## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Ellipses

- Understand the meaning and significance of the major and minor axes.
- Understand the fundamental equation a
^{2}= b^{2}+ c^{2}and use it frequently. - Use symmetry to help you graph an ellipse.
- Understand the standard formula for the equation of an ellipse.
- Know how to put an equation in standard form by completing the square.

### Ellipses

^{2}+ 4y

^{2}− 2x − 16y − 19 = 0

- Group the x's with the x's and the y's with the y's, move the constant term to the right of the equation
- (x
^{2}− 2x) + (4y^{2}− 16y) = 19 - Factor out a 4 from the y's
- (x
^{2}− 2x) + 4(y^{2}− 4y) = 19 - Complete the square by adding [(b
^{2})/4] - (x
^{2}− 2x + [(b^{2})/4]) + 4(y^{2}− 4y + [(b^{2})/4]) = 19 + [(b^{2})/4] + 4( [(b^{2})/4] ) - (x
^{2}− 2x + [(( − 2)^{2})/4]) + 4(y^{2}− 4y + [(( − 4)^{2})/4]) = 19 + [(( − 2)^{2})/4] + 4( [(( − 4)^{2})/4] ) - (x
^{2}− 2x + [4/4]) + 4(y^{2}− 4y + [16/4]) = 19 + [4/4] + 4( [16/4] ) - (x
^{2}− 2x + 1) + 4(y^{2}− 4y + 4) = 19 + 1 + 16 - (x − 1)
^{2}+ 4(y − 2)^{2}= 36 - Divide left and right side of equation by 36. Simplify
- [((x − 1)
^{2})/36] + [(4(y − 2)^{2})/36] = [36/36]

^{2})/36] + [((y − 2)

^{2})/9] = 1

^{2}+ 4y

^{2}− 2x − 16y − 19 = 0

- Write the ellipse in Generic Conic Form
Horizontal [((x − h) ^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1Vertical [((y − k) ^{2})/(a^{2})] + [((x − h)^{2})/(b^{2})] = 1- Notice how this problem is the same as 1. Use the resultfrom 1 to graph. Notice how
- this ellipse will be a horizontal ellipse.
- [((x − 1)
^{2})/36] + [((y − 2)^{2})/9] = 1 - Find the center (h,k)
- Center = (h,k) = (1,2)
- Find the horizontal major axis, and graph it
- a
^{2}= 36 - a = √{36} = 6
- Starting from the center, go 6 units to the left and 6 units to the right.
- Find the Vertical Minor Axis
- b
^{2}= 9 - b = √9 = 3
- Starting from the center, go 3 units up and 3 units down.
- Draw a smooth ellipse that passes through the end - points of the Horizontal Major and Vertical Minor axis.

^{2}+ 9y

^{2}− 72x − 126y + 189 = 0

- Group the x's with the x's and the y's with the y's, move the constant term to the right of the equation
- (4x
^{2}− 72x) + (9y^{2}− 126y) = − 189 - Factor out a 4 from the x's and 9 from the y's
- 4(x
^{2}− 18x) + 9(y^{2}− 14y) = − 189 - Complete the square by adding [(b
^{2})/4] - 4(x
^{2}− 18x + [(b^{2})/4]) + 9(y^{2}− 14y + [(b^{2})/4]) = − 189 + 4( [(b^{2})/4] ) + 9( [(b^{2})/4] ) - 4(x
^{2}− 18x + [(( − 18)^{2})/4]) + 9(y^{2}− 14y + [(( − 14)^{2})/4]) = − 189 + 4( [(( − 18)^{2})/4] ) + 9( [(( − 14)^{2})/4] ) - 4(x
^{2}− 18x + [324/4]) + 9(y^{2}− 14y + [196/4]) = − 189 + 4( [324/4] ) + 9( [196/4] ) - 4(x
^{2}− 18x + 81) + 9(y^{2}− 14y + 49) = − 189 + 4( 81 ) + 9( 49 ) - 4(x − 9)
^{2}+ 9(y − 7)^{2}= 576 - Divide left and right side of equation by 36. Simplify
- [(4(x − 9)
^{2})/576] + [(9(y − 7)^{2})/576] = [576/576]

^{2})/144] + [(9(y − 7)

^{2})/64] = 1

^{2}+ 9y

^{2}− 72x − 126y + 189 = 0

- Write the ellipse in Generic Conic Form
Horizontal [((x − h) ^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1Vertical [((y − k) ^{2})/(a^{2})] + [((x − h)^{2})/(b^{2})] = 1- Notice how this problem is the same as 3. Use the result from 3 to graph. Notice how
- this ellipse will be a horizontal ellipse.
- [((x − 9)
^{2})/144] + [(9(y − 7)^{2})/64] = 1 - Find the center (h,k)
- Center = (h,k) = (9,7)
- Find the horizontal major axis, and graph it
- a
^{2}= 144 - a = √{144} = 12
- Starting from the center, go 12 units to the left and 12 units to the right.
- Find the Vertical Minor Axis
- b
^{2}= 64 - b = √{64} = 8
- Starting from the center, go 8 units up and 8 units down.
- Draw a smooth ellipse that passes through the end - points of the Horizontal Major and Vertical Minor axis.

^{2}+ 4y

^{2}− 108x + 40y + 280 = 0

- Group the x's with the x's and the y's with the y's, move the constant term to the right of the equation
- (9x
^{2}− 108x) + (4y^{2}+ 40y) = − 280 - Factor out a 9 from the x's and 4 from the y's
- 9(x
^{2}− 12x) + 4(y^{2}+ 10y) = − 280 - Complete the square by adding [(b
^{2})/4] - 9(x
^{2}− 12x + [(b^{2})/4]) + 4(y^{2}+ 10y + [(b^{2})/4]) = − 280 + 9( [(b^{2})/4] ) + 4( [(b^{2})/4] ) - 9(x
^{2}− 12x + [(( − 12)^{2})/4]) + 4(y^{2}+ 10y + [((10)^{2})/4]) = − 280 + 9( [(( − 12)^{2})/4] ) + 4( [((10)^{2})/4] ) - 9(x
^{2}− 12x + [144/4]) + 4(y^{2}+ 10y + [100/4]) = − 280 + 9( [144/4] ) + 4( [100/4] ) - 9(x
^{2}− 12x + 36) + 4(y^{2}+ 10y + 25) = − 280 + 9( 36 ) + 4( 25 ) - 9(x − 6)
^{2}+ 4(y + 5)^{2}= 144 - Divide left and right side of equation by 144. Simplify
- [(9(x − 6)
^{2})/144] + [(4(y + 5)^{2})/144] = [144/144] - [((x − 6)
^{2})/16] + [((y + 5)^{2})/36] = 1 - or

^{2})/36] + [((x − 6)

^{2})/16] = 1

^{2}+ 4y

^{2}− 108x + 40y + 280 = 0

- Write the ellipse in Generic Conic Form
Horizontal [((x − h) ^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1Vertical [((y − k) ^{2})/(a^{2})] + [((x − h)^{2})/(b^{2})] = 1- Notice how this problem is the same as 5.Usetheresultfrom 5 to graph. Notice how
- this ellipse will be a vertical ellipse.
- [((y + 5)
^{2})/36] + [((x − 6)^{2})/16] = 1 - Find the center (h,k)
- Center = (h,k) = (6, − 5)
- Find the Vertical Major Axis , and graph it
- a
^{2}= 36 - a = √{36} = 6
- Starting from the center, go 6 units up and 12 units down
- Find the Horizontal Minor Axis
- b
^{2}= 16 - b = √{16} = 4
- Starting from the center, go 4 units right and 4 units left
- Draw a smooth ellipse that passes through the end - points of the Vertical Major and Horizontal Minor axis.

Endpoints of major axis:(4,7),( − 6,7)

Foci:(3,7),( − 5,7)

- Notice how this ellipse is going to be a Horizontal Ellipse in the form [((x − h)
^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1 - Find the length of the major axis
- 2a = | − 6 − 4| = | − 10| = 10
- a = 5
- Find the center of the ellipse given that a = 5 and one enpoint is (4,7).
- You may also find the center by finding the mid - poing of the major axis.
- Center = (h,k) = (h = 4 − 5,7) = ( − 1,7)
- So far you have
- [((x + 1)
^{2})/(5^{2})] + [((y − 7)^{2})/(b^{2})] = 1 - Find b using the foci and the relationship a
^{2}= b^{2}+ c^{2} - The center is located at ( − 1,7) and one foci located at (3,7), therefore recall that
- c is the distance between the center and the foci, so
- c = | − 1 − 3| = | − 4| = 4
- To find b, then
- b
^{2}= a^{2}− c^{2} - b = √{a
^{2}− c^{2}} = √{5^{2}− 4^{2}} = √{25 − 16} = √9 = 3 - Complete the equation of the ellipse
- [((x + 1)
^{2})/(5^{2})] + [((y − 7)^{2})/(b^{2})] = 1 - [((x + 1)
^{2})/(5^{2})] + [((y − 7)^{2})/(3^{2})] = 1

^{2})/25] + [((y − 7)

^{2})/9] = 1

Endpoints of major axis:(1,8),( − 9,8)

Foci:( − 1,8),( − 7,8)

- Notice how this ellipse is going to be a Horizontal Ellipse in the form [((x − h)
^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1 - Find the length of the major axis
- 2a = | − 9 − 1| = | − 10| = 10
- a = 5
- Find the center of the ellipse given that a = 5 and one enpoint is (1,8).
- You may also finbd the center by finding the mid - poing of the major axis.
- Center = (h,k) = (h = 1 − 5,8) = ( − 4,8)
- So far you have
- [((x + 4)
^{2})/(5^{2})] + [((y − 8)^{2})/(b^{2})] = 1 - Find b using the foci and the relationship a
^{2}= b^{2}+ c^{2} - The center is located at ( − 4,8) and one foci located at ( − 7,8), therefore recall that
- c is the distance between the center and the foci, so
- c = | − 7 − ( − 4)| = | − 3| = 3
- To find b, then
- b
^{2}= a^{2}− c^{2} - b = √{a
^{2}− c^{2}} = √{5^{2}− 3^{2}} = √{25 − 9} = √{16} = 4 - Complete the equation of the ellipse
- [((x + 4)
^{2})/(5^{2})] + [((y − 8)^{2})/(b^{2})] = 1 - [((x + 4)
^{2})/(5^{2})] + [((y − 8)^{2})/(4^{2})] = 1

^{2})/25] + [((y − 8)

^{2})/16] = 1

Center = ( − 10,1)

Vertical Major axis = 12 units

Horizontal Minor axis = 8

- Notice how this ellipse is going to be a Vertical Ellipse in the form [((y − k)
^{2})/(a^{2})] + [((x − h)^{2})/(b^{2})] = 1 - Find a using length of the vertical major axis
- 2a = 12
- a = 6
- Find b using length of the horizontal minor axis
- 2b = 8
- b = 4
- Fill in the formula
- [((y − 1)
^{2})/(6^{2})] + [((x + 10)^{2})/(4^{2})] = 1

^{2})/36] + [((x + 10)

^{2})/16] = 1

Center = ( − 3, − 3)

Horizontal Major axis = 24 units

Vertical Minor axis = 10

- Notice how this ellipse is going to be a Vertical Ellipse in the form [((x − h)
^{2})/(a^{2})] + [((y − k)^{2})/(b^{2})] = 1 - Find a using length of the horizontal major axis
- 2a = 24
- a = 12
- Find b using length of the vertical minor axis
- 2b = 10
- b = 5
- Fill in the formula
- [((x + 3)
^{2})/(12^{2})] + [((y + 3)^{2})/(5^{2})] = 1

^{2})/144] + [((y + 3)

^{2})/25] = 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Ellipses

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- What Are Ellipses? 0:11
- Foci
- Properties of Ellipses 1:43
- Major Axis, Minor Axis
- Center
- Length of Major Axis and Minor Axis
- Standard Form 5:33
- Example: Standard Form of Ellipse
- Vertical Major Axis 9:14
- Example: Vertical Major Axis
- Graphing Ellipses 12:51
- Complete the Square and Symmetry
- Example: Graphing Ellipse
- Equation with Center at (h, k) 19:57
- Horizontal and Vertical
- Difference
- Example: Center at (h, k)
- Example 1: Equation of Ellipse 24:05
- Example 2: Equation of Ellipse 27:57
- Example 3: Equation of Ellipse 32:32
- Example 4: Graph Ellipse 38:27

### Algebra 2

### Transcription: Ellipses

*Welcome to Educator.com.*0000

*So far in conic sections, we have discussed parabolas and circles.*0002

*The next type of conic section we are going to cover is the ellipse.*0006

*First of all, what is an ellipse? Well, an ellipse is formally defined as the set of points in a plane*0012

*such that the sum of the distances from two fixed points is constant.*0019

*Well, what does that mean? First of all, this is the general shape of an ellipse.*0024

*And these two points here are the foci of the ellipse: we will call them f _{1} and f_{2}.*0029

*And looking back at this definition, it is the set of points in the plane...*0036

*and if you pick any of these points (say right here) and measure the distance from this point to one focus*0041

*(we will call that d _{1}), and then we measure the distance from that same point to f_{2},*0051

*to the other focus, we will call that distance d _{2}.*0059

*This definition states that, if I add up these two distances, d _{1} + d_{2}, they will equal a constant.*0063

*I could pick any point; I could then pick this point and say, "OK, here is another distance," calling that, say, d _{3}, and this one d_{4}.*0074

*And if I added up this distance and this distance, d _{3} + d_{4}, I could get that same constant.*0087

*And I could do that with any point on the ellipse.*0094

*Continuing on with some properties of the ellipse: an ellipse actually has two axes of symmetry.*0104

*One is called the major axis, and the other is the minor axis, and these intersect at the center of the ellipse.*0110

*Here, we are going to look at an ellipse that is centered at the origin (it has a center at (0,0)).*0117

*And again, I have foci f _{1} and f_{2}.*0122

*This is one of the vertices of the ellipse; here is a vertex; this is the second vertex of the ellipse.*0127

*And the major axis runs from one vertex to the other vertex.*0135

*And you can see that it passes through both of the foci; this is the major axis.*0140

*And what we are looking at here is an ellipse that has a horizontal major axis.*0149

*In a few minutes, we will look at ellipses that are oriented the other way--ellipses that are oriented with a vertical major axis.*0154

*So, that does exist; but right now we will just focus on this for the general discussion.*0163

*Here we have the major axis; and intersecting at the center is a second axis called the minor axis.*0167

*Looking more closely at the relationships between the major axis, the minor axis, the foci, and the distances relating them:*0182

*let's call the distance from one vertex to the center A: this distance is A.*0193

*Therefore, the length of the major axis is 2A.*0200

*And this is going to become important later on, when we are working with writing equations for ellipses,*0209

*or taking an equation and then trying to graph the ellipse.*0214

*That is my first distance: this is actually called the semimajor axis.*0218

*The length of the semimajor axis is A; the length of the major axis is 2A.*0226

*Now, looking at the minor axis: from this point to the center, this length is B.*0230

*Therefore, the length of the minor axis is equal to 2B.*0237

*Now, looking at the foci: the distance from one focus to the center is going to be C.*0246

*Therefore, the distance from one focus to another, or the distance between the foci, is equal to 2C.*0255

*There is also an equation relating these A, B, and C; and that is A ^{2} = B^{2} + C^{2}.*0276

*So, keep this equation in mind: again, it becomes important, because you might be given A and B, but not C;*0287

*or you may be given a drawing, a sketch of the ellipse, and then asked to write an equation based on it.*0294

*And sometimes you need to find this third component in order to write that equation.*0300

*And you can do that by knowing that A ^{2} = B^{2} + C^{2}.*0304

*Again, A is the length of the semimajor axis; B is the distance from here to the center of the minor axis*0309

*(2B is the length of the minor axis); and C is the distance between one focus and the center (2C gives the distance between the two foci).*0319

*Those are important relationships to understand when working with the ellipse.*0328

*Standard form: looking at what the standard form of the equation of an ellipse with a center at (0,0) and a horizontal major axis is:*0333

*it is x ^{2}/A^{2} + y^{2}/B^{2} = 1.*0343

*This is the standard form; and again, we already discussed what A is and B is.*0353

*And if you figure out those from the graph, and are given those, then you can go ahead and write your equation.*0359

*Or, given the equation, you can graph the ellipse.*0365

*Let's take a look at an example: A ^{2}/9 + y^{2}/4 = 1.*0368

*So again, this is just the equation for an ellipse with a horizontal major axis, so it is sketched out here that way.*0379

*We can make this more precise by saying, "OK, A ^{2} = 9; therefore, A equals √9, or 3."*0388

*That means that the distance from here to here is going to be 3.*0405

*And since this is centered at the origin, this will actually be at the point (3,0).*0409

*So, let's write that as a coordinate pair, (0,3).*0415

*OK, that means that this length of A is 3; over here, the other vertex is going to be at (0,-3).*0420

*So, I have one vertex at (0,3) and the other at (0,-3); and I have f _{1} here and f_{2} here.*0431

*And the length of the major axis is actually 6: it is going to be equal to 2A.*0437

*2A equals 6, and that is the length of the major axis.*0443

*B ^{2} = 4: since B^{2} = 4, B = √4, which equals 2.*0452

*Therefore...actually, this needs to be written the other way; this is actually (3,0), and (-3,0).*0461

*Now, up here, we have (0,2) and (0,-2).*0474

*And 2, then, is the length; that is B--that is the length of half of the minor axis.*0485

*2B = 4, and this is the length of the minor axis.*0492

*You can get a lot of information just by looking at the equation of the ellipse in standard form.*0498

*If I needed to, I could figure out the distance between the foci, and I could figure out what C is.*0505

*And remember, C is this length; because recall that A ^{2} = B^{2} + C^{2}.*0512

*And I know that A ^{2} is 9; I know that B^{2} is 4; so I am looking for C^{2}.*0518

*9 - 4 gives me 5, so C ^{2} = 5; therefore, C = √5, and that is approximately equal to 2.2.*0524

*So, this distance here is roughly 2.2; and the distance between these foci would be about 4.4.*0535

*So, just by having this equation, I could graph the ellipse, and I could find this third component that was missing.*0544

*OK, so that was standard form for an ellipse that has a horizontal major axis.*0555

*For an ellipse that has a vertical major axis, you are going to see that the A ^{2} is associated with the y^{2} term.*0560

*In the horizontal major axis, we had x ^{2}/A^{2}.*0567

*Now, if you were given an equation, how would you know what you are dealing with--*0572

*if you were dealing with a vertical major axis or a horizontal major axis?*0578

*Well, A ^{2} is the larger number; so let's say I was given something like y^{2}/16 + x^{2}/9 = 1.*0581

*When I look at this, I see that the larger number is associated with y ^{2}; so that tells me that I have a vertical major axis.*0594

*If it had been x ^{2} associated with 16, then I would have said, "OK, that is a horizontal major axis."*0602

*Again, I can graph this ellipse by having this equation written in standard form.*0609

*I know that A ^{2} equals 16; therefore, A equals the square root of 16; it equals 4.*0614

*This time, I am going to go along here for the major axis; and that makes sense,*0622

*because I have a focus here and another focus here, and the major axis passes through the two foci.*0628

*OK, so (0,4) is going to give me one vertex; (0,-4) will give me the other vertex.*0638

*And remember that 2A = 8, and that is going to be the length of the major axis.*0647

*And again, right now, we are limiting our discussion to ellipses with a center at (0,0).*0658

*Later on, we will expand the discussion to talk about the graphs of ellipses with centers in other areas of the coordinate plane.*0664

*OK, so now I have B ^{2} = 9; therefore, √9 is going to equal B; this tells me that B equals 3.*0672

*So, since B equals 3, the length of half of the minor axis is going to be 3.*0689

*So, right here at (3,0) is going to be one point, and I am going to have the other point right here at (-3,0).*0696

*And the length of the minor axis...2B = 6, and that is the length of the minor axis.*0704

*Again, I can use the relationship A ^{2} = B^{2} + C^{2}*0719

*to figure out what C is, and to figure out where the foci are located.*0725

*I know that A ^{2} is 16; I know that B^{2} is 9; and I am trying to figure out C^{2}.*0730

*So, I take 16 - 9; that gives me 7 = C ^{2}; therefore, C = √7.*0738

*And the square root of 7 is approximately equal to 2.6, so this is going to be up here at about (0,2.6).*0746

*And f _{2} is going to be down here at about (0,-2.6).*0755

*So again, using standard form, you can graph the ellipse, and you can find where the foci are, based on the values of A ^{2} and B^{2}.*0758

*We talked a little bit about graphing ellipses; but sometimes you are not given the equation in standard form.*0771

*If the equation is not in standard form, you have to put it that way.*0776

*And working with other conic sections, we have learned that you can put an equation into standard form for a conic section by completing the square.*0780

*You can also use symmetry, just as we did when graphing parabolas or circles.*0788

*So, let's say you are given an equation like this: 3x ^{2} + 4y^{2} - 18x - 16y = -19; and you are asked to graph it.*0794

*You are going to put it in standard form; but it is nice, first of all, to know what kind of shape you are working with.*0807

*And you can tell that just by looking at this, even though it is not in standard form yet.*0811

*And the reason is: I see that I have an x ^{2} term and a y^{2} term*0815

*on the same side of the equation, with the same sign, with different coefficients; that tells me that I am working with an ellipse.*0820

*And we are going to go into more detail in the lecture on conic sections.*0827

*We are going to review how to tell apart the equations for various conic sections.*0831

*But just briefly now: recall that a parabola would have either an x ^{2} term or a y^{2} term, not both.*0835

*With a circle, the coefficients of x ^{2} and y^{2} are the same; that is for a circle.*0845

*For an ellipse, the coefficients of the x ^{2} and y^{2} terms are different.*0861

*Now, you see that these have the same sign; so with an ellipse,*0876

*it is important to note that the x ^{2} and y^{2} terms have the same sign.*0881

*If they don't have the same sign, that is actually a different shape.*0884

*They have the same sign; but this coefficient (the x ^{2} coefficient) is 3, and the y^{2} coefficient is 4.*0887

*That tells me I have an ellipse, not a circle.*0894

*My next step is to complete the square and write this in standard form,*0898

*so that, if I wanted to graph it, I would have all of the information that I need.*0903

*The first thing to do is to group the x variable terms and the y variable terms.*0906

*This gives me x ^{2} - 18x + 4y^{2} - 16y = -19.*0911

*Now, when I am looking at this, I remember that, to complete the square, I want to end up with a leading coefficient of 1.*0931

*I am going to factor out this 3 to get 3(x ^{2} - 6x); and I am going to need to add*0937

*something else over here to complete the square--a third term.*0944

*Here, factor out the 4 to give me (y ^{2} - 4y) = -19.*0949

*Recall that, to complete the square, you are going to add b ^{2}/4 to each set of terms.*0957

*So, for this x group, we have b ^{2}/4 = -(6)^{2}/4 = 36/4 = 9.*0963

*So, I am going to add a 9 here; and very importantly, to the right side, I am going to add 9 times 3, which is 27.*0978

*Add that to the right, because if I don't, the equation won't be balanced anymore.*0995

*Now, the y variable terms: I need to add something here to complete the square.*1003

*And I will work over here for this: b ^{2}/4 equals (-4)^{2}/4 = 16/4 = 4.*1013

*So, I am going to add 4 here; but to the right, what I need to add is 4 times 4; so I need to add 16.*1026

*Now that I have this, my next step is to write this in standard form: 3...and this is x - 3, the quantity squared;*1038

*if I squared this, I would get this back; plus 4y - 2, the quantity squared, equals...*1049

*-19 plus 16 gives me, let's see, -3; 27 minus 3 gives me 24 on the right.*1057

*Well, recall that for standard form, I want a 1 over here on the right; I want this to equal 1.*1069

*So, this gives me...I need to divide both sides by 24; this is going to give me 3(x - 3) ^{2}/24 + 4(y - 2)^{2}/24 = 24/24.*1076

*This finally leaves me with (x - 3) ^{2}...this 3 will cancel, and I am going to get 8 in the denominator;*1101

*this 4 will cancel, and I will get 6 in the denominator; equals 1.*1109

*Now, I have the bigger term associated with x; that tells me that I have a horizontal major axis.*1115

*So, the major axis is horizontal; and let me go ahead and write that up here.*1126

*x ^{2}/a^{2} + y^{2}/b^{2} = 1.*1144

*OK, now you can see that this looks slightly different; and we have these extra terms here.*1154

*And what that tells us is something that we are going to discuss in a moment.*1160

*And what we are going to discuss is situations where the center of the ellipse is not at (0,0).*1164

*But we have the same information that we would have with the center being at (0,0).*1174

*And that is that I have A ^{2}, which equals 8; and I know I have a horizontal major axis; and I have B^{2} = 6.*1180

*Looking, now, at equations for ellipses with centers at (h,k), at somewhere other than 0:*1198

*if you look at the situation where we have a horizontal major axis, versus a vertical major axis,*1206

*you can again see that it is very similar to when the center is at (0,0).*1212

*The x is associated with the A ^{2} term when the major axis is horizontal.*1217

*The y is associated with the A ^{2} term when the major axis is vertical.*1221

*The only difference is that we now have these terms telling us where the center is.*1227

*And if the center was going to be 0, all that would happen is that you would have x - 0 and y - 0,*1231

*which then gives you back the equation we worked with before, x ^{2}/A^{2} + y^{2}/B^{2},*1237

*or y ^{2}/A^{2} + x^{2}/B^{2}, all equal to 1.*1244

*So, let's talk about an example for this: (x - 4) ^{2} + (y + 6)^{2} = 1.*1252

*This is all over 100, and this is all divided by 25.*1264

*What this tells me is that the center is at (h,k); so this is h (that is 4).*1269

*You need to be careful here, because what you have is a positive; but the standard form is a negative.*1278

*And it is perfectly fine to write this like this, but you need to keep in mind that this is really saying...*1284

*if you think about it, y + 6 is the same as y - -6.*1291

*So, when I look at it this way, I realize that, if it is in this form, k is actually going to be -6.*1296

*And if you are not careful about that, you can end up putting your center in the wrong place.*1302

*Let's let this be 2, 4, 6, -2, -4, -6; the center is at (4,-6), right here.*1309

*A ^{2} = 100; my larger term is my A^{2}term, and I have a horizontal major axis.*1323

*Therefore, A = √100, which is 10.*1331

*So, since A equals 10 and the length of the major axis is horizontal, then I am going to need to go 10 over from 4.*1337

*It is going to be all the way out here at 14.*1351

*So, this is where one vertex would be--this is going to be at 4 + 10 gives me (14,-6).*1357

*That is one vertex: -2, 4, 6, 8, 10, 12, 14...so -14 is going to be about here.*1368

*And I am going to have the other vertex here at (-14,-6).*1376

*I have a minor axis: B ^{2} = 25; therefore B = 5.*1384

*So, what that tells me is that I come here at -6; I add 5 to that; and that is going to bring me right here at -1; that is going to be right about here.*1389

*And then, -8, -10, -12...-(6 + 5) is going to be down here at -11; so this ellipse is going to roughly look like this.*1400

*OK, and standard form allowed me to determine that the center is at (4,-6);*1419

*that I have a vertex; if I add 10, that is going to give me this vertex right over here*1427

*(let's draw this more at the vertex); that I have another vertex here; that the major axis is going to pass through here;*1433

*and then, I am going to have a minor axis passing through here.*1441

*All right, in the first example, we are asked to find the equation of the ellipse that is shown.*1447

*And I am going to go ahead and label some of these points.*1453

*This is going to be f _{1}; and let's say we are given f_{1} as being at (0,3).*1457

*This is going to be 1, 2, 3: each mark is going to stand for one.*1463

*Down here, I have f _{2}; that should actually be down here a little bit lower; so that is f_{2} at (0,-3).*1467

*Let's say we are also given a vertex at (0,5), and the other vertex at (0,-5).*1480

*All right, so I am asked to find the equation of the ellipse shown.*1492

*And the one thing I see is that there is a vertical major axis.*1495

*Since there is a vertical major axis, it is going to be in the general form (y - k) ^{2}/A^{2} + (x - h)^{2}/B^{2} = 1.*1503

*Now, I notice that the center is at (0,0); so that tells me that h and k are (0,0).*1520

*So, this is actually going to become the simpler form, y/A ^{2} + x/B^{2} = 1.*1528

*The next piece of information I have is the length of A.*1537

*So, this line is going to give me A; and I know that, since the center is here at (0,0), the length of this is 5; therefore, A = 5.*1541

*Since A equals 5, A ^{2} is going to equal 5^{2}, or 25.*1553

*I don't know B; this is going to be B, but I don't know what it is.*1561

*However, I do know C; if I look here, this is C--from -3 to the center is 3, so C = 3.*1567

*Therefore, C ^{2} = 3^{2}, or 9.*1583

*The other piece of information I have is the equation we have been working with,*1587

*that states that A ^{2} = B^{2} + C^{2}.*1591

*Therefore, I can find B ^{2}, which I need in order to write my equation.*1602

*So, I know that A ^{2} is 25; I know that C^{2} is 9; and I am looking for B^{2}.*1606

*25 - 9 is 16; therefore, B ^{2} = 16, and B = 4.*1620

*Now, I can go ahead and write my equation; I have all the information I need, so let's put it all together right here.*1628

*y, divided by A ^{2}...I know that A^{2} is 25...(that is actually y^{2} right here--*1633

*let's not forget the squares)...plus x ^{2} divided by B^{2}...B^{2} is 16...equals 1.*1647

*So again, this had a center at (0,0); so remember that this is the general form of the equation.*1655

*When you have a center at (0,0), it becomes this form.*1660

*I had a vertical major axis, so I am using this form, where the A ^{2} is associated with the y^{2} term.*1663

*And by having A ^{2} and C^{2}, I was able to determine what B^{2} is.*1671

*Example 2: let's find the equation of the ellipse satisfying a major axis 10 units long and parallel to the y-axis; minor axis 4 units long; center at (4,2).*1678

*Let's start with the center--the center is at (4,2)--the center is right here at (4,2).*1690

*And it says that the major axis is parallel to the y-axis.*1699

*What that tells me is that I have a vertical major axis.*1703

*Since this is a vertical major axis and I know the center (I know (h,k)), I am going to be working with the general form*1711

*(y - k) ^{2}/A^{2} + (x - h)^{2}/B^{2} = 1.*1719

*I have h; I have k; I need to find A ^{2} and B^{2}.*1728

*The other piece of information I have is that the major axis is 10 units long; and I know that it is vertical.*1732

*Since it is 10 units long, that means that the major axis length is equal to 2A, and I know that that length is 10;*1738

*therefore, I just take 10 divided by 2, and I get that A = 5.*1751

*So, I am starting here; and if I take 5 + 2, I am going to get that I am going to have the vertex up here at 7.*1757

*5 + 2 is going to give me 7, right there; so I go to (4,7); that is where this is going to be.*1775

*And I got that by keeping the x where it was, and then adding 2 where the center was and adding 5 to that, which is the length of A.*1793

*Then, I am going to go down; again, it is going to be at 4.*1803

*But then, if I take 2 and subtract 5 from it, I am going to get -3.*1810

*So, right here at (4,-3) is the other vertex.*1815

*Now, I was not asked to graph this; but I am graphing it so that I have an understanding of what each of these means,*1823

*so that I can go ahead and write the equation.*1829

*What I have now is...I know what A is, which means I can figure out A ^{2}; and I know the center.*1832

*The last thing I know is that the minor axis length equals 2B: I am given that that is 4 units long; therefore, B = 2.*1844

*So, if I started here at 4, and I added 2 to that, I am going to end up with (6,2).*1857

*And I am going to end up with...if I start at 4 and I subtract 2, I am going to end up with (2,2).*1867

*And this is B; this is A.*1881

*I have a minor axis that has a length of 2B, which tells me that B = 2.*1887

*From this information, I can go ahead and write this equation.*1891

*I know A equals 5, so A ^{2} equals 5^{2}, which is 25.*1897

*B = 2; B ^{2}, therefore, equals 2^{2}, or 4.*1904

*So, I have everything I need to write this: (y - k)...well, k is 2; the quantity squared, divided by A ^{2};*1911

*A ^{2} is 25; plus (x - h)...h is 4...the quantity squared, divided by B^{2}, which is 4, equals 1.*1920

*So, this standard form describes the ellipse with the major and minor axis here.*1936

*And you could finish that out by just connecting these points and drawing the ellipse if you wanted to finish graphing it.*1941

*Find the equation of the ellipse satisfying endpoints of the major axis at (11,3) and (-7,3) and foci at (7,3) and (-3,3).*1953

*All right, endpoints of the major axis: let's do 2, 4, 6, 8, 10, 12, and -2, 4, -6...OK.*1962

*The endpoint is at (11,3): 11 is right here, and then we will have 3 be right here.*1979

*The other endpoint is at -7, which is going to be here, 3.*1991

*And that tells me the major axis: since the major axis is horizontal, we are going to be working with the general form*1999

*(x - h) ^{2}/A^{2} + (y - k)^{2}/B^{2} = 1.*2010

*So, that is my first piece of information.*2018

*I can also figure out the length of the major axis.*2020

*So, since the major axis goes from -7 to 11, if I take -7 minus 11, I am going to get -18.*2024

*And a length is going to be an absolute value, so I am just going to take 18; the length is going to be the absolute value of this difference.*2045

*All right, I know that the major axis length is 18; the other thing I know is that the major axis length equals 2A, as we have discussed.*2053

*So, 2A = 18; therefore, A = 9; so I know that the distance from the center to this endpoint is 9.*2061

*Therefore, I could just say, "OK, 9 minus 11 gives me 2; and I know that the y-coordinate will be 3; so that is (2,3)."*2076

*Another way to solve this, without using all this graphing, would have simply been to find the midpoint.*2087

*I am going to go ahead and do that, as well, because the midpoint of the major axis is the center of the ellipse.*2093

*Let's try that, as well--the center, using the midpoint formula.*2100

*Recall the midpoint formula: we are going to take x _{1} + x_{2} (that is 11 + -7);*2105

*and we are going to take the average of that (we will divide it by 2); that is going to give me the x-coordinate.*2112

*For the y-coordinate, I am going to take 3 + 3, and I am going to divide that by 2.*2117

*And this will give you a center at...11 - 7 gives you 4; 4 divided by 2 is 2.*2126

*3 + 3 is 6, divided by 2 gives you 3; and that is exactly what I came up with using the graphing method.*2138

*So, just to show you: you could have figured this out algebraically (where the center is); or you could have figured it out using graphing.*2144

*This gives me (h,k), so I have (h,k), and I have A, which is 9, so I can get A ^{2}.*2154

*The next thing I need to do is figure out B, and they don't give me B; but what they do give me are the foci.*2162

*There are foci at (7,3) (7 is here--focus at (7,3)--we will call this one f _{1} at (7,3)); and this is the center right here.*2167

*There is another focus at (-3,3): f _{2} is going to be at (-3,3).*2187

*Recall that the distance from one focus to the other is 2C; the distance from one focus to the center is C.*2201

*Let's just work with this; this is C; 7 - 2 is 5; therefore, C = 5.*2210

*So, I have A = 9; I have C = 5 (again, that is the distance from the center to a focus);*2221

*or I could have figured out the distance from one focus to another--that is 2C--and then divided by 2;*2229

*So, I have A, and I have C, and I know that A ^{2} = B^{2} + C^{2}.*2233

*So, this gives me 9 ^{2} = B^{2} + 5^{2}: 9^{2} is 81, equals B^{2} + 25.*2240

*If I take 81 - 25, I am going to get 56 = B ^{2}.*2254

*And I don't even need to take the square root of that, because to put this into standard form, I actually need B ^{2}.*2259

*So, I get (x - h); remember, the center is (h,k), so that is 2; the quantity squared, divided by A ^{2}...*2265

*recall that A ^{2} is 9^{2}, so it is 81; plus (y - k)^{2}...k is 3, the quantity squared;*2273

*divided by B ^{2}...I determined that B^{2} is 56; all of this equal to 1.*2284

*Just by knowing the endpoints of the major axis and the location of the foci, I could figure out A ^{2}*2289

*and B ^{2}, as well as the center, and then write this equation for the ellipse in standard form.*2295

*Finally, we are asked to graph an ellipse that is not given to us in standard form.*2307

*We have some extra work to do: we are going to actually have to complete the square in order to even graph this.*2311

*So, let's go ahead and start by grouping the x terms together and y terms together.*2318

*Also note that, since this has an x ^{2} term and a y^{2} term on the same side of the equation,*2329

*with the same sign (they are both positive), but different coefficients, I know I have an ellipse.*2335

*It is not a circle, because for a circle, these coefficients would be the same.*2341

*Grouping the terms together gives me 14x ^{2} - 56x + 6y^{2} - 24y = -38.*2347

*Looking at this, I see that I have a common factor of 2.*2368

*If I divide both sides by 2, I can simplify this equation; so the numbers I will work with will be smaller.*2372

*So, let's divide both sides by 2 to get 7x ^{2} - 28x + 3y^{2} - 12y = -19.*2378

*The next thing to do is to factor out the leading coefficient, since it is not 1.*2391

*I am going to factor out a 7 to get x ^{2} - 4x, plus...*2397

*over here, I am going to factor out a 3; that gives me y ^{2} - 4y, all equals -19.*2402

*I need to complete the square, so I need to get b ^{2}/4.*2411

*In this case, that is going to give me 4 ^{2}/4 = 16/4 = 4.*2415

*So, over here, I am going to add a 4; very important--on the right, I have -19, and I am adding to the left 7 times 4; that is 28.*2423

*So, I need to add 28 to the right.*2440

*All right, over here, for the y terms: b ^{2}/4: again, we have a b term that is 4, so I am going to end up with the same thing, 4.*2446

*Now, here I am actually adding 4 times 3 (is 12), so I need to add that to the right, as well, to keep the equation balanced.*2462

*This is the easiest step to mess up on: you are focused here on completing the square,*2470

*and then you sometimes don't remember that you have to add the same thing to both sides.*2474

*Now, working on writing this in standard form: this is going to give me (x - 2) ^{2} + 3(y - 2)^{2} = -19 + 12...*2479

*that is going to leave me with 28; -19 + 12 is going to be -7; 28 minus 7 equals 21.*2493

*To get this into standard form, I need to have a 1 on the right, so I am going to divide both sides by 21.*2504

*This is going to give me 7(x - 2) ^{2}/21 + 3(y - 2)^{2}/21 = 21/21.*2510

*This cancels to (x - 2) ^{2}/3; this becomes (y - 2)^{2}/7; and this just becomes 1.*2528

*Now, I have standard form; I can do some graphing.*2547

*I have a center at, let's see, (2,2); that is right here; the center is at (2,2), so that is h and k.*2550

*And I notice, actually, that the larger term is under the y; it is associated with the y.*2577

*That tells me that this has a vertical major axis; and therefore, I am going to keep that in mind--that it is going to be oriented this way.*2587

*The ellipse is going to end up like this, instead of like this.*2600

*I have my center at (2,2); therefore, A ^{2} = 7; A = √7.*2604

*The square root of 7 is about 2.6, so it gets messy, as always, when you are working with radicals.*2620

*But it is about 2.6; so what I have to do is say, "OK, my vertex up here is going to be at x = 2, and then y is going to equal 2 + 2.6, which is 4.6."*2626

*So, (2,4.6): and again, I got that by saying the length of A (the length from the vertex to the center) is 2.6.*2643

*So, 2 + 2.6 gives me 4.6.*2655

*Over here, I am going to take 2, and I am going to subtract 2.6 from it; so x is still going to be 2; now y is going to be about here, which is (2, -.6).*2658

*Again, 2 is here; the length of A is 2.6, so it is going to be 2 - 2.6; this gives me my major axis.*2674

*I know, from vertex to vertex, where this ellipse is going to land.*2685

*Now, the minor axis, B: I know that B ^{2} = 3; therefore, B = √3, which is approximately equal to 1.7.*2690

*This is not to do the same thing, but going along the x direction, the horizontal direction.*2700

*So again, this is A; OK, now to get B, I am going to have 2, and I am going to add 1.7 to it.*2706

*That is going to give me 3.7; it is going to land about here.*2715

*In this direction, I am going to subtract; I am going to say that I have 2 - 1.7, so that is going to land here, at about .3.*2722

*And in the y direction (in the vertical direction) it is still going to be up at 2.*2739

*So again, to get this, I said 2 + 1.7; that brought me to (3.7,2)--that is this point.*2742

*This point is at 2 minus 1.7, so that is (.3,2).*2753

*So, this gives me the general shape of this ellipse, like this; so I can get a good sketch, based on this equation.*2761

*I took this equation, and I recognized that it was an ellipse.*2782

*I completed the square to get it in standard form, and saw that it had a vertical major axis and that it had a center at (2,2).*2787

*I then found A to determine where the vertices would be, and then B to determine the width of the ellipse here; and then I could get a good sketch.*2795

*That concludes this session of Educator.com on ellipses; thanks for visiting!*2806

1 answer

Last reply by: Dr Carleen Eaton

Sun Jul 8, 2018 1:20 PM

Post by John Stedge on June 27 at 03:44:59 PM

for completing the square the number you add would just be the x/y term before you factor out the leading coefficient of the x^2/y^2 terms correct? I saw this in example 4 where the x/y terms before factoring were the same as the numbers you added to the right side of the equation in the completeing the square step.

0 answers

Post by Shiden Yemane on March 20 at 10:57:07 AM

Thanks, it was very helpful!

1 answer

Last reply by: Dr Carleen Eaton

Wed Jan 14, 2015 7:41 PM

Post by Saadman Elman on December 19, 2014

It was very helpful. Thanks.

1 answer

Last reply by: Rafael Mojica

Thu May 22, 2014 10:22 AM

Post by Rafael Mojica on May 22, 2014

Hello,

on example 4 you made a mistake on the equation that you used. Since major axis is parallel to y-axis. So, (x-h)^2/b^2-(y-k)^2/a^2=1 is our formula.

1 answer

Last reply by: Dr Carleen Eaton

Sat Sep 14, 2013 2:44 PM

Post by Anwar Alasmari on August 11, 2013

the video around 7:00, I think the point has to be (3,0), not (0,3) as well as (-3,0).

2 answers

Last reply by: Dr Carleen Eaton

Mon Jan 23, 2012 5:01 PM

Post by Yi Zhang on July 22, 2011

The left vertex on the major axis is supposed to be at V(-6,-6).