Sal introduces the standard equation for hyperbolas, and how it can be used in order to determine the direction of the hyperbola and its vertices. Created by Sal Khan.
Let's see if we can learn a thing or two about the hyperbola. And out of all the conic sections, this is probably the one that confuses people the most, because it's not quite as easy to draw as the circle and the ellipse. You have to do a little bit more algebra. But hopefully over the course of this video you'll get pretty comfortable with that, and you'll see that hyperbolas in some way are more fun than any of the other conic sections. So just as a review, I want to do this just so you see the similarity in the formulas or the standard form of the different conic sections. If you have a circle centered at 0, its equation is x squared plus y squared is equal to r squared. And we saw that this could also be written as-- and I'm doing this because I want to show that this is really just the same thing as the standard equation for an ellipse. If you divide both sides of this by r squared, you get x squared over r squared plus y squared over r squared is equal to 1. And so this is a circle. And once again, just as review, a circle, all of the points on the circle are equidistant from the center. Or in this case, you can kind of say that the major axis and the minor axis are the same distance, that there isn't any distinction between the two. You're always an equal distance away from the center. So that was a circle. An ellipse was pretty much this, but these two numbers could be different. Because your distance from the center could change. So it's x squared over a squared plus y squared over b squared is equal to 1. That's an ellipse. And now, I'll skip parabola for now, because parabola's kind of an interesting case, and you've already touched on it. So I'll go into more depth in that in a future video. But a hyperbola is very close in formula to this. And so there's two ways that a hyperbola could be written. And I'll do those two ways. So it could either be written as x squared over a squared minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it. And the second thing is, not only will you forget it, but you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. If you multiply the left hand side times minus b squared, the minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. I'm solving this. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. Let me do it here-- actually, I want to do that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily close to the asymptote. It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away, as x gets infinitely large. So in order to figure out which one of these this is, let's just think about what happens as x becomes infinitely large. So as x approaches infinity. So as x approaches infinity, or x approaches negative infinity. So I'll say plus or minus infinity, right? It doesn't matter, because when you take a negative, this gets squared. So this number becomes really huge as you approach positive or negative infinity. And you'll learn more about this when we actually do limits, but I think that's intuitive. That this number becomes huge. This number's just a constant. It just stays the same. So as x approaches positive or negative infinity, as it gets really, really large, y is going to be approximately equal to-- actually, I think that's congruent. I always forget notation. Approximately. This just means not exactly but approximately equal to. When x approaches infinity, it's going to be approximately equal to the plus or minus square root of b squared over a squared x squared. And that is equal to-- now you can take the square root. You couldn't take the square root of this algebraically, but this you can. This is equal to plus or minus b over a x. So that tells us, essentially, what the two asymptotes are. Where the slope of one asymptote will be b over a x. This could give you positive b over a x, and the other one would be minus b over a x. And I'll do this with some example so it makes it a little clearer. But we still know what the asymptotes look like. It's these two lines. Because it's plus b a x is one line, y equals plus b a x. Let's say it's this one. This asymptote right here is y is equal to plus b over a x. I know you can't read that. And then the downward sloping asymptote we could say is y is equal to minus b over a x. So those are two asymptotes. But we still have to figure out whether the hyperbola opens up to the left and right, or does it open up and down? And there, there's two ways to do this. One, you say, well this is an approximation. This is what you approach as x approaches infinity. But we see here that even when x approaches infinity, we're always going to be a little bit smaller than that number. Because we're subtracting a positive number from this. We're subtracting a positive number, and then we're taking the square root of the whole thing. So we're always going to be a little bit lower than the asymptote, especially when we're in the positive quadrant. Right? So to me, that's how I like to do it. I think, we're always-- at least in the positive quadrant; it gets a little more confusing when you go to the other quadrants-- we're always going to be a little bit lower than the asymptote. So we're going to approach from the bottom there. And since you know you're there, you know it's going to be like this and approach this asymptote. And then since it's opening to the right here, it's also going to open to the left. The other way to test it, and maybe this is more intuitive for you, is to figure out, in the original equation could x or y equal to 0? Because when you open to the right and left, notice you never get to x equal to 0. You get to y equal 0, right here and here. But you never get to x equals 0. And actually your teacher might want you to plot these points, and there you just substitute y equals 0. And you can just look at the original equation. Actually, you could even look at this equation right here. Can x ever equal 0? If you look at this equation, if x is equal to 0, this whole term right here would cancel out, and you'd just be left with a minus b squared. Which is, you're taking b squared, and you put a negative sign in front of it. So that's a negative number. And then you're taking a square root of a negative number. So we're not dealing with imaginaries right now. So you can never have x equal to 0. But y could be equal to 0, right? You can set y equal to 0 and then you could solve for it. So in this case, actually let's do that. If y is equal to 0, you get 0 is equal to the square root of b squared over a squared x squared minus b squared. If you square both sides, you get b squared over a squared x squared minus b squared is equal to 0. I know this is messy. So then you get b squared over a squared x squared is equal to b squared. You could divide both sides by b squared, I guess. You get a 1 and a 1. And then you could multiply both sides by a squared. You get x squared is equal to a squared, and then you get x is equal to the plus or minus square root of a. So this point right here is the point a comma 0, and this point right here is the point minus a comma 0. Now let's go back to the other problem. I have a feeling I might be running out of time. So notice that when the x term was positive, our hyperbola opened to the right and the left. And you could probably get from detective reasoning that when the y term is positive, which is the case in this one, we're probably going to open up and down. And let's just prove that to ourselves. So let's solve for y. You get y squared over b squared. We're going to add x squared over a squared to both sides. So you get equals x squared over a squared plus 1. Multiply both sides by b squared. y squared is equal to b squared over a squared x squared plus b squared. You have to distribute the b squared. Now take the square root. I'll switch colors for that. So y is equal to the plus or minus square root of b squared over a squared x squared plus b squared. And once again-- I've run out of space-- we can make that same argument that as x approaches positive or negative infinity, this equation, this b, this little constant term right here isn't going to matter as much. You're just going to take the square root of this term right here. Which essentially b over a x, plus or minus b over a x. And once again, those are the same two asymptotes, which I'll redraw here, that line and that line. But in this case, we're always a little bit larger than the asymptotes. So in the positive quadrant, that tells us we're going to be up here and down there. Another way to think about it, in this case, when the hyperbola is a vertical hyperbola, where it opens up and down, you notice x could be equal to 0, but y could never be equal to 0. And that makes sense, too. Because if you look at our original formula right here, x could be equal to 0. If x was 0, this would cancel out and you could just solve for y. But if y were equal to 0, you'd have minus x squared over a squared is equal to 1, and then you would have, if you solved this, you'd get x squared is equal to minus a squared. And we're not dealing with imaginary numbers, so you can't square something, you can't get a negative number. So once again, this would be impossible. So that's this other clue that tells you it opens up and down. Because in this case y could never equal 0. Anyway, you might be a little confused because I stayed abstract with the b's and the a's. In the next couple of videos I'll do a bunch of problems where we draw a bunch of hyperbolas, ellipses, and circles with actual numbers. See you soon.