k, a and b
◦ We know a = 9 and b = 4
◦ How do we find center???
◦ Use midpoint formula
◦ (h , k) = (-2 , 5)
 Plug into formula
( x  2)2 ( y  5)2

1
92
42

C = (-2,9)
minor
major
D = (-2,1)
B = (7,5)
Let’s Try One
Find an equation of an ellipse centered at the origin that is 20 units wide
and 10 units high.
Since the largest part of the ellipse is horizontal and the width is 20
units, place the vertices at (±10, 0).
Place the co-vertices at (0, ±5).
So, a = 10, b = 5, a2 = 100, and b2 = 25.
x2
a2 +
y2
b2 = 1
x2
y2
100 + 25
= 1
Standard form for an ellipse with a
horizontal major axis.
Substitute 100 for a2 and 25 for b2.
x2
y2
An equation of the ellipse is 100 + 25 = 1.
Write the equation of the ellipse given…
foci are at (2,5) and (-2,5)
vertex at (-3,5)



Draw a graph with given info
Use given info to get measurements
◦ Find the center first.
◦ The center is in the middle of the foci. Use
midpoint formula to find (h , k)
◦ (h , k) = (0 , 5)
◦ Then c = distance from center to foci
◦ So c = 2
◦ Then a = distance from center to vertex
◦ so a = 3
Use standard form
( x  h)2 ( y  k )2

1
a2
b2

◦ Need values for h,k, a and b
◦ We know a = 3, c = 2, (h,k) = (0,5)
◦ How do we find b???
◦ Use c² = a² – b²
◦
4 = 9 – b²
◦
we get b² = 5
Plug into formula
( x ) 2 ( y  5 )2

1
9
5
a=3
c=2
(h,k) = (0,5)
Working Backwards
Find the foci of the ellipse with the equation 9x2 + y2 = 36. Graph the
ellipse.
9x2 + y2 = 36
y2
x2
+
=1
4
36
Write in standard form.
Since 36 > 4 and 36 is with y2, the major axis is vertical, a2 = 36, and b2 = 4.
c2 = a2 – b2
= 36 – 4
Find c.
Substitute 4 for a2 and 36 for b2.
= 32
c=±
32 = ± 4
2
The major axis is vertical, so the coordinates of the foci are (0, ±c).
The foci are: (0, 4 2 ) and (0, – 4 2).
Let’s Try One
Write an equation of the ellipse with foci at (0, ±4) and covertices at (±2, 0).
Since the foci have coordinates (0, ±4), the major axis is vertical.
Since c = 4 and b = 2, c2 = 16, and b2 = 4.
c2 = a2 – b2
Use the equation to find a2.
16 = a2 – 4
Substitute 16 for c2 and 4 for b2.
a2 = 20
Simplify.
x2 y2
+
=1
4
20
An equation of the ellipse is
Substitute 20 for a2 and 4 for b2.
y2
x2
+
4
20
= 1.
Application of foci in ellipses

You may think that most objects in space that orbit something else
move in circles, but that isn't the case. Although some objects follow
circular orbits, most orbits are shaped more like "stretched out"
circles or ovals. Mathematicians and astronomers call this oval
shape an ellipse.
The Sun isn't quite at the center of a planet's elliptical orbit. An
ellipse has a point a little bit away from the center called the
"focus". The Sun is at the focus of the ellipse. Because the Sun is at
the focus, not the center, of the ellipse, the planet moves closer to
and further away from the Sun every orbit
More on orbits…


Orbits are ellipses. An ellipse can be like a circle, or it can be long and
skinny. Mathematicians and astronomers use the term "eccentricity" to
describe the shape of an orbit. Eccentricity = c/a. An orbit shaped almost
like a circle has a low eccentricity close to zero. A long, skinny orbit has a
high eccentricity, close to one
http://windows2universe.org/physical_science/physics/mechanics/orbit/orbi
t_shape_interactive.html
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