Transcript Slide 1

Ellipse

Conic Sections

Ellipse The plane can intersect one nappe of the cone at an angle to the axis resulting in an

ellipse

.

Ellipse - Definition An ellipse is the set of all points in a plane such that the sum of the distances from two points (foci) is a constant. d 1 + d 2 = a constant value.

Finding An Equation

Ellipse

Ellipse - Equation To find the equation of an ellipse, let the center be at (0, 0). The vertices on the axes are at (a, 0), (-a, 0), (0, b) and (0, -b). The foci are at (c, 0) and (-c, 0).

Ellipse - Equation According to the definition. The sum of the distances from the foci to any point on the ellipse is a constant.

Ellipse - Equation The distance from the foci to the point (a, 0) is 2a. Why?

Ellipse - Equation The distance from (c, 0) to (a, 0) is the same as from (-a, 0) to (-c, 0).

Ellipse - Equation The distance from (-c, 0) to (a, 0) added to the distance from (-a, 0) to (-c, 0) is the same as going from (-a, 0) to (a, 0) which is a distance of 2a.

Ellipse - Equation Therefore, d 1 + d 2 = 2a. Using the distance formula, (

x

c

) 2 

y

2  ( ) 2 

y

2  2

a

Ellipse - Equation Simplify: (

x

c

) 2 

y

2  ( ) 2 

y

2  2

a

( ) 2 

y

2  2

a

 (

x

c

) 2 

y

2 Square both sides.

( ) 2 

y

2  4

a

2  4

a

(

x

c

) 2 

y

2 Subtract y 2 and square binomials.

x

2  2

xc

c

2  4

a

2  4

a x c

) 2 

y

2 (

x

c

) 2 

y

2 

x

2  2

xc

c

2

Ellipse - Equation Simplify:

x

2  2

xc

c

2  4

a

2  4

a

(

x

c

) 2 

y

2 

x

2  2

xc

c

2 Solve for the term with the square root.

 4

xc

 4

a

2   4

a

(

x

c

) 2 

y

2

xc

a

2 

a

(

x

c

) 2 

y

2 Square both sides.

xc

a

2  2  

a

( ) 2 

y

2 2 

Ellipse - Equation Simplify: 

xc

 2

x c

2 2 2

x c a

2    2  

a

2

xca

2 2

xca

2 (  

a a

4 4 ) 2   

y

2

a

2 2

a x

2 2  

x

 2  2

xc

2

xca

2 

c

2   2 2

a c y

2   2

a y

2 2 2

x c

a

4  2

a x

2  2 2

a c

 2

a y

2  Get x terms, y terms, and other terms together.

2

x c

2  2

a x

2   2

a y

2  2

a c

2 

a

4

Ellipse - Equation Simplify:  2

x c

2 

c

2  2

a x

2   

a

2 

x

2 2

a y

2  2

a y

2  

a

2 

c

2 2

a c

2 

a

2  

a

4 Divide both sides by a 2 (c 2 -a 2 ) 

a

2

c

2  

c

2

a

2  

a

2

x

2  

a

2 

c

2  2

a

2  

a

2

a

2  

c

2

c

2 

a

2 

a

2  

x

2

a

2  

c

2

y

2 

a

2   1

Ellipse - Equation

x

2

a

2  

c

2

y

2 

a

2   1 Change the sign and run the negative through the denominator.

x

2

a

2

y

2  

a

2 

c

2   1 At this point, let’s pause and investigate a 2 – c 2 .

Ellipse - Equation d 1 + d 2 must equal 2a. However, the triangle created is an isosceles triangle and d 1 d 2 = d 2 . Therefore, d 1 for the point (0, b) must both equal “a”.

and

Ellipse - Equation This creates a right triangle with hypotenuse of length “a” and legs of length “b” and “c”. Using the pythagorean theorem, b 2 + c 2 = a 2 .

Ellipse - Equation We now know…..

x

2

a

2  

a

2

y

2 

c

2   1 and b 2 + c 2 = a 2 b 2 = a 2 – c 2 Substituting for a 2 - c 2

x

2

a

2 

y b

2 2  1 where c 2 = |a 2 – b 2 |

Ellipse - Equation The equation of an ellipse centered at (0, 0) is ….

x

2

a

2 

y b

2 2  1 where c 2 = |a 2 – b 2 | and c is the distance from the center to the foci.

 Shifting the graph over h units and up k units, the center is at (h, k) and the equation is

y

k

 2  1

a

2

b

2 where c 2 = |a 2 – b 2 | and c is the distance from the center to the foci.

Ellipse - Graphing 

y

k

 1 where c 2 = |a 2 – b 2 | and c is the distance from the

a

2

b

2  2 center to the foci.

b Vertices are “a” units in the x direction and “b” units in the y direction.

a c c a b The foci are “c” units in the direction of the longer (major) axis.

Graph - Example #1

Ellipse

Ellipse - Graphing Graph: 

x

 2 16 

y

 3  2  1 25 Center: (2, -3) Distance to vertices in x direction: 4 Distance to vertices in y direction: 5 Distance to foci: c 2 =|16 - 25| c 2 = 9 c = 3

Ellipse - Graphing Graph: 

x

 2 16 

y

 3  2  1 25 Center: (2, -3) Distance to vertices in x direction: 4 Distance to vertices in y direction: 5 Distance to foci: c 2 =|16 - 25| c 2 = 9 c = 3

Graph - Example #2

Ellipse

Ellipse - Graphing Graph: 5

x

2  2

y

2  10

x

 12

y

 27  0 Complete the squares.

5 5 5 5

x

  

x x x

2 2 2     10 1  2 2 2

x x x

  2 ??

y

 2 2  12

y

2  2 

y

 3

y

 2 2   6

y

y

 6 50 

y

27 9   ??

  

x

 10 1  

y

 25 3  2  1  27

Ellipse - Graphing Graph: 

x

 10 1  

y

 25 3  2  1 Center: (-1, 3) -5 -2 -4 8 6 4 2 5 Distance to vertices in x direction: 10 Distance to vertices in y direction: 5 Distance to foci: c 2 =|25 - 10| c 2 = 15 c = 15

Find An Equation

Ellipse

Ellipse – Find An Equation Find an equation of an ellipse with foci at (-1, -3) and (5, -3). The minor axis has a length of 4. The center is the midpoint of the foci or (2, -3). The minor axis has a length of 4 and the vertices must be 2 units from the center. Start writing the equation.

Ellipse – Find An Equation 

y

k

 2  1

a

2

b

2 

x

 2

a

2 c 2 = |a 2 

y

 4 3  2 – b 2 |. Since  1 the major axis is in the x direction, a 2 > 4 9 = a 2 – 4 a 2 = 13 Replace a 2 in the equation.

Ellipse – Find An Equation The equation is: 

x

 2 13 

y

 3  2  1 4

Ellipse – Table 

a

2

y

k

 2

b

2  1 Center: Vertices:  (h, k)   Foci: If a 2 > b 2 If b 2 > a 2 c 2 = |a 2 – b 2 |   

c

  

b