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## Precalculus

# Intro to hyperbolas

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.

## Want to join the conversation?

- At7:40, Sal got rid of the -b^2, how is that possible?(13 votes)
- As
`x`

approaches infinity, the`b²`

term becomes less and less significant.

For example, consider`x = 1,000,000`

and`b = 5`

.(23 votes)

- I have actually a very basic question. The variables a and b, do they have any specific meaning on the function or are they just some paramters?(11 votes)
- Yes, they do have a meaning, but it isn't specific to one thing. You may need to know them depending on what you are being taught. If you are learning the foci (plural of focus) of a hyperbola, then you need to know the Pythagorean Theorem:

a^2 + b^2 = c^2

The foci are +-c

Even if you aren't learning the foci, you still need them for the asymptotes.(7 votes)

- Is a parabola half an ellipse? Also can the two "parts" of a hyperbola be put together to form an ellipse?(10 votes)
- They look a little bit similar, don't they? But no, they are three different types of curves.(7 votes)

- At4:19how does it become b squared over a squared x squared?(8 votes)
- Well what'll happen if the eccentricity of the hyperbolic curve is equal to infinity?

Will it be equal to a pair of two parallel straight lines?

copyd this question quz cant see answers!(5 votes)- its a bit late, but an eccentricity of infinity forms a straight line. This is because eccentricity measures who much a curve deviates from perfect circle. So circle has eccentricity of 0 and the line has infinite eccentricity.(6 votes)

- Hang on a minute why are conic sections called conic sections.(1 vote)
- circle equation is related to radius.how to hyperbola equation ?(4 votes)
- 2y=-5x-30

y=-5x/2-15(0 votes)

- at about7:20, won't the second line's slope be -a/b instead of -b/a? i learned that perpendicular lines were negative reciprocals, not just negative opposites.(3 votes)
- the asymptotes are not perpendicular to each other.(6 votes)

- At about10:30Sal states (in purple) that for x²=a² x = +/- sqrt(a). I don't get this? If you square root both sides you get x=a, right?(5 votes)
- This is a mistake in the video. I'm not sure why it has not been corrected. (Thanks for the time-stamp: they really help with a question like this :-). Notice that Sal doesn't use this mistaken result: when he locates his solutions on the x-axis, immediately after making his mistake, he labels them, correctly, as "a" and "-a"(2 votes)

- What is the intuition behind having a negative sign in hyperbola equations X^2/a^2 - Y^2/b^2 = 1. This is the only difference between an ellipse and a hyperbola equation. Can anyone please explain me the reason of the sign intuition changing the shape of a ellipse curve to a hyperbola ?(3 votes)
- As I'm learning/reviewing this myself, I would take this answer with a grain of salt.

The shape of a hyperbola, as opposed to an ellipse, has to do with the behaviors of x and y as they approach infinity. A hyperbola's equation will result in asymptotes reflected across the x and y axis, while the ellipse's equation will not.

In order to understand why, let's have an equation of a hyperbola and an ellipse, respectively: x^2/9 - y^2/4 = 1; x^2/9 + y^2/4 = 1.

When solving for values of y for the hyperbola, we first rearrange its equation to isolate y:

-y^2/4 = 1 - x^2/9

y^2 = 4/9x^2 - 4

y = +/- the square root of 4/9x^2 - 4 (x can be as large or as negative as you want and still output a real solution).

As x approaches +/- infinity, y approaches +/- 2/3x. If the denominators were equal, then y would ~x, but never reach it, resulting in the asymptotes.

Using the same steps for the ellipse equation, y is +/- 2 - 2/3x, or +/- the square root of 4 - 4/9x^2. Here, x cannot be greater than 3 or less than -3, since that would output an imaginary number.(4 votes)

## Video transcript

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.