0 energy points

# Foci of a hyperbola from equation

Sal discusses the foci of hyperbolas and shows how they relate to hyperbola equations. Created by Sal Khan.
Video transcript
In the last video, we learned that an ellipse can be defined as the locus of all points where the sum of the distances to two special points, called foci-- and let me draw this all out, so that's my x-axis-- the sum of the distance to these two special points, called focuses or foci, is a constant. So if this is my ellipse-- I'll draw it out, that looks about where I want it to be right around, that looks about right-- it's centered at the origin, doesn't have to be, but for our purposes, let's make it centered on the origin. If this is one focus point right here, and this is the other focus point, this ellipse could be defined as the set of all points, or the locus of all points, where if I take the distance of any one of these points that exist on the ellipse and take the distance to each of the locuses-- sorry, each of the focuses-- if I take that-- nope, I don't want to do that-- if I take that distance and add it to this distance, so let's call this d1-- nope, too thick-- let's call that d1, this is d2-- that that's going to be equal to a constant number along the whole ellipse. So if I take a random point along the ellipse, say I take this point right here, and if I were to sum this distance to that distance-- so let's call this d3, this is d4-- the sums of these distances to the focus along with this ellipse are going to be a constant. So in this case, d2 plus d1, this plus that, is going to be equal to d3 plus d4. And this would be true wherever you go along the whole ellipse, and we learned in the last video that this quantity is actually going to be equal to 2a, where a is the distance of the semi-major radius. If this is the formula for the ellipse, this is where the a comes from. x squared over a squared plus y squared over b squared is equal to 1. And we learned that the focus, the focal distance-- or the distance from the central of the ellipse, which is this distance right here-- that focal distance is just the square root of the difference of these two numbers. If this is the focal distance from here to here, it's just equal to the square root of-- if a is larger then it would be a squared minus b squared, which is the case in this ellipse. If we have a vertical ellipse, and I really didn't cover it in the last video, but let me just show you what it would look like. Let's say that the ellipse looks something like this. I'll use blue. Let's say the ellipse looks like that. In this case, our semi-major radius is now in the y direction, so this is b, this is a, and in this case, b is greater than a, because the ellipse is tall and skinny. In this case the, focuses are always going to lie along the major axis. In this case the major axis is the vertical axis, so the focuses are going to sit here and here, and in this case, the focal lengths are going to be vertical down from the origin and vertical up from the origin, and we get, instead of it being a squared minus b squared, now since b is larger than a, the focal length, which is this, is going to be equal to b squared minus a squared. Fair enough. Now I did all of that to kind of compare it to what we're going to cover in this video, which is the focus points or the foci of a hyperbola. And a hyperbola, it's very close to an ellipse, you could probably guess that, because if this is the equation of an ellipse, this is the equation of a hyperbola. x squared over a squared minus y squared over b squared is equal to 1. Or we could switch these around, where the minus is in front of the x instead of the y. And we could cover that in a second. But this hyperbola looks something like this. Let me see if I can draw it. If I draw the axes, and then I want to draw the asymptotes-- you could prove it, you could look at the some of the previous videos, but the asymptotes for this hyperbola are going to be y is equal to plus or minus b over ax, so it's going to look-- I'll just draw them as kind of tilted lines, so it would look something like that, something like-- nope. I want to make it something like that. Those are the as-- but this one's centered at the origin, because it hasn't been shifted, and then that's those two lines right there. And then, this is what I call kind of a horizontal hyperbola. The way you can think about that is, well, if you solve for y, you'll see that you're always going to be a little bit lower than the asymptote. The other option is you say, OK, can x or y equal 0? Well if y is equal to 0, that puts us along the x-axis, and you get x squared over a squared is equal to 1, which means that x squared is equal to a squared, which means that x is equal to plus or minus a. So the points a,0, and the point minus a,0, are both on this hyperbola. And since they have to kind of be contained by these asymptotes, never go through it, you know that this is going to be a hyperbola that opens to the left and the right, so it'll look something like this. Let me use its color. So it'll look something-- this is the hard part-- it'll get closer and closer on that side, and then you kind of view this as one of the vertexes or vertices of the hyperbola, and it'll go like that. Where this distance-- and notice the similarity here with the ellipse-- this distance right here-- let me do it in a more vibrant color-- this distance right here, between these two I guess you could call them the elbows of the two ellipses-- that distance right there, that is a, and this is also a. So you have a 2a distance, which is very similar to this situation, where this distance is a and this distance is a. So your distance between the two left and right points in a horizontal ellipse is the same as the distance between the two left and right points on a hyperbola. It's just the hyperbola opens outward while the ellipse opens inward. Fair enough. But the whole point of this video is to discuss foci, and you might have guessed-- and I did touch on it in the last video-- that hyperbola also have foci, but they open. They're going to be to the right and the left of these two points, so this is a-- let me do them in a bright color, because I want you see them-- let's say that those are the two foci-- and a hyperbola, this is-- and notice the difference. An ellipse, one of the definitions of an ellipse, was the locus of all points or the set of all points where the distance from each of those points, the sum of the distance from each of those points to the two foci is a constant. Now the definition of a hyperbola, one of the definitions of a hyperbola, can be the locus of all points where you take the difference-- not the sum, you take the difference of the distances between the two foci, so if-- let me write that down. So this is d1, and this is d2, so you have a situation-- and we could take the absolute value of the difference, because they might be, they might at some points d1 will be longer than d2 if you're on this curve. If you're on this curve, d1 will be shorter than d2-- So d1 minus d2, the absolute value is going to be equal to constant. In the ellipse situation, d1 plus d2 was a constant. So they're very closely related. In the ellipse, you're taking the sum of the distances to the focus points and saying it's a constant. In a hyperbola, you're taking the difference of the distances to the focus points and saying that's a constant. So this number right here is going to be the exact same thing as if I took a point right here-- and I'm picking these points arbitrarily, as long as they're on the hyperbola, and if I called these two points d3 and d4, the difference between d1 and d2 is the same thing as the difference, between d3 minus d4. This is going to be a constant the entire way around the ellipse. And so the next question is, what is this constant going to be equal to? And this is where it's useful to find a point where you can kind of get the intuition, and we did it with the ellipses, where we said if we take these points, we used logic in the last video to say oh, the sum of the distance between this and this, that sum, is going to be equal to, we saw, is going to be equal to 2a, or the distance of the semi-major axis. Because this distance was the same as this distance, so this plus this is the same thing as this plus that, which is 2a. So the entire time, the contents, the sum of the distances to the two foci was equal to 2a. Now in the hyperbola, what is the difference of the distances to the two foci? So let's take this point right here on the hyperbola, and we're saying so what is-- let me do a good color-- what is the magenta distance-- that's the distance to that foci-- minus this-- let me pick another color-- minus this light blue distance? This magenta distance minus this light blue distance. So we can make a very similar argument that we made in the ellipse situation. This light blue distance is the same, the distance from this vertex or from this leftmost point of this rightward opening hyperbola to this focus is the same as this distance. Because a hyperbola is symmetric around the origin or the focal length is the same on either side of the center of the hyperbola depending on how you may view it, but I think that's not too much of a stretch of a statement for you to for you to accept. So if this distance is the same as this distance, then the magenta distance minus this blue distance is going to be equal to this green distance. And this green distance is what? That's 2a. We saw that at the beginning of this video. So this, once again, is also equal to 2a. Anyway, I'll leave you there right now. Actually, let's actually just do one problem, just because I like to make one concrete. Because I told you at the beginning that if you wanted to find the-- so if you have an ellipse-- so if you have-- this is an ellipse, x squared over a squared plus y squared over b squared is equal to 1, we learned that the-- that's over b squared-- this is an ellipse. We learned that the focal length is equal to the square root of a squared minus b squared. Now for a hyperbola, you kind of see that there's a very close relation between the ellipse and the hyperbola, but it is kind of a fun thing to ponder about. And a hyperbola's equation looks like this. x squared over a squared minus y squared over b squared, or it could be y squared over b squared minus x squared over a square is equal to 1. It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. So if I were to give you-- so notice the difference. It's just a difference in sign. You're taking the difference of those two denominators, and now you're taking the sum of the two denominators. So if I were to give you the following hyperbola. x squared over 9 plus y squared over 16 is equal to 1. Well, the first thing you do is-- though I've never-- well, we could just figure out the focal length just by plugging into the formula. The focal length is equal to the square root of a squared plus b squared. This is squared, right? a is three. b is 4. So 9 plus 16 is 25, which is equal to 5. And so if we were to graph this-- that's my y-axis, that's my x-axis-- and the focal length is the distance to, in this case, to the left and the right of the origin. If it was kind of an up and down opening hyperbola, it would be above and below the origin, so this is a-- oh sorry, this should be a plus. We're doing with a hyperbola, that should be a minus. Don't want to confuse you. What I had written before, with a plus, that would have been an ellipse. A minus is the hyperbola. So the two asymptotes-- this is centered at the origin, it hasn't been shifted-- are going to be 16 over 9, so it's going to be fairly steep asymptotes, it's going to look something like that and that, those are the two asymptotes-- The two vertex points are at 2 times a. a is three, right? a squared is equal to 9, b squared is equal to 16, so this is the center, so the two vertex points, this is 3 minus 3, and then the focal points are going to be at, from the center, 5 to the right, so that's going to be right here, so that's 5,0, and minus 5,0. This is minus 3, and this is 3, so if we were to graph it, it would look something like this. There you go. And if you were to take an arbitrary point on that hyperbola and take this distance and subtract that from that distance, that will be a constant number that would be exactly equal to 2a or exactly equal to 6, in this particular example. Anyway. Hopefully I didn't confuse you too much with that little sign error near the end of the video, but in the next video, I'll prove to you this formula, which is a little bit of hairy algebra but it's fun to do regardless.