Adv. Algebra D

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

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.

d1 + d2 = a constant value.

Ellipse

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).

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

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

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

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.

Therefore, d1 + d2 = 2a. Using the distance formula,

2 2 2 2( ) ( ) 2x c y x c y a

Simplify:

2 2 2 2( ) ( ) 2x c y x c y a

2 2 2 2( ) 2 ( )x c y a x c y

Square both sides.

2 2 2 2 2 2 2( ) 4 4 ( ) ( )x c y a a x c y x c y Subtract y2 and square binomials.2 2 2 2 2 2 22 4 4 ( ) 2x xc c a a x c y x xc c

Simplify:

2 2 2 2 2 2 22 4 4 ( ) 2x xc c a a x c y x xc c Solve for the term with the square root.

2 2 24 4 4 ( )xc a a x c y

2 2 2( )xc a a x c y Square both sides.

222 2 2( )xc a a x c y

Simplify:

222 2 2( )xc a a x c y

2 2 2 4 2 2 2 22 2x c xca a a x xc c y 2 2 2 4 2 2 2 2 2 2 22 2x c xca a a x xca a c a y 2 2 4 2 2 2 2 2 2x c a a x a c a y

Get x terms, y terms, and other terms together. 2 2 2 2 2 2 2 2 4x c a x a y a c a

Simplify:

2 2 2 2 2 2 2 2 4x c a x a y a c a

2 2 2 2 2 2 2 2c a x a y a c a

Divide both sides by a2(c2-a2)

2 2 2 2 2 22 2

2 2 2 2 2 2 2 2 2

c a x a c aa y

a c a a c a a c a

2 2

2 2 21

x y

a c a

At this point, let’s pause and investigate a2 – c2.

2 2

2 2 21

x y

a c a

Change the sign and run the negative through the denominator.

2 2

2 2 21

x y

a a c

d1 + d2 must equal 2a. However, the triangle created is an isosceles triangle and d1 = d2. Therefore, d1 and d2 for the point (0, b) must both equal “a”.

This creates a right triangle with hypotenuse of length “a” and legs of length “b” and “c”. Using the pythagorean theorem, b2 + c2 = a2.

We now know…..

2 2

2 2 21

x y

a a c

and b2 + c2 = a2

b2 = a2 – c2

Substituting for a2 - c2

2 2

2 21

x y

a b where c2 = a2 – b2

2 2

2 21

x h y k

a b

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

2 2

2 21

x y

a b

where c2 = a2 – b2 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

where c2 = a2 – b2 andc is the distance from the center to the foci.

Major Axis is vertical when…..

Vertices:Co-vertices:

Major Axis is horizontal when….

Vertices:Co-vertices:

12

2

2

2

b

y

a

x1

2

2

2

2

b

x

a

y

0,a

),0( b

),0( a

)0,( b

Write an equation of an ellipse in standard form that has a vertex at (-4, 0), a co-vertex at (0, 3), and is centered at the origin.

Find an equation of an ellipse centered at (2, -4) that is 20 units high and 10 units wide.

Find the center, vertices, co-vertices, and foci of the ellipse with the equation x2+9y2=36. Then graph the ellipse.

In “whispering galleries” a sound made at one focus can be clearly heard at the other focus, even though very little can be heard by someone in between. Suppose the distance between the foci are 100 feet apart and the length of the room is 150 feet. Find the equation of the ellipse. How high is the ceiling at it’s highest point?

10.4 p. 571 #3-7, 19-21, 29, 30, 33, 34,

45, 47, 70, (76 graph)Review for cumulative test over probability and rational

functions