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Foci of an ellipse from equation

Sal explains how the radii and the foci of an ellipse relate to each other, and how we can use this relationship in order to find the foci from the equation of an ellipse. Created by Sal Khan.
Video transcript
Let's say we have an ellipse formula, x squared over a squared plus y squared over b squared is equal to 1. And for the sake of our discussion, we'll assume that a is greater than b. And all that does for us is, it lets us so this is going to be kind of a short and fat ellipse. Or that the semi-major axis, or, the major axis, is going to be along the horizontal. And the minor axis is along the vertical. And let's draw that. Draw this ellipse. I want to draw a thicker ellipse. Let's say, that's my ellipse, and then let me draw my axes. OK, this is the horizontal right there. And there we have the vertical. And we've studied an ellipse in pretty good detail so far. We know how to figure out semi-minor radius, which in this case we know is b. That's the same b right there. And that's only the semi-minor radius. Because b is smaller than a. If b was greater, it would be the major radius. And then, of course, the major radius is a. And that distance is this right here. Now, another super-interesting, and perhaps the most interesting property of an ellipse, is that if you take any point on the an ellipse, and measure the distance from that point to two special points which we, for the sake of this discussion, and not just for the sake of this discussion, for pretty much forever, we will call the focuses, or the foci, of this ellipse. And these two points, they always sit along the major axis. So, in this case, it's the horizontal axis. And they're symmetric around the center of the ellipse. So let's just call these points, let me call this one f1. And this is f2. And it's for focus. Focuses. f2. So the super-interesting, fascinating property of an ellipse. And it's often used as the definition of an ellipse is, if you take any point on this ellipse, and measure its distance to each of these two points. So, let's say that I have this distance right here. Let's call this distance d1. And then I have this distance over here, so I'm taking any point on that ellipse, or this particular point, and I'm measuring the distance to each of these two foci. And this is d2. We'll do it in a different color. So this is d2. This whole line right here. That's d2. So when you find these two distances, you sum of them up. So this d2 plus d1, this is going to be a constant that it actually turns out is equal to 2a. But it turns out that it's true anywhere you go on the ellipse. Let me make that point clear. And I'm actually going to prove to you that this constant distance is actually 2a, where this a is the same is that a right there. So, just to make sure you understand what I'm saying. So let me take another arbitrary point on this ellipse. Let's take it right there. And if I were to measure the distance from this point to this focus, let's call that point d3, and then measure the distance from this point to that focus -- let's call that point d4. d4. If I were to sum up these two points, it's still going to be equal to 2a. Let me write that down. d3 plus d4 is still going to be equal to 2a. That's just neat. And, actually, this is often used as the definition for an ellipse, where they say that the ellipse is the set of all points, or sometimes they'll use the word locus, which is kind of the graphical representation of the set of all points, that where the sum of the distances to each of these focuses is equal to a constant. And we'll play with that a little bit, and we'll figure out, how do you figure out the focuses of an ellipse. But the first thing to do is just to feel satisfied that the distance, if this is true, that it is equal to 2a. And the easiest way to figure that out is to pick these, I guess you could call them, the extreme points along the x-axis here and here. We're already making the claim that the distance from here to here, let me draw that in another color. That this distance plus this distance over here, is going to be equal to some constant number. And using this extreme point, I'm going to show you that that constant number is equal to 2a, So let's figure out how to do that. So one thing to realize is that these two focus points are symmetric around the origin. So, whatever distance this is, right here, it's going to be the same as this distance. Right there. Because these two points are symmetric around the origin. So this, right here, is the same distance as that. And, of course, we have -- what we want to do is figure out the sum of this distance and this longer distance right there. Well, what's the sum of this plus this green distance? Well, this right here is the same as that. So this plus the green -- let me write that down. So let me write down these, let me call this distance g, just to say, let's call that g, and let's call this h. Now, if this is g and this is h, we also know that this is g because everything's symmetric. So what's g plus h? Well, that's the same thing as g plus h. Which is the entire major diameter of this ellipse. Which is what? Well, we know the minor radius is a, so this length right here is also a. So the distance, or the sum of the distance from this point on the ellipse to this focus, plus this point on the ellipse to that focus, is equal to g plus h, or this big green part, which is the same thing as the major diameter of this ellipse, which is the same thing as 2a. Fair enough. Hopefully that that is good enough for you. Now, the next thing, now that we've realized that, is how do we figure out where these foci stand. Or, if we have this equation, how can we figure out what these two points are? Let's figure that out. So, the first thing we realize, all of a sudden is that no matter where we go, it was easy to do it with these points. But even if we take this point right here and we say, OK, what's this distance, and then sum it to that distance, that should also be equal to 2a. And we could use that information to actually figure out where the foci lie. So, let's say I have -- let me draw another one. So that's my ellipse. And then we want to draw the axes. For clarity. Let me write down the equation again. Just so we don't lose it. x squared over a squared plus y squared over b squared is equal to 1. Let's take this point right here. These extreme points are always useful when you're trying to prove something. Or they can be, I don't want to say always. Now, we said that we have these two foci that are symmetric around the center of the ellipse. This is f1, this is f2. And we've already said that an ellipse is the locus of all points, or the set of all points, that if you take each of these points' distance from each of the focuses, and add them up, you get a constant number. And we've figured out that that constant number is 2a. So we've figured out that if you take this distance right here and add it to this distance right here, it'll be equal to 2a. So we could say that if we call this d, d1, this is d2. We know that d1 plus d2 is equal to 2a. And an interesting thing here is that this is all symmetric, right? This length is going to be the same, d1 is is going to be the same, as d2, because everything we're doing is symmetric. These two focal lengths are symmetric. This distance is the same distance as this distance right there. So, d1 and d2 have to be the same. There's no way that you could -- this is the exact center point the ellipse. The ellipse is symmetric around the y-axis. So if d1 is equal to d2, and that equals 2a, then we know that this has to be equal to a. And this has to be equal to a. I think we're making progress. And the other thing to think about, and we already did that in the previous drawing of the ellipse is, what is this distance? This distance is the semi-minor radius. Which we already learned is b. And this of course is the focal length that we're trying to figure out. This should already pop into your brain as a Pythagorean theorem problem. So we have the focal length. And we could do it on this triangle or this triangle. I'll do it on this right one here. This focal length is f. Let's call that f. f squared plus b squared is going to be equal to the hypotenuse squared, which in this case is d2 or a. Which is equal to a squared. And now we have a nice equation in terms of b and a. We know what b and a are, from the equation we were given for this ellipse. So let's solve for the focal length. The focal length, f squared, is equal to a squared minus b squared. So, f, the focal length, is going to be equal to the square root of a squared minus d squared. Pretty neat and clean, and a pretty intuitive way to think about something. So you just literally take the difference of these two numbers, whichever is larger, or whichever is smaller you subtract from the other one. You take the square root, and that's the focal distance. Now, let's see if we can use that to apply it to some some real problems where they might ask you, hey, find the focal length. Or find the coordinates of the focuses. So let's add the equation x minus 1 squared over 9 plus y plus 2 squared over 4 is equal to 1. So let's just graph this first of all. This could be interesting. So I'll draw the axes. That's the x-axis. This is the y-axis. And we immediately see, what's the center of this? The center is going to be at the point 1, negative 2. And if that's confusing, you might want to review some of the previous videos. Center's at 1, x is equal to 1. y is equal to minus 2. That's the center. And then, the major axis is the x-axis, because this is larger. And so, b squared is -- or a squared, is equal to 9. And the semi-minor radius is going to be equal to 3. So, if you go 1, 2, 3. Go there. Than you have 1, 2, 3. No. 1, 2, 3. 1, 2, 3. I think this -- let's see. 1, 2, 3. You go there, roughly. And then in the y direction, the semi-minor radius is going to be 2, right? The square root of that. So b is equal to 2. So you go up 2, then you go down 2. And this ellipse is going to look something like -- pick a good color. It's going to look something like this. And what we want to do is, we want to find out the coordinates of the focal points. So, the focal points are going to sit along the semi-major axis. And we need to figure out these focal distances. And then we can essentially just add and subtract them from the center. And then we'll have the coordinates. What we just showed you, or hopefully I showed you, that the the focal length or this distance, f, the focal length is just equal to the square root of the difference between these two numbers, right? It's just the square root of 9 minus 4. So the focal length is equal to the square root of 5. So, if this point right here is the point, and we already showed that, this is the point -- the center of the ellipse is the point 1, minus 2. The coordinate of this focus right there is going to be 1 plus the square root of 5, minus 2. And the coordinate of this focus right there is going to be 1 minus the square root of 5, minus 2. And all I did is, I took the focal length and I subtracted -- since we're along the major axes, or the x axis, I just add and subtract this from the x coordinate to get these two coordinates right there. So, anyway, this is the really neat thing about conic sections, is they have these interesting properties in relation to these foci or in relation to these focus points. And in future videos I'll show you the foci of a hyperbola or the the foci of a -- well, it only has one focus of a parabola. But this is really starting to get into what makes conic sections neat. Everything we've done up to this point has been much more about the mechanics of graphing and plotting and figuring out the centers of conic sections. But now we're getting into a little bit of the the mathematical interesting parts of conic sections. Anyway. See you in the next video.