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Lecture Comments (9)

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).

Ellipses

  • Understand the meaning and significance of the major and minor axes.
  • Understand the fundamental equation a2 = b2 + c2 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

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

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 plane0012

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 f1 and f2.0029

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 focus0041

(we will call that d1), and then we measure the distance from that same point to f2,0051

to the other focus, we will call that distance d2.0059

This definition states that, if I add up these two distances, d1 + d2, 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, d3, and this one d4.0074

And if I added up this distance and this distance, d3 + d4, 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 f1 and f2.0122

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 A2 = B2 + C2.0276

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 A2 = B2 + C2.0304

Again, A is the length of the semimajor axis; B is the distance from here to the center of the minor axis0309

(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 x2/A2 + y2/B2 = 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: A2/9 + y2/4 = 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, A2 = 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 f1 here and f2 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

B2 = 4: since B2 = 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 A2 = B2 + C2.0512

And I know that A2 is 9; I know that B2 is 4; so I am looking for C2.0518

9 - 4 gives me 5, so C2 = 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 A2 is associated with the y2 term.0560

In the horizontal major axis, we had x2/A2.0567

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, A2 is the larger number; so let's say I was given something like y2/16 + x2/9 = 1.0581

When I look at this, I see that the larger number is associated with y2; so that tells me that I have a vertical major axis.0594

If it had been x2 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 A2 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 B2 = 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 A2 = B2 + C20719

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

I know that A2 is 16; I know that B2 is 9; and I am trying to figure out C2.0730

So, I take 16 - 9; that gives me 7 = C2; 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 f2 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 A2 and B2.0758

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: 3x2 + 4y2 - 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 x2 term and a y2 term0815

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 x2 term or a y2 term, not both.0835

With a circle, the coefficients of x2 and y2 are the same; that is for a circle.0845

For an ellipse, the coefficients of the x2 and y2 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 x2 and y2 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 x2 coefficient) is 3, and the y2 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 x2 - 18x + 4y2 - 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(x2 - 6x); and I am going to need to add0937

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

Here, factor out the 4 to give me (y2 - 4y) = -19.0949

Recall that, to complete the square, you are going to add b2/4 to each set of terms.0957

So, for this x group, we have b2/4 = -(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: b2/4 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

x2/a2 + y2/b2 = 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 A2, which equals 8; and I know I have a horizontal major axis; and I have B2 = 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 A2 term when the major axis is horizontal.1217

The y is associated with the A2 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, x2/A2 + y2/B2,1237

or y2/A2 + x2/B2, 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

A2 = 100; my larger term is my A2term, 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: B2 = 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 here1427

(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 f1; and let's say we are given f1 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 f2; that should actually be down here a little bit lower; so that is f2 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/A2 + (x - h)2/B2 = 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/A2 + x/B2 = 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, A2 is going to equal 52, 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, C2 = 32, or 9.1583

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

that states that A2 = B2 + C2.1591

Therefore, I can find B2, which I need in order to write my equation.1602

So, I know that A2 is 25; I know that C2 is 9; and I am looking for B2.1606

25 - 9 is 16; therefore, B2 = 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 A2...I know that A2 is 25...(that is actually y2 right here--1633

let's not forget the squares)...plus x2 divided by B2...B2 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 A2 is associated with the y2 term.1663

And by having A2 and C2, I was able to determine what B2 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 form1711

(y - k)2/A2 + (x - h)2/B2 = 1.1719

I have h; I have k; I need to find A2 and B2.1728

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 A2; 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 A2 equals 52, which is 25.1897

B = 2; B2, therefore, equals 22, or 4.1904

So, I have everything I need to write this: (y - k)...well, k is 2; the quantity squared, divided by A2;1911

A2 is 25; plus (x - h)...h is 4...the quantity squared, divided by B2, 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 form1999

(x - h)2/A2 + (y - k)2/B2 = 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 x1 + x2 (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 A2.2154

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 f1 at (7,3)); and this is the center right here.2167

There is another focus at (-3,3): f2 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 A2 = B2 + C2.2233

So, this gives me 92 = B2 + 52: 92 is 81, equals B2 + 25.2240

If I take 81 - 25, I am going to get 56 = B2.2254

And I don't even need to take the square root of that, because to put this into standard form, I actually need B2.2259

So, I get (x - h); remember, the center is (h,k), so that is 2; the quantity squared, divided by A2...2265

recall that A2 is 92, so it is 81; plus (y - k)2...k is 3, the quantity squared;2273

divided by B2...I determined that B2 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 A22289

and B2, 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 x2 term and a y2 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 14x2 - 56x + 6y2 - 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 7x2 - 28x + 3y2 - 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 x2 - 4x, plus...2397

over here, I am going to factor out a 3; that gives me y2 - 4y, all equals -19.2402

I need to complete the square, so I need to get b2/4.2411

In this case, that is going to give me 42/4 = 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: b2/4: 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, A2 = 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 B2 = 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