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# Conic section from expanded equation: hyperbola

## Video transcript

Let's do another conic section identification problem. So, I have 4y squared minus 50x is equal to 25x squared plus 16y plus 109. So, the first thing I like to do is to group all of the x and y terms onto one side of the equation and leave all the constants on the other side. So let's do that. So, on the left-hand side I'll put the 4y squared. 4yy squared. And, actually, I'm also going to group all the x and y terms in this step. So the 4y squared. Let's move this 16y onto the left-hand side. So if I subtract 16y from both sides of this equation, I get minus 16y, minus 16y on the left-hand side and of course it will disappear on the right-hand side. And then I want to subtract the 25x squared from both sides of this equation. So I get minus 25x squared minus 50x. That's that right there. And then I'll leave this 109 on the right-hand side. It's equal to 109. And now that we have the x's and the y's on the same side of the equation, we know what type of -- we know the general direction we're going to go in. Because they're on the same side. They have different coefficients. And one is positive and one is negative. So that lets us know that we're dealing with a hyperbola. So let's complete the square and get it into the standard form. So, the easiest way to complete the square is if you have a 1 coefficient on the y squared and the x squared terms. So let's factor out a 4, in this case. So you get 4 times y squared minus 4y. I'm going to add something later, when I complete the square. Minus 25 times x squared, plus, let's see, minus 50 divided by minus 25 is 2. Plus 2x, I'm going to add something later. Is equal to 109. And the things we're going to add, those are what complete the square. Make these things a perfect square. So, if I take this, you have a minus 4 here. I take half of that number. This is just completing the square, I encourage you to watch the video on completing the square where I explain why this works. But I think I have a minus 4. I take half of that, it's minus 2. And then minus 2 squared is plus 4. Now, I can't do one thing to one side of the equation without doing it to the other. And I didn't add a 4 to the left-hand side of the equation. I actually added a 4 times 4, right? Because you have this 4 multiplying it out front. So I added at 16 to the left side of the equation, so I have to also add it to the right-hand side of the equation. Right? This is equivalent to also having a plus 16 here. That might make a little bit clearer, right? When you factor it out, and it becomes a 4. And we would have added a 16 up here as well. Likewise, if we take half of this number here. Half of 2 is 1. 1 squared is 1. We didn't add a 1 to the left-hand side of the equation, we added a 1 times minus 25. So we want to put a minus 25 here. And, likewise, this would have been the same thing as adding a minus 25 up here. And you do a minus 25 over here. And now, what does this become? The y terms become 4 times y minus 2 squared. y minus 2 squared. Might want to review factoring a polynomial, if you found that a little confusing, that step. Minus 25 times x plus 1 squared. That's that, right there. x plus 1 squared, is equal to, let's see, 109 plus 16 is 25 minus 25, it equals 100. We're almost there. So we want a 1 here, so let's divide both sides of this equation by 100. So, you will get y minus 2 squared. 4 divided by 100 is the same thing as 1/25, so this becomes over 25. Minus, let's see, 25/100 is the same thing as 1/4, so this becomes x plus 1 squared over 4 is equal to 1. And there you have it. We have it in standard form and, yes, indeed, we do have a hyperbola. Now, let's graph this hyperbola. So the first thing we know is where the center of this hyperbola is. Is the center of this hyperbola is at the point x is equal to minus one. So it's an x is equal to minus 1. y is equal to 2. And let's figure out the asymptotes of this hyperbola. So if this was -- this is the way I always do it, because I always forget the actual formula. If this was centered at 0 and it looked something like this. y squared over 25 minus x squared over 4 is equal to 1. I do this to figure out what the asymptotes would have been if we were centered at 0. Because it's a lot easier to deal with these equations than to deal with these. So we could solve -- we multiply both sides by 100. We're kind of unwinding what we just did. So if you -- actually, let's multiply both sides by 25. So then you get y squared minus 2 over 4x squared is equal to 25. And then, I'll just go right here. And then if I add 25 over 4x squared to both sides, I get y squared is equal to 25 over 4x squared plus 25. And so, y is equal to the plus or minus square root of 25 over 4x squared plus 25. And like always, the asymptotes, the hyperbola will never equal the asymptotes or intersect the asymptotes, but it's what the graph approaches as x approaches positive and negative infinity. So, as x approaches positive and negative infinity, and you'll learn the concept of limits later on. But I think you get it at this point. Because that's what the idea of an asymptote even is. Is that as x gets really larger, approaching this line -- so as x approaches positive or negative infinity, as we've done in the previous videos, this term starts to matter a lot less. Because this term is huge. So then y is approximately equal to plus or minus the square root of just this term. Now, the square root of just this term is 5/2 x. So those would be our asymptotes if we were centered at 0. But, of course, we're centered at negative 1, 2. So let's graph that. And then we could figure out if it's a upward-opening or downward-opening graph. We're centered at negative 1, 2. So I want to be in the, that's my y axis. This is my x axis. And we're centered at minus 1, 1, 2. That's the center. And this would've been the two lines of the asymptotes if we were centered at 0. But now this tells us the slope of the two asymptotes. So the asymptotes are going to intersect at the center of our hyperbola, so to speak. So these are the slopes of the two asymptotes. And one is positive 5/2. So, positive 5/2 means if we go over 2, so 1, 2, in x, we go up 5. So, 1, 2, 3, 4, 5. So we'll end up right over there. So I can draw that line, I just need two points for a line. So that line would look like that. And the other asymptote is minus 5/2. So for every 2 we go over to the right, we go down 5. So, 1, 2. 1, 2, 3 4, 5. So we end up right about there. And so that line would look like that. Good enough. So those are the two asymptotes, and they go on forever in those directions. And now we can think of it two ways. We could either say, OK, if we look at -- actually, look at this one. If it was centered at 0, could x equals 0? Well, sure x could equal 0. If x is 0, then y squared over 25 equals 1. y squared equals 25. y would be plus or minus 5. So, in this case, this term could be equal to 0. So we could say that x could equal negative 1. If x is equal to negative 1, then y minus 2 squared over 25 will equal -- let's do that. Let's set. If x is equal to negative 1, x is equal to negative 1, then what does this expression become? I don't want to lose it, so I'll write it right there. So then you get y minus 2 squared over 25. This becomes 0 minus 0 is equal to 1. So you get y minus 2 squared over 25 is equal to 1. y minus 2 squared is equal to 25. Just multiplied both sides by 25. y minus 2 is equal to plus or minus, I'm just taking the square root of both sides, 5. So y minus 2 is equal to positive 5 or y minus 2 is equal to minus 5. Add 2 to both sides of this, you get y is equal to 7. Add 2 to both sides of this, you get y is equal to minus 3. So we know that the points minus 1, 7 and minus 1, minus 3, are both on this graph. So, minus 1 is here. 1, 2, 3, 4, 5, 6, 7, minus 1, 7, and minus 1, 1, 2, 3, are both on this graph. So that lets us know, since we're inside here, this tells us that this is kind of a vertical asymptote, and another way to guess it is, if you see that y squared term is positive. Or the other way to think about it is, when you take the positive square root, when you take the positive square root, you're always going to be a little bit above the asymptote. That's the other way to think about this. That we're always going to be a little bit -- and this is the positive square root. The positive square root is the top line. So we're always going to be a little bit above the asymptote. This is the asymptote. But we're always a little bit above it. And obviously as this number gets larger, this starts to matter a lot less, so the graph is going to look something like this. It's going to come down and then go off, and never quite touch the asymptote, but approach it. So it's going to get really close to the asymptote, and then go off, and go off in that direction. Anyway, hope you found that helpful. This was a slightly hairier problem, so it should be instructive.