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### Course: Precalculus > Unit 5

Lesson 6: 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: At2:00Sal 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?(144 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)
- Ottslater,

When the formula for the hyperbola is in the form of

y²/a² - x²/b² = 1

The equation for the asymptotes are

y= (a/b)x and

y=(-a/b)x

https://www.khanacademy.org/cs/ya-xb-1-with-asymototes/6330435208478720

I hope that is of help.(11 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)

- I'm a bit confused, What exactly is he solving for when he gets to3:30? 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)

- 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)
- So basically, this is just a translation to every point being slid over one right and slid over one down.(2 votes)
- Yep, and you can use this as a template for other translations.(1 vote)

- how do you find the asymtotes of a rational function like (X^2+5X-4)/(2X-2), or the reciprocal of that, (those may be unsolvable i just made them up on the spot) but if you had a quadratic divided by a linear or vice versa. Because in my class we end up with one vertical asymtote and one oblique(diagonal) one. ?(2 votes)
- The vertical asymptote occurs when the bottom equals 0 (at x = 1). A slant asymptote can be found by dividing out the top by the bottom (using polynomial division) and ignoring the remainder. This would be y = x/2+3.(1 vote)

## 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.