1. 30. Ellipses

2. Ellipse
An ellipse is a set of points in a plane the sum of whose distances from two fixed points, called foci, is a constant.
For any point P that is on the ellipse , d2 + d1 is always the same. P d2 d1 F1 F2

4. Standard Form Equation of an Ellipse
(x – h) (y – k)2
= 1 a2 b2
The center of the ellipse is at the point (h, k)
a is ½ the length of the horizontal axis
b is ½ the length of the vertical axis
Where c is the distance from the center to a focus point.
c2 = a2 – b if a2 > b2
c2 = b2 – a if b2 > a2

5. Graphing an Ellipse Graph: x2 y2 4 9 + = 1 Center: (0, 0)
= 1
Center: (0, 0) V
Minor Axis: 4 (horizontal) F
Major Axis: 6 (vertical)
Vertices: (0, 3) and (0, - 3) F
c2 = 9 – 4 = 5 V
c = Ö5 = 2.24
Foci: (0, 2.24) and (0, -2.24)

6. Graphing an Ellipse Graph: (x – 2)2 (y + 1)2 16 9 + = 1
= 1
Center: (2, -1)
Major Axis: 8 (horizontal)
Minor Axis: 6 (vertical) V
Vertices: (6, -1) and (-2, -1) V
c2 = 16 – 9 = 7
c = Ö 7 = 2.65
Foci: (4.65, -1) and (-0.65, -1)

7. Finding an Equation of an Ellipse
Find the equation of the ellipse given that
Vertices are (0, 4), (0, -4)
Foci are (0, 3), (0, -3) V
Center: (0, 0)
Major Axis: 8 (vertical)
b = 4 and b2 = 16
Since c = 3, and c2 = b2 – a2
9 = 16 – a2
a2 = 7 V
Equation:
(x – h) (y – k) a b2
= 1
x y
= 1

8. Finding an Equation of an Ellipse
Find the equation of the ellipse given the graph
Then locate the foci of the ellipse
Center: (-1, 1) V1
Major Axis: 6 (horizontal), so a = 3
Minor Axis: 2 (vertical), so b = 1
Equation:
(x + 1) (y – 1)
= 1
c2 = a2 – b2 V2
c2 = 9 – 1 = 8
c = Ö8 = 2.83
(x – h) (y – k) a b2
= 1
Foci are (1.83, 1), (-3.83, 1)

9. Finding an Equation of an Ellipse
Find the equation of the ellipse given that
Foci are (-2, 0), (2, 0)
y-intercepts: -3, 3
Major Axis must be horizontal since the foci are on the major axis
Center: (0, 0)
c= 2 and b = 3 F F
c2 = a2 – b2
4 = a2 – 9
a2 = 13
Equation:
(x – h) (y – k) a b2
= 1
x y
= 1

10. Converting an Equation
Convert the following equation to standard form
Then graph the ellipse and locate its foci
9x2 + 25y2 – 36x + 50y – 164 = 0
9(x2 – 4x + ___ ) + 25(y2 + 2y + ___) = ___ + ___ 4 1 36 25
9(x – 2)2 + 25(y + 1)2 = 225
(x – 2) (y + 1)2
= 1
c2 = 25 – 9 = 16
c = 4 F F
Foci: (-2, -1) and (6, -1)

11. Converting an Equation
Convert the following equation to standard form
Then graph the ellipse and locate its foci
4x2 + 9y2 – 48x + 72y + 144 = 0.
4(x2 – 12x + ___ ) + 9(y2 + 8y + ___) = ___ + ___ 36 16
144
144
4(x – 6)2 + 9(y + 4)2 = 144
(x – 6) (y + 4)2
= 1
c2 = 36 – 9 = 27
c = 5.2
Foci: (.8, -4) and (11.2, -4) F F

13. Application Problems
A semielliptical archway has a height of 20 feet at its midpoint and a width of 50 feet.
Can a truck that is 14 ft high and 10 ft wide drive under the archway without moving into the oncoming lane? P
The real question is “What is the value of y at point P when x = 10” ? 20 10
Equation of the ellipse:
x y2
= 1 50
16x2 + 25y2 = 10,000
Yes, the truck will be able to drive under the archway without moving into the oncoming lane.
When x = 10,
y2 = 10,000
25y2 = 8400
y2 = 336
y = 18.3