Math 143 Section 7.1 The 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.

F1 F2

d1

d2

PFor any point P that is on the ellipse , d2 + d1 is always the same.

(x – h)2 (y – k)2

a2 b2+ = 1

The center of the ellipse is at the point (h, k)

c2 = a2 – b2 if a2 > b2

Where c is the distance from the center to a focus point.

c2 = b2 – a2 if b2 > a2

a is ½ the length of the horizontal axis

b is ½ the length of the vertical axis

Vertices: (0, 3) and (0, - 3)

Graph: x2 y2 4 9

c2 = 9 – 4 = 5c = 5 = 2.24

Foci: (0, 2.24) and (0, -2.24)

+ = 1

Center: (0, 0)

Minor Axis: 4 (horizontal)

Major Axis: 6 (vertical)

V

V

F

F

Vertices: (6, -1) and (-2, -1)

Graph: (x – 2)2 (y + 1)2 16 9

c2 = 16 – 9 = 7c = 7 = 2.65

Foci: (4.65, -1) and (-0.65, -1)

+ = 1

Center: (2, -1)

Minor Axis: 6 (vertical)

Major Axis: 8 (horizontal)

V

V

Find the equation of the ellipse given that

9 = 16 – a2

Center: (0, 0)

Major Axis: 8 (vertical)

Since c = 3, and c2 = b2 – a2

Vertices are (0, 4), (0, -4)Foci are (0, 3), (0, -3)

V

V

(x – h)2 (y – k)2 a2 b2

+ = 1

b = 4 and b2 = 16

a2 = 7

Equation:

x2 y2 7 9

+ = 1

Find the equation of the ellipse given the graph

c2 = 9 – 1 = 8

Center: (-1, 1) Major Axis: 6 (horizontal), so a = 3

c2 = a2 – b2

Then locate the foci of the ellipse

Foci are (1.83, 1), (-3.83, 1)

V1

V2

(x – h)2 (y – k)2 a2 b2

+ = 1c = 8 = 2.83

Equation:

(x + 1)2 (y – 1)2 9 1

+ = 1

Minor Axis: 2 (vertical), so b = 1

Find the equation of the ellipse given that

4 = a2 – 9

Center: (0, 0)

Major Axis must be horizontal since the foci are on the major axis

c2 = a2 – b2

y-intercepts: -3, 3Foci are (-2, 0), (2, 0)

(x – h)2 (y – k)2 a2 b2

+ = 1

c= 2 and b = 3

a2 = 13

Equation:

x2 y2 13 9

+ = 1

F F

Convert the following equation to standard formThen graph the ellipse and locate its foci

+ = 1

F F

9x2 + 25y2 – 36x + 50y – 164 = 0

9(x2 – 4x + ___ ) + 25(y2 + 2y + ___) = 164 + ___ + ___4 361 25

9(x – 2)2 + 25(y + 1)2 = 225

(x – 2)2 (y + 1)2 25 9

c2 = 25 – 9 = 16 c = 4

Foci: (-2, -1) and (6, -1)

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?

50

20

P

10

The real question is “What is the value of y at point P when x = 10” ?

Equation of the ellipse:x2 y2 625 400

+ = 1

1600 + 25y2 = 10,000When x = 10, 25y2 = 8400

y2 = 336y = 18.3

16x2 + 25y2 = 10,000Yes, the truck will be able

to drive under the archway without moving into the oncoming lane.