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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition)>Unit 2

Lesson 10: Hyperbolas not centered at the origin

# Equation of a hyperbola not centered at the origin

Sal analyzes the hyperbola whose equation is (x-1)^2/16-(y+1)^2/4=1, and graphs it. Created by Sal Khan.

## Want to join the conversation?

• One thing that is a bit unclear to me in these types of equations: At Sal has (x-1)^2 /16 - (y+1)^2/4 = 1 so the center is (1,-1). But if you plug both those coordinates back into the equation you get 0 = 1 which makes no sense. What detail am I missing here?
(31 votes)
• Bill,

That's an excellent observation, and you'll find the same thing happens in the equations of the circle and the ellipse, too. So let's use the equation of a circle first, and I think we'll talk about what happens with it… and I think that will help make the hyperbola make more sense too.

So here is an equation of a circle:
(x-h)^2+(y-k)^2=r^2

In this equation the center of the circle is at (h,k), and the circle has a radius equal to r. So let's throw in some numbers really quick:

(x-3)^2+(y-4)^2=25

Ok, in this circle, the center is at (3,4) and the radius is 5, because r^2 is 25 and 5 is the square root of 25. Now, notice what happens if you plug in 3 for x and 4 for y? You get:
(3-3)^2 + (4-4)^2 =25
or: (0)^2 +(0)^2 =25
or: 0=25

Now, clearly 0 doesn't equal 25, and it's the same problem that you noticed on the hyperbola. So what is going on?

Well, remember that the point (3,4) is the CENTER of the circle, and the equation gives us the formula for the circle itself. The center of the circle isn't actually ON the circle, though, which is why when we plug in the coordinates for the center of the circle the equation doesn't evaluate to a true statement. The equation ONLY evaluates to a true statement if we plug in points that are ON the circle. If we plug in any points that aren't on the circle, the equation doesn't evaluate to a true statement.

The same thing happens in the equation for a hyperbola. The center of the hyperbola isn't actually on either of the two curves that make up the hyperbola. So if we plug in the numbers for the center of the hyperbola, the equation doesn't evaluate to a true statement. That doesn't mean that we don't have the correct numbers for the center… it just means that the center of the hyperbola isn't actually on the curves that make up the hyperbola just like the center of a circle isn't on the curve that draws the circle.

Does that help this make sense?
(142 votes)
• How do you know where the foci are on the graph?
(19 votes)
• There is an equation to find the foci of an ellipse. Using f^2=p^2-q^2, with p as the length of the major radius and a as the length of the minor radius, we can find the distance from the center along the major radius that the foci are. I hope this answers your question. :D
(1 vote)
• Can somebody say what the equation to find the asymptotes is? He never really says, he does this in a really weird manner that for me at least is really hard to understand, i'd rather it be shown more simply
(9 votes)
• Why does that make a difference when x approaches infinity we forget about the +or- 4 for us now that x is basically in the area of + or - 10 for arguments sake +or - 4 should make a HUGE difference?
(6 votes)
• When X = 10, yes, the -4 under the square root sign does make a big difference. However, for argument's sake, let's say that X = 1,000 in the equation from the video y = +or- sqrt(x^2 /4 -4) Then: sqrt (1,000,000/4 - 4) which can simplify to +or- sqrt (249,996) or +or- 499.996 If you take away the -4 from under the square root, you get +or- sqrt (250,000) or +or- 500 which gives a difference of .004. A very small difference, and that's just at X = 1,000. Imagine how small the difference is at X = 1,000,000 or X = 100,000,000,000,000...
(10 votes)
• Is it possible for a hyperbola to have no x and y intercepts whatsoever?
(2 votes)
• Yes, it is possible for a hyperbola to have neither an x nor a y intercept. Specifically, hyperbola of the form y = k/x where k is constant that is a non-zero real number. In this form, there is no y intercept because the function is undefined at x=0. There is no y interept because k/x cannot equal 0 for any real value of x.
(4 votes)
• isnt the asymptote of the real equation slightly different than the one he wrote?
(2 votes)
• I'm a bit confused, What exactly is he solving for when he gets to ? And what exactly did he do by getting there?
(2 votes)
• I am trying to understand why the hyperbolas formula is equal to 1? Why not for example equal to 0!
(1 vote)
• Well, the standard formula for the hyperbola is an equation, so if it is a number not equal to 0 then you can just divide by that number on both sides to simplify the equation to the point where it does equal 1.
And when the formula is equal to 0, you actually get the asymptotes of the hyperbola! The hyperbola equation equal to 0 can be shown as
(x^2)/(a^2)-(y^2)/(b^2)=0
where any transformations are applied to the asymptotes as well. Let's try simplifying it!
Add to both sides and get:
(x^2)/(a^2)=(y^2)/(b^2)
Take the square root:
x/a= plus or minus y/b
multiply by b (or -b) on both sides:
y= plus or minus bx/a
which can be written as
f(x)=plus or minus (b/a)x
which is actually the equation(s) for the asymptotes!
(2 votes)
• Didn't Sal forget to shift the asymptote?
(2 votes)
• What happens to the hyperbola as the numbers underneath the x and y get larger, does it just make the middle of the hyperbola further away from the place where the asymptotes meet?
(2 votes)

## Video transcript

Let's see if we can tackle a slightly more difficult hyperbola graphing problem. Let's add the hyperbola. Make this up on the fly x minus 1 squared over 16 minus y plus 1 squared over 4 is equal to 1. So the first thing to recognize is that this is a hyperbola and we'll in a few videos, do a bunch of problems where the first point is just to identify what type of conic section we have and then the second step is actually graph the conic section. I already told you that we're going to be doing a hyperbola problem, so you know it's a hyperbola. But the way to recognize that is that you have this minus of the y squared term and then we actually have it shifted. The classic or the standard non-shifted form of a hyperbola or a hyperbola centered at 0 would look something like this. Especially if it has the same asymptotes just shifted, but centered at 0 it would look like this: x squared over 16 minus y squared over 4 is equal to 1. And the difference between this hyperbola and this hyperbola the center of this hyperbola is at the point x is equal to 1 y is equal to minus 1. And the way to think about it is x equals 1 makes this whole term 0, and so that's why it's the center. And y equal to minus 1 makes this whole term 0. And on here, of course, the center is the origin. Center is 0, 0. So the easy way to graph this is to really graph this one, but you shift it so you use the center being 1 minus 1 instead of the center being 0, 0. So let's do that. So let's figure out the slope of the two asymptotes here and then we can shift those two slopes so that it's appropriate for this hyperbola right here. So if we go with this one, let's just solve for y. That's what I always like to do whenever I'm graphing a hyperbola. So we get minus y squared over 4. Subtracting x squared over 16 from both sides minus x squared over 16 plus 1. I'm working on this hyperbola right here, not this one, and then I'm going to just shift it later. And then let's say multiply both sides by minus 4 and you get y squared is equal to-- see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared over 4 minus 4. And to figure out the asymptotes you just have to think about well what happens as x approaches positive or negative infinity. As x gets really positive or x gets really negative. And we've done this a bunch of times already. I think this is important. This is more important than just memorizing the formula, because it gives you an intuition of where those equations for the lines of the asymptote actually come from. Because these are what this graph or this equation or this function approaches as x approaches positive or negative infinity. As x approaches positive or negative infinity, what is y approximately equal to, in this case? Well once again, this term is going to dominate. This is just a 4 right here. You could imagine when x is like a trillion or a negative trillion, this is going to be huge number and this is going to be just like you know you almost view it like the ground off area. You take the square root of that and so this is going to dominate. So as you approach positive or negative infinity, y is going to be approximately equal to the square root, the positive and negative square root, of x squared over 4. So y would be approximately equal to positive or negative x over 2, or 1/2x. Let's do that. Let's draw our asymptotes. And remember, these are the asymptotes for this situation. But now of course, we're centered at 1 negative 1. So I'm going to draw two lines with these slopes, with positive 1/2 and negative 1/2 slopes, but they're going to be centered at this point. I just got rid of the shift just so I could figure out the asymptotes but of course this is the real thing that we're trying to graph, so let me do that. This is my y-axis this is my x-axis and the center of this is at 1 negative 1. So x is equal to 1 y is equal to minus 1. And then the slopes of the asymptotes were positive and negative 1/2. So let's do the positive 1/2. So that means for every 2 you run over, so if you go positive in the positive x direction 2, you move up 1. So you go to the right 2 and up 1. So that's the first one. Let me draw that asymptote. Looks something like that, and then we draw it from this point to that point. Got to have a steady hand. And then the other asymptote is going to have a minus 1/2 slope. Remember this is our center 1 minus 1, so if I go down 1 and over. So when I go over 2, I go down 1, so that will be right there, Let me draw that asymptote. And then just to continue it in the other direction I want to make the lines overlap. It's going to look something like that. So we've drawn our asymptotes for this function, and now we have to figure out if it's going to be a vertical hyperbola or a horizontal hyperbola. And the easy way to think about it is to try and make-- and we can do it two ways. I mean if you just look at this equation right here. When you're taking the positive square root, we're always going to be slightly below the asymptote. The asymptote is this thing, but we're always going to be slightly below it. So that tells us that were always going to be slightly below the asymptote on the positive square root, and we're always going to be slightly above the asymptote on the negative square root. Because it's going to be little less, and it's negative. But I'll let you think about that. My intuition is that it's going to be there and there. It's more than intuition. I know that we're going to be a little bit less than the negative square root, but I'll do it the other way. I'll do it the way I did in the last video. So the other way to think about it is what happens when this term is 0? For this term to be 0, x has to be equal to 1. And does that ever happen? Can x be equal to 1? If x is equal to 1 here this term is 0. And then you have a situation where-- and then you have a minus y squared over 4 would have to equal 1, or this would have to be a negative number. So x could not be equal to 1. So y could be equal to negative 1. Let's try that out. If y is equal to negative 1, this term right here disappears. So when y is equal to negative 1, you're just left with-- x minus 1 squared over 16 is equal to 1. I just canceled out this term, because I'm saying what happens when y is equal to negative 1. You multiply both sides by 16. Let me do it over here. These get messy. x minus 1 squared is equal to 16. Take the square root of both sides. x minus 1 is equal to positive or negative 4. And so if x is equal to positive 4, if you add 1 to that x would be equal to 5. And then if x minus 1 would be minus 4 and you add 1 to that you will have x is equal to 3. So our 2 points or our 2 points closest to our center are the points 5 comma negative 1 and 3 comma negative 1. Let's plot those 2. So 5, 1 2 3 4 5, negative 1 and 3, negative 1. Is that right? No, minus 3, because x minus 1 could be minus 4. That's what happens when you skip steps. If you have the minus 4 situation, then x is equal to minus 3. You go 1 2 3 minus 3, minus 1. So those are both points on this hyperbola. And then our intuition was correct, or it was what I said. That-- the positive square root is always going to be slightly below the asymptote, so we get our curve. It's going to look something like this. It's going to get closer and closer, and then here it's going to get closer and closer in that direction. It keeps getting closer and closer to that asymptote. And here, it's going to keep getting closer and closer to the asymptote on that side and then on that side. And of course these asymptotes keep going on forever and forever. If you want you could try out some other points just to confirm. You could plot that point there, or that point there just to confirm that that's the case. The hard part really is just to identify the asymptotes and just to figure out do we sit on the left and the right, or do we sit on the top and the bottom. And then you're done. You can graph your hyperbola. See you in the next video.