 Well, welcome back to episode 29 on ellipses. You remember this is the second of three conic cross sections, what we call conic sections, for a double cone. If you'll go to the green screen just for a minute, let me introduce our objectives for today. Last time, you remember that we were discussing a double cone. And we were going to cut the double cone, not horizontally, but we were going to cut it with a plane at a slight angle so that it comes out the other side. And therefore, this cross section won't be a circle, but it'll be sort of an elongated circle that we call an ellipse. And this episode is about ellipses. So let's go to our objectives for today. First of all, we want to look at the fundamental equations for ellipses. There are two fundamental equations, like there were two fundamental equations for parabolas last time, depending on whether they were along the x or the y-axis. Then we'll look at transformations of these ellipses. And finally, we'll look at the reflective property of ellipses and several applications that go with that. OK, let me just demonstrate what I mean by an ellipse. If you'll go to the overhead camera, what I've done is to take a string that's a yellow string, so it doesn't show up very well, but it's fixed at two points. Now, what I'm going to do is put this marker. Let me switch to a different marker for this one. I'm going to put this marker inside the string and pull it tight. And you see what happens is I get a curve. I'll have to turn this over to get the bottom half. And so I get a curve that is an ellipse. And this demonstrates. It's not working so well on the bottom half, I'm afraid. Yeah. OK, so this demonstrates the property of an ellipse. And it is this. I have two fixed points that I'm going to call foci. So this is like focus number one, and I'll call this focus number two. So the plural is foci. And when I put a marker inside of it, like this pencil, I pull it out to a point. And this point has the property that the distance of the first focus plus the distance to the second focus has a constant sum of the two distances, which is in fact the length of the string that lies in between. So as one of the distances gets bigger, the other one gets smaller. But the sum remains the same. OK, here's a formal definition of an ellipse that demonstrates what we were last showing there. It says an ellipse is the set of points in the plane, the sum of whose distances to two fixed points, which we call foci, the sum is constant. We can draw an ellipse by suspending a string at the foci and tracing the curve by holding a pencil tight against the string. The center of the ellipse is the midpoint between the foci. Now let's go over here to the big board, and we'll take a look at what this definition means for us. OK, now if we take that definition that we just looked at, we can put it on a coordinate system, and we can derive the equation of an ellipse. And once we have the equation of ellipse, we won't have to refer to that definition so much. So let's say here we have our x, y coordinate system, and there were those two fixed points where we had the string nailed down. I'm going to call those focus number 1 and focus number 2. And then we had a string that we hung between the two foci. So let's say this is the string that hangs between it. And I'm going to put the focus number 1 with coordinates, let's say, c, 0. In other words, from the origin, I'm going to go over c units to get to focus number 1. And for focus number 2, I'm going to go back c units, so that'll make this negative c, 0. And for the length of the string, let me just write this up here. For the length of the string, I'm going to call it 2a. You might say, why are you putting a 2 in there? Actually, in the derivation, that 2 is going to cancel out, so it's going to make our answer turn out to be a little bit simpler. So I'm going to call that 2a. And so 2a is the total length of that string. Now what we did is we put our marker inside the string, we pulled it taut. So when I pull it down tight, it would come down maybe to here. And I get a curve that goes around the foci and comes back like that. Let me just remove that string now. So this is sort of an illustration of what we were just doing there with that marker board. Now if I were to choose a random point on the ellipse, let's say I pick a point right here. It could be any point around there. We don't know specifically what that point is, so I'm going to call it the point xy. And what I know is that the distance from xy to focus number 1 plus the distance from xy to focus number 2, that total distance has to equal 2a, because that's the total length of the string. The distance to focus number 1, I'm going to call that d1. And the distance to focus number 2 is d2. And so this leads us to a mathematical equation, or at least the beginnings of an equation, that says that d1 plus d2 is equal to the length of the string, which was equal to 2a. OK class, now what I'm going to need from you is a way of expressing distance d1 using the distance formula. Who can tell me how to write d1 using the distance formula? Matt, what would you say? Well, d1 would, let's see, that's the distance from xy to the point c0 down here. x minus c squared. OK, let's see. There's going to be a square root over there. Yeah, and then x minus c, quantity squared, because that's the difference in the 2x coordinates. OK, plus what? Plus y minus 0 squared. Yeah, I think he's having trouble reading it, because he's pretty far away. y minus 0 squared, exactly. OK, so that's using the distance formula to find distance d1. Now, I could do something similarly to find the distance d2. Let's see, Lene, what would you say is distance d2? x minus, or x plus c squared. Yeah, because it's minus a negative. So x plus c squared. And x minus 0 squared. How about y minus 0 squared? Oh, that's a minute. And what's something we still have to put on that? And put a root. Put a square root on that, yeah, because we're using the distance formula. Now that's equal to 2a. Now, you know, at this moment, if I knew the value of c, and if I knew the value of a, so that I could find the length of the string, if I substituted those numbers in for a and for c, then essentially I could say this is the equation of an ellipse. But boy, is that messy. So what we want to do is to simplify this a bit. And I have two square roots in this equation. So if you think back to intermediate algebra, you remember that when you have a sum or difference of radicals that you want to reduce, what you generally do is to isolate one of the radicals in square root. So I'm going to isolate one of these. Let's say we isolate the one right here on the left. And while I'm at it, I'll just call that y squared to save a little space. So this says the square root of x minus c squared plus y squared is equal to 2a minus the other square root. The square root of x plus c squared plus y squared. OK, now that I have a radical isolated, I'm going to square both sides. And the square on the left is x minus c squared plus y squared. Now to square the expression on the right, I'm going to use a formula. Let me just write it just below here. If I have capital A minus capital B squared, who can tell me what that is when you expand it? A squared minus 2ab plus b squared. Plus b squared. OK, now I'm going to use that same formula right here. But in place of A, I have 2 times little a. And in place of b, I have that square root. And I want to square this just like I squared this. So I'm going to get the first term squared. That's the 2a squared, or 4a squared altogether, minus 2 times this product. Well, 2 times that product would be a 4a times the square root. And we said minus. I'll put a minus in front of that. Plus the last term squared. And the last term squared, well, I just throw the radical away. And I get the quantity x plus c squared plus y squared. OK, now let's remove this formula. Now that we have used that. And so my next step is to simplify this as best we can. Well, I can cancel off a y squared right now. And I think if I square x minus c, and if I square the x plus c, I'll get some more things to cancel. Let's do that next. x squared minus 2cx plus c squared is equal to 4a squared minus 4a times this square root. x plus c squared plus y squared plus. And now if I square x plus c, I get x squared plus 2cx plus c squared. Now that gives me an opportunity to cancel some more terms. I can cancel off an x squared. And I can cancel off a c squared. And while I'm at it, I can actually combine the cx's. I have a negative 2cx, and I have a positive 2cx. So I think those are going to add together. But my biggest problem is I have the square root. I have one square root left over. So once again, I'm going to isolate this and square it. Because there's a negative in front, I think what I'll do is move that term to the other side and make it into a positive 4a times the square root of x plus c squared plus y squared. And everything else I'll collect on the other side. Let's see, we have a 4a squared. And we have all together, we're going to have 4cx. And I don't think there are any other terms left over. Now, before I square this, why don't we get rid of the 4? I have a 4 here, a 4 here, and a 4 here. So we can reduce that. And this says a times the square root of x plus c squared plus y squared equals a squared plus cx. Now, if we wanted, I suppose we could also divide out the a. But since it won't cancel out, I think I'll leave the a over here. And when I square both sides, not only will I square the radical, but I'll square the a. OK, let's do that right now. When I square the left-hand side, I now have a squared times the quantity x plus c squared plus y squared. And on the right-hand side, when I square this sum, I have to square a binomial just like I did before. My first term will be a to the fourth. Then my next term will be 2a squared cx. And my last term will be c squared x squared. Now, you might think at this point that this is just getting totally out of hand, that things are getting too complex. But this is now going to start reducing quite a bit. For example, if I substitute right here in place of x plus c, if I write x squared plus 2cx plus c squared, I'm going to multiply through by a there and set it equal to this. Now, I'm sort of running out of space. So I'm going to go back up to the top to continue that argument. I'll leave that on the bottom so we can still refer to it. OK, so if I multiply through by a squared, I have a squared x squared plus 2a squared cx plus a squared c squared plus a squared y squared. I have to multiply that in. Is equal to, on the other side, a to the fourth plus 2a squared cx plus c squared x squared. At this point, I expect to see some more terms canceling out. Do you see anything that'll cancel there? How about the 2a squared cx and 2a squared cx? I can get rid of those terms. But I don't think anything else cancels, although there are some things that'll add together. For example, I have some x-squares here and I have some x-squared terms there. So my next step will be to get all the variables on one side, the x's and the y's, and all the constant terms, that is the things with only a's and c's. I'll get those on the other side. So if I move these guys over here, I have a squared x squared minus c squared x squared. And here's another term with a variable in it, plus a squared y squared. And on the other side, I have a to the fourth. That's a constant term. And then I have a minus a squared c squared. I call those constant terms because they don't contain any x's or y's. OK, now you might say, well Dennis, how do you know to do this? I mean, how do you know that these are the steps that are required to get wherever you're trying to go? Well, this isn't the first time that I've done this. And I know what answer is I want to create, so I'm just filling in the details that'll get me to where I'd like to be. Now in the first term here, I'm going to factor out the x squared. And this gives me a squared minus c squared times x squared. That's just combining those two terms. Plus a squared y squared. And on the other side, I'm going to factor out the common factor a squared times a squared minus c squared. OK, now I could say this is the equation of an ellipse. If you know the a, that is if you know what the length of the string is, so you can get a. And if you know the value of c, or the foci r, you could plug in those numbers. And this would be the equation of an ellipse. But there's actually still a simpler form that lies ahead. It's only a couple steps away. What I'm going to do is to divide by this product so that I get a 1 right there. And I'm going to divide that product into the first term. So we're dividing by a squared times a squared minus c squared. And I'm going to divide that product into the second term. So I have a squared times a squared minus c squared here. And you see, this allows me to reduce the first term. What is that going to be when I reduce that, this first term? x squared over a squared? x squared over a squared, yeah. And what will the second term be when I reduce it? y squared over a squared minus c squared. y squared over a squared minus c squared. And that's equal to 1. OK, now, we would have been given the c because we would know where the foci are. We'd know the value of a because we knew the length of the string, which was 2a. So the one thing I'm going to do to change this to make it look even simpler is I'm going to change this difference to a new term. I'm going to call it b squared. And I'm going to let b squared be that difference, just so I can write that denominator more simply. So if I make this substitution, then I have x squared over a squared plus y squared over b squared is equal to 1. Now, after all that work, look at what we came through. This is what we conclude is the equation of an ellipse. The a squared on the bottom here, this is half the length of the string squared. And the denominator here is just the difference between a squared minus c squared. And I've just chosen to call it b squared. And this is what we might call the standard equation of an ellipse. OK, now, let's go back over to the greenboard. And while we're at it, let's go to the next graphic while I go over there. OK, here's a summary of what we've just arrived. It says an ellipse drawn using a string with length 2a and suspended at foci at plus or minus c0, those are our points on the x-axis, that the equation is x squared over a squared plus y squared over b squared is 1. This is where a is bigger than b. And the reason a is bigger than b is because you see b squared is a squared minus c squared. So a squared has to be bigger than b squared. Now, the center of that ellipse is at the origin. That's the point halfway between the foci. We call the major axis that the foci are resting on. We call that the major axis. And in this case, that's the x-axis. And the minor axis or the other axis that the foci are not on, that would be the y-axis. OK, now, if we had located the foci on the y-axis, if we go to the next graphic, we will see the second version, the second form of an ellipse. Here we are. It says if the string, again, length 2a, is suspended at foci, this time the foci are on the y-axis at 0 plus or minus c, then the equation, if we derive it in the very same way, I just did that one. But I doubt that anybody wants to see all that work again. It ends up being x squared over b squared plus y squared over a squared is 1. And once again, a is bigger than b. So you notice the denominator under the y is bigger than the denominator under the x's. And again, this is where b squared is a squared minus c squared. The center of the ellipse is at the origin, because it's the midpoint between the foci. But now the major axis is the y-axis. That's where the foci lie. And it's where the elongation takes place. And the minor axis is the x-axis. OK, so if we go to the green screen, let me write those two formulas down here that we can apply in the next example. OK, now that was quite a bit of information for me to load on you there. So let me summarize this so we don't miss some of the subtleties between these two formulas. If you go to the green screen, I think I can point out what we need to know. If the foci are along the x-axis, then that means that the larger square will be under the x-term. You remember in the derivation that I did on the board, I made a substitution under the y-term. And that said that b squared was a squared minus c squared. I'll just write that down here. b squared was a squared minus c squared. And that forces a to be larger than b. So you can tell that the ellipse is stretched along the x-axis if the larger square is under the x. Now on the other hand, if the foci were on the y-axis, if it went through the same derivation, I would find that a squared is under the y-squared term. And that difference that I substituted for previously would now be under the x-squared term. And I'd be substituting, once again, b squared equals a squared minus c squared. a is still bigger than b, but the larger square is under the y. So when we see an ellipse given in one of these two forums, I know which is the major axis by looking for the larger square. OK, let's go to the next example. And we'll apply this to solving a sort of a typical homework problem that you'll be seeing. This says the equation of an ellipse, or find the equation of an ellipse with foci at plus or minus 4, 0 drawn using a string 10 inches long. OK, so let's see. The important information there is the foci. The foci are at plus or minus 4, 0. And the length of the string is 10. Well, let's see. If the foci are at plus or minus 4, 0, that tells me that c is equal to 4. And you remember, the length of the string was 2a. And therefore, a is going to be 5. So now I know c, and I know a. And from that, I can calculate b, because b squared is a squared minus c squared. There's an a, a b, and a c. So b squared is equal to 25 minus 16, or 9. And that tells me then that b, if I take the positive square root, that I'll choose b to be 3. So a is 5, b is 3, c is 4. OK, now the foci are on the x-axis. So that tells me the basic equation is going to be x squared over a squared plus y squared over b squared is 1. And for a squared, I'll substitute in 25 since a is 5. So x squared over 25 plus y squared over b squared, b was 3, b squared was 9. So I'll put a 9 here, and this is equal to 1. So that's the equation of the ellipse. Let me erase the left-hand half, and I'll draw that ellipse. And we'll see how it looks now that we have its equation. OK, here's the x-axis, the y-axis. The foci were at plus or minus 4, 0. So if I go out four units, there's the first focus. And if I go back four units, there's the other focus. So focus 1, focus 2. Now, when I draw the ellipse, the easiest way to sketch it is to find the intercepts. So to find out where this ellipse crosses the x-axis, what I should do is let y be 0. If I let y be 0, if I eliminate the y term, what that tells me is that x squared over 25 is 1. So when y is 0, x would be plus or minus 5. So at 5 and at negative 5, my ellipse crosses the x-axis. So I expect to see the ellipse passing through this point, passing through this point, just outside the foci. How would I find out where this ellipse crosses the y-axis? Is that x equal to 0? Just let x be 0, exactly. And if I let x be 0, then I have the equation y squared over 9 is 1. And that says that y is going to be plus or minus 3. So if I go up 3 and go down 3, I get a point right here and a point right here. And that's where the ellipse crosses the minor axis. So this ellipse is going to come around. It's going to cross at negative 3, come back up, and go through positive 3. And this is a graph of the ellipse. Now you see, if I were to take a string and suspend it from these two foci, and if the string were 10 inches long, then when I stretch the string, it should go to the ellipse and back to there at any given point, say up here. It should stretch up to there and back down to here. So this is the length of the string that we see as we pass around drawing that ellipse. But now I've been able to sketch the ellipse without using the string idea, but just plotting intercepts. OK, let's go to the next example. OK, now here's an example where I've actually given you the equations of the ellipse, and I ask you to give information about the ellipse, including the graph. In the first case, we have the equation x squared over 4 plus 4y squared is equal to 1. And we're asked to sketch the graph of the ellipse by finding its center, the focus, the intercepts, and the length of the string associated with this. Now, when you look at this, you might say, well, Dennis, that doesn't quite look like the fundamental form, because you have a product here, not a quotient. You remember, we're expecting to see something like, something like, let me try a different marker. x squared over a squared plus y squared over b squared is 1, although the a squared and the b squared could be reversed depending on which is the major axis, which is the minor axis. But we don't have a quotient right here. Now, can anyone tell me how we could make that into a quotient? Y squared over 1 fourth? Yeah, you see, I think if we get rid of the 4 on top in this term, if we just divide by 4, then this is going to say x squared over 4. Now, if I divide by 4 on top and bottom, there's a 1. There's a 1 on the bottom right there. So if I divide by 4, that'll be y squared over. And what Stephen has suggested is make that 1 fourth is equal to 1. You see, if I invert and multiply, that would kick the 4 back up into the numerator. So I do have an a squared and a b squared. And how do I know that this is the a squared and this is the b squared? a squared's 4. How do I know that in this equation? Could it be the other way around? Could this be a squared and this be b squared? a's always bigger. a's always bigger. So a squared's always bigger. So this tells me that a is equal to 2. And it tells me that b is equal to 1 half. I'm taking positive square roots here. OK, so when I go to draw the graph, I'll just put the graph over here on the side. When I go to draw the graph, it tells me that the major axis intercepts are plus or minus 2. Now, by the way, are the major axis intercepts on the x-axis or the y-axis? The x-axis. Yes, because the larger square is under the x term. So I should go out 2 in the x direction and negative 2. And in the y direction, I should go up 1 half and go down 1 half. So to sketch the ellipse, it looks like this. It's a very long, thin-looking ellipse in that case. OK, so we've been able to sketch the graph. But there were some other questions that they asked. They said finding the center, well, because of the, this is in that standard form, the center is at the origin, halfway between the foci. So the center is at the origin. Find the foci. Well, would you say the foci are on the x or the y-axis? At the x-axis. They're on the x-axis because that's the major axis. That's the longer axis in this case. But to find the foci, I'm going to need to know the value of c. So we know that a is 2 and b is 1 half. So let's use this space to calculate c. Now we said we know that a is equal to 2 and b is equal to 1 half. And we have our fundamental substitution that b squared is a squared minus c squared. We know a and b. We want to know what is c, so I can locate the foci. So b squared is 1 fourth. a squared is 4 minus c squared. So c squared is equal to 4 minus 1 fourth, which is 15 over 4. And that says that c then must be the square root of 15 over 2. I'm taking the square root on top and on bottom. Now the square root of 15 over 2, that's almost 4 over 2. Square root of 16 over 2 would be what I could reduce to be 2. This number is a little bit less than 2. So the foci are just inside. And I can't really do adequate justice to locate them here. But they're just inside. And the foci are at plus or minus the square root of 15 halves comma 0. The other question was to find the intercepts. Well, we found the intercepts when we would have graphed this. The intercepts are at 2,0 and at negative 2,0. And the minor axis intercepts are at 0,1 half. And let's see, that one is 0 and minus 1 half. So we can find the intercepts. And finally, it says, what is the length of the string associated with each? What's the length of the string associated with this ellipse? See, what was the formula for the length of the string? You remember? Two. Two times what? Two times a. Two times a, that's what I'm going to say. Yeah, two times a. And so two times a would be 4. So the length of the string is 4. You know, by the way, that's also the distance between the major axis intercepts. We go from plus 2 to minus 2. That's a total of four units across there. That's always the length of the string. 2a is, shall we say, the longest diameter in the ellipse. And 2a is the length of the string, of course. So we get 4 in this case. OK, now let's go to the other example, part b. In this one, we don't have the standard equation either, because I'm expecting to see constants underneath x squared and y squared. But also, I'm expecting to see a 1 over here. So let's just make it a 1 by dividing by 100. So I get a 1 there. And if I divide by 100 over here, I have 16x squared over 100 plus 25y squared over 100. Now, I can reduce each of these. The first one, I can divide top and bottom by 4. And I get 4x squared over 25. And the second one, I get y squared over 4 is equal to 1. But I have that 4 on top. So let's use the same trick that Stephen suggested from over here. Let's divide top and bottom by 4. So this is x squared over 25 fourths plus y squared over 4 is equal to 1. Now, which one of these fractions has the larger denominator? Which one of these rational expressions? The y. Well, let's see. Now, the y denominator is 4. How much is 25 over 4? It's a little over 6, isn't it? So actually, this still has the larger denominator. So that tells me that the a squared is 25 over 4. So a, the a down here, must be 5 halves. And the b must be 2. Let's go ahead and calculate c while we're at it. I know that b squared is a squared minus c squared. So 25 fourths is equal to, oops, no, wrong number there. b is 2, so 4 is equal to 25 fourths minus c squared. So c squared is 25 over 4 minus 4. That's going to be, let's see, 25 fourths, take away 16 fourths is 9 fourths. And so c is going to be 3 halves. So we know a, we know b, and now we know c. c is equal to 3 halves. OK, I'll just erase this so I can use it to record some more information. Let's see. If I draw the graph, the larger square, again, is under the x term. So that says that I should go out a in the x direction. So I'll go out 2 and 1 half or 5 halves in each direction. There's 2 and 1 half. Here's negative 2 and 1 half. And I should go up 2 in the y direction. Go up 2 and go down 2. So when I draw this ellipse, it looks like this. OK, this one's a little bit easier to draw because it's not quite so thin and narrow. And the foci are at 3 halves on the x-axis. There's a focus there, focus number 1. And there's a focus over here at negative 3 halves, focus number 2. So this is the intercept at 5 halves, 0. And over here, we have negative 5 halves, 0. And the minor intercepts are at 0, 2. And at 0, negative 2. Now, if I record the information as well as the graph that we were asked for back over here, first of all, the center is at the origin because the foci are on opposite sides of the origin, same distance. The foci are at plus or minus 3 halves, 0. So if you look in the back of your textbook on the homework problems, you'll see this information listed separately like I'm doing here. The intercepts, well, the major intercepts, those would be the x-intercepts are at plus or minus 5 halves, 0. And the minor intercepts are at 0 and plus or minus 2. And the length of the string, well, who can tell me, what is the length of the string? Would be 5, yeah, it's 2 times a. a was 5 halves, so it's 2 times 5 halves are 5 units long. OK, now let's go to the next graphic on eccentricity. There's a term called eccentricity to describe the differences between these ellipses. You know, some of these ellipses have been rather roundish, and others are very long and thin. In the two examples that I just worked, the first one, part A, was a rather thin one, stretched horizontally. And the second one was almost round. Now, if you look at this graphic, it says ellipses vary in size, some being nearly circular, while others are rather thin. These variations depend on the values of A and C. And the ratio C over A is called the eccentricity of the ellipse. And it's denoted by the letter E. Now, you have to be a little careful here, because where did the number E, where did the value, the letter E, come up earlier in this course? The natural exponential base. Natural exponentials. Yeah, you remember it was about 2.718. So here we're using the letter E to represent something else, something other than that, called the eccentricity. That's because we're a long way from logarithms. We're talking about conic sections right now. But it's not surprising that eventually the same letters might be used more than once. So E is the ratio C over A. And it says it should be clear that for an ellipse that E has to be between 0 and 1. Now, the reason that should be, quote, so clear is you see C is always smaller than A, because the foci are inside of the ellipse. And therefore, C over A would have to be smaller than 1. But on the other hand, C and A are both positive, so the ratio is positive. Now, also it says it should be clear that, oh, no. Then it says that for E approximately 1, the ellipse is very thin. Now, that's because if E is almost 1, what can you tell me about the relationship between C and A? If the ratio is almost 1. They're very close together. They're almost the same value if the ratio is 1. So if you put the foci out very close to the ends of the string, in other words, you're pulling the string almost tight, you tend to get a very thin ellipse. On the other hand, if E is almost 0, which says that C is relatively small compared to A, it says the foci are close together, and the string is relatively long. So you tend to get an ellipse that's almost round. Now let's go to the next graphic, and I'll just show you some illustrations of this. On the left, we have an ellipse whose eccentricity is 1-tenth. And so that's fairly close to 0. And so I tend to get an almost round ellipse. You know, actually, this rendering of the ellipse is not quite right. This is sort of an elliptical egg shape, but it should have been sort of sideways. So if you imagine rotating that 90 degrees, then I'm thinking of this ellipse in the same relationship with these two that are stretched sideways, not vertically. But you see, this is almost round, and its eccentricity, even as it's shown, its eccentricity is very small. Now if the eccentricity is about a half, what that means is that C is about half the value of A, and the ellipse tends to be, well, what can we say, more elliptical? These are all ellipses, but it's not quite as roundish as the one over here was. And then finally, when the ratio is approximately 1, say 0.9, then that says that C is very far out, so the string can't stretch very much, and you get an elongated, thinner looking ellipse. So that's referred to as the eccentricity. And the eccentricity is merely a way of relating to someone, the relative shape of the ellipse. If the eccentricity is small, it's rather roundish. If the eccentricity is closer to 1, it's rather thin, like you see here. OK, time for the next example. In this next example, we have a change of coordinate system. We're going to make a translation of an ellipse. In this case, we have x minus 2 squared over 4 plus y minus 4 squared over 36 is 1. We're asked to sketch this ellipse, finding the center, the foci, the axes, the intercepts, and now the eccentricity. OK, let's go to the green screen, and I'll just write that problem up here. The problem said x minus 2 squared over 4 plus y minus 4 squared over 36 is equal to 1. What makes this problem different is you see I've subtracted something from the x and subtracted something from the y. But you know, I'm thinking, so over here is sort of my thinking cloud, I'm thinking this looks a lot like x squared over 4 plus y squared over 36 equals 1. Now, the ellipse that I'm thinking of over here, which is its major axis, the x-axis or the y-axis? The y-axis. The y-axis, why is that? It's bigger. Because it has a bigger denominator, yeah. And it looks like we would go up and down 6 on the y-axis. I'd go left and right 2 on the x-axis, because a is 6, and it looks like b is 2. And those are the intercepts at the a's and at the b's. Well, I have to do the same thing here, except I have to shift the ellipse over 2 to the right and shift it up 4. So that's going to move the center of the ellipse off a bit. So here's the x-axis. Here's the y-axis. And it looks like in the ellipse that I have here, a is equal to 6 and b is equal to 2. And we're going to be shifting it by moving over 2 and up 4. Now, I'm going to keep these units rather small, because I think this is going to be a rather large ellipse. So if I go over 2 and if I go up 4, then I get this point as the center, because that's where the origin is moved to. And the origin was the original center. Now, I'm just going to kind of dot in a little coordinate system along here to help me locate the other points. These are not asymptotes that I'm drawing. I've previously drawn dotted lines for asymptotes or for directrix. This is just merely to help me locate where the new points will be. This ellipse has been stretched in the y direction. So I'm going to go up 6 units. That would put that up at 10. And I'm going to go down 6 units. I was at 4. So that would put me right here at negative 2. And those are the major axis intercepts. Then I'm going to go to the left and right 2 units. And because I went over 2 units to get to the center, when I go back 2 units, I'm just touching the y axis. And so my ellipse is going to look like this. I think my just erased a point there. OK, so I get this ellipse right here. Now, this is a graph of the ellipse that we have up here in this equation. But there's some information that we're asked about this problem. First of all, where's the center of the ellipse? Who can tell me what's the center of this ellipse? We just did it. Yeah, we just did that. It's 2, 4. Yeah, 2, 4. OK, now where are the major axis intercepts? 2, 10. 2, 10. And 2, negative 2. 2, negative 2, yeah. 2, 10, and 2, and negative 2. You see, it's not quite as easy as just putting a plus or minus on these coordinates, because the center is no longer at the origin. The minor axis intercepts are at 4, 4, and at 0, 4. What's something else we'd need to know about this ellipse? We found the intercepts. The foci. The foci, yeah, the foci. Now, let me just make some room to right here. To find the foci, I'm going to use our fundamental identity that says that b squared is a squared minus c squared. So b squared is a squared minus c squared. b squared is equal to 4, because here's b squared right there. a squared is equal to 36 minus c squared. This tells me that c squared is equal to 32. And so c is equal to 4 times the square root of 2, which is almost, it's almost 6. You know, the square root of 36 is 6. So this would be a little bit smaller, a little bit smaller than 6 would be. So to find the foci, well, let's see, we were at this point 2, 4, and we went up. We want to go up c units to get to that focus. That'll be at 2, and then 4 plus 4 square root of 2. But I'll put a plus or minus in there, because I want to either go up 4 square root of 2 or go down 4 square root of 2. And this gives me a formula for the two foci. OK, now let's go to the next graphic. And this says that we want to recall from the previous episode that we said that if you write the equation ax squared plus by squared plus cx plus dy plus e equals 0, you always get a conic. Now let's look at this example. Suppose we have x squared plus 4y squared minus 2x plus 24y plus 29 equals 0. Now you notice this is in the form of ax squared plus by squared, et cetera. a is equal to 1, b is equal to 4. Now to find out what sort of a curve, what sort of a conic this is, I'm going to complete the squares. So I'm going to group the x terms together, x squared minus 2x. And I'm going to group the y terms together, 4y squared plus 24y. And the 29, I'll just move to the other side, call it a minus 29. Now to complete the square, I need to add on a 1 there. So I'll need to add on a 1 over here. Let's see, I'm almost off the boards. I'll just put it right underneath it plus 1. And oh, you know what, before I do that, I need to factor out a 4 here. So let me just back up. Let's factor out a 4 so we'll know what to add to each one of these terms. So let's leave out that plus 1. I'll write this as x squared minus 2x. And here I want to factor out a 4. 4 times y squared plus 6y equals minus 29. Now if I add a 1 on the left, I'll add a 1 on the right. And if I add a 1 on the left, we have to be careful here. If I won't be adding a 1, I'll be adding a 9 to complete the square. How much should I add on the right? 36. 36, yes. And that's because we're adding in 4 nines because we actually have 4 times this expression. So I have to add on 36 there. So this gives me x minus 1 squared plus 4 times y plus 3 squared equals a total of 8 is what we get over there. OK, now what am I going to have to do to put this in standard form? Divide both sides by 8. Divide by 8, yeah. So that'll give me a 1 here. And that's what I'm expecting to see in standard form. But I'll have to divide by 8 in each of these terms. Divide by 8 here. And when I divide by 8 here, I'll cancel off the 4. And that'll be over a 2. OK, well this problem is a little different because of just the values of a and b and c. Can anyone tell me what does a equal in this problem? Root 8. Yeah, you see this is a squared. It's the larger of the two squares. So we have to put the square root of 8 here. And the square root of 8 is 2 times the square root of 2. And the value for b is the square root of 2 because b squared is 2. And the value for c is, well, let's see. We know that b squared is a squared minus c squared. So we have 2 equals 8 minus c squared. And so c squared would be, let's see if I bring the c squared over, 8 minus 2 is 6. So now we know that c is the square root of 6. Now, here's a problem where a, b, and c are all irrational numbers. We don't always get integers. We don't always get fractions. But we can still evaluate a, b, and c. OK, and then also the center of the ellipse is at 1, negative 3. So let me erase this middle portion right here. And we'll draw an ellipse centered at 1, negative 3 with these values for a, b, and c. OK, if we go back to the overhead, we have the center at 1 and negative 3. OK, there's the center. And the major axis was the x-axis because I had the larger square under the x. So if I go forward 2 square roots of 2, and if I go back 2 square roots of 2, and if I go up the square root of 2, which is a little over 1, and if I go down the square root of 2, then my ellipse will be roughly right here. I realize this is sort of a crude sketch, but that's all we need. When you're drawing these and I ask you to sketch these, you don't have to make a very, very accurate sketch. But one that should at least convey the information. And so we have that the center is at 1, negative 3. We have the foci at, let's see, now the foci would lie along the x-axis. That's going to have to be 1 plus or minus 2 square roots of 2, negative 3, because the foci are going to be going out of the x direction from 1. The major and minor intercepts, I won't bother writing down their coordinates because I want to move on to the next example. But you can see here, we're going to be going up the square root of 2 and down the square root of 2 from the point 1, negative 3. OK, let's go to an application of ellipses that's on the next graphic. This one's a little different from some of the other applications that we've seen. I called it Westward Ho. Now we have a wagon train traveling 10 miles each day. And the wagon train has a scout on horseback and the scout can travel 20 miles every day. So each morning the scout sets out from the traveling party to look for danger before returning for vitals. But of course, when the scout comes back, the wagon train has moved forward. So when the scout comes back, he catches up with the wagon train 10 miles down the road. So write an equation for the curve that is the outer boundary of his wanderings. OK, so if you would, let's come to the screen here. And I'm going to put a coordinate system on this problem. So you see the way it's worded, there's no coordinate system in there. It's just a wagon train moving. So I'm going to put an x and a y axis so that the wagon train starts off over here at plus 10. And at the end of the day, no, see, that would be plus 5, wouldn't it? And at the end of the day, the wagon train's gone 10 miles. That'll be negative 5. Now we have a scout who rather than sticking with the wagon train going 10 miles, the scout can actually go out further, and then at the end of the day, come back and catch up with the train. And what's the total distance that the scout could travel? 20. Would be 20, yeah. So this is a total length of 20. Now you know, it doesn't have to be that the scout goes out as far as the scout can go. The scout could maybe come up here and stop and whatever, and then later come back to the train. But this represents the furthest out the scout could go in order to get back to the train at the end of the day. So the scout couldn't go any further than that. Well, you see this is beginning to look a little bit like an ellipse because the distance that the scout travels has a total of 20 and the distance between the two foci is only 10. So it looks to me like what we're describing as the outer boundary is an ellipse. And the ellipse would look something like, well, let's see, probably wouldn't be quite that distorted. But there would be an ellipse with foci at plus or minus 5. And instead of the length of the string, it would be the distance the scout can travel. That's going to be 20. So can anyone tell me the value of A if these two together, if these two together make 20? Would be 10, yeah, because you remember before the length of the string would be 2A. So A is 10. And what about the 5? What is that going to tell me about the ellipse? A is 10, but what is 5? What is that? It's the focal part. Yeah, so which letter would that be? C. C. So C is equal to 5. Now the question was to find the equation of the ellipse that represents the outer boundary of the scout that the outer boundary that the scout can move. So we have A, we have C. What we need is B. So B squared is A squared minus C squared. So that's going to be 100 minus 25. So B squared is equal to 75. So from this we can get A squared. From this I have B squared. So the equation of this ellipse is X squared over A squared. That'll be 100. Plus Y squared over 75 is equal to 1. So that's the equation of the outer boundary that the scout can reach before coming back to camp at the end of the day. OK, as our other application for ellipses, if we stay on the green screen, let me just draw an illustration of what's going to come up on the next graphic. If you have an ellipse on a coordinate system, and it has two foci, one here and one over here, and ellipse has the property that if you had, for example, a light source at this focus, and the light were to go out and reflect off a reflective elliptical mirror, it would reflect to the other focus. So it has a reflective property different but sort of parallel to what we saw for a parabola. And if the light goes backwards over here, it'll still be reflected to that focus. OK, let's go to the next graphic, and we'll see this illustrated or discussed. The reflective property for ellipses as follows. Every conic has a unique reflective property. For ellipses, the property says that light directed from a light source at one focus will be reflected to the other focus. Now, one application of this is in a whispering chamber, and it says it is this property that is the basis for a whispering chamber, such as the capital rotunda in Washington, DC. If you've been to Washington, DC, you may know that in the rotunda there, I think it was designed by Thomas Jefferson, there's an elliptical dome, shall we say half an elliptical dome, and basically an ellipse that's been revolved around its central axis. And then chopped off, so you have only the upper half of the dome. And at the two foci, you can stand at one focus, and supposedly you can whisper, and the person at the other focus can hear what you're saying. And that's because the sound that goes out is reflected off the elliptical dome and reflected to the other focus. So that uses this reflective property. You know, if we had an elliptical pool table that was, say, look like this, and if it had one hole in it, one pocket that was at a focus, then you could make a shot into that pocket by simply taking a pool ball, shooting it right across the focus, and it would be reflected to the other focus point where the pocket is. So if you know where this focus is, all you do is shoot a ball toward that focus, and it'll be reflected to the other focus point. That's probably why they don't make elliptical pool tables. It would be easy to play on them, plus they'd be difficult to construct. Well, so far we've seen now discussions of two of the three conics. We've looked at parabolas, and we've looked at ellipses. Next time in episode 30, we'll look at hyperbolas. See you then.