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## Precalculus

# Vertices & direction of a hyperbola

Sal picks the graph of y²/9-x²/4=1 based on the hyperbola's center, direction, & vertices.

## Want to join the conversation?

- I understand that b² is nine because the distance between the center and the vertex of the hyperbola is three. However, I am confused by the a² amount, four. In an ellipse this would mean two was the length of the minor radius, but there is not minor radius in a hyperbola as far as I know, so what does the four mean?(15 votes)
- Good question. The 2 relates to the change in x on the asymptote. If you look at these graphs you can imagine diagonal lines going through the origin that the graph would get close to but never touch. These are asymptotes. The equations of the lines for the hyperbola on the left are y=3/2x and y=-3/2x. The 3 comes from the a² value being 9, and the 2 comes from the b² value being 4.(16 votes)

- i am confused the equation of a ellipse looks exactly similar to the one shown in this video ..... so how to spot the difference ?(3 votes)
- Similar, but not exactly the same.

The equation of an ellipse is x²/a² + y²/b² = 1

But for a hyperbola we have x²/a² - y²/b² = 1

One has a plus: the ellipse, the other has a minus: the hyperbola(16 votes)

- At1:34in the upper right corner would a^2 be 9 and b^2 be 4?(2 votes)
- Yes, but there is no standard hyperbola form. This problem has a positive y term and a negative x term, so if you assign a to the positive term it is. Some people will always assign a to the x term.(1 vote)

- I guess 'the vertices of hyperbola and the vertex of parabola has the opposite meaning to that of ellipse'. In ellipse the point(s) which is farthest from the center is the vertex . And the point(s) closest to the center is it's co-vertex.

But here , in case of hyperbola and parabola,**why do we call the points closer to the center as vertices not co-vertices**?. That's really confusing to me.

All these are conic sections but why the naming is different?(4 votes)- It is because there is only one point closest as it extends to infinity there is no farthest point. If everything had vertices and co-vertices, every point on a circle would be both. So, they are conic sections only because we can cut them from a cone. But for naming schemes, they are almost independent.(1 vote)

- Hi, I was wondering if the equation y=7/x would be a hyperbola, since all of these examples aren't diagonal.(1 vote)
- Yes! Any function with the equation A/x (where A is a constant) is considered to be a rectangular hyperbola.(2 votes)

- Where do the b^2 and a^2 come from? I understand the way the equation in manipulated but not the way they show in the graph. Please help.(1 vote)
- So based on the equation of a vertical hyperbola, would 9 = a^2 and 4 = b^2?

"2a" is the major axis?

"2b" is the minor axis?

Where is the major and minor axis on a hyperbola graph?(1 vote) - What if option C had also it's center at 3, how can I figure out which one is correct? choosing a point and see if its correct?(1 vote)
- Yeah, choosing a point and checking which is closer to the graph I think is the simplest way of doing it. You may have to use approximations though.(1 vote)

- Why do you set the x term to equal 0?(1 vote)
- Its so you can solve for y which give you the vertices, sorry for the really late response lol(1 vote)

- sal khan's voice is so hot...(0 votes)

## Video transcript

- [ Voiceover] Which
of the following graphs can represent the hyperbola, y squared over nine minus x squared
over four is equal to one. And we have our four choices here. Choices A and C open up to the top and the bottom or up and down. Choices B and D, you could see D here, open to the left and the right. And you can see within the ones that open up to the left, to the right or the up, down ones, they
have different vertices. So I encourage you to pause the video and see if you can figure out, which of the following graphs represent the equation of a hyperbola, or the graphs of this equation right over here. Alright, so there's a bunch
of ways to think about it. One thing you might say, well, what's the center of this hyperbola? And since in our equation, we just have a simple y squared or a simple x squared we know that the center is
going to be at zero, zero. If the center was anywhere else, if the center was at the point h comma k, then this equation
would be y squared minus the y-coordinate of the center. Sorry, not y-squared,
it would be y minus k, so y minus the y-coordinate of
the center squared over nine minus x minus the x-coordinate of the center squared
over four is equal to one. And in this case, well,
this is just a case, where k and h, or h and
k, are both equal to zero. So we just get, you could view this as y minus zero squared
and x minus zero squared. So the center in this case
is going to be at zero, zero. And you see that for all of them. Now the next question you might ask, well, is this going to
be opening up and down, or is it going to be
opening left and right? And the key thing to realize is that you just need to look at whichever term, and when it's written in
standard form like this, when you have y minus k
squared over something squared minus x minus h squared over something squared is equal to one. Or it could be the other way around. The x-term might be positive and then the y-term would be negative, if we're dealing with the hyperbola. So the key is to just look at
whichever term is positive. That will tell you which
direction the hyperbola opens in. Since the y-term here is
the one that is positive, it tells us that this hyperbola is going to open up and down. Now you could just memorize that but I, that's never too satisfying. I always want to know why does that work. And the key thing to
realize is, if the y-term is positive than you can set
the other term equal to zero. And the way you would set the other term equal to zero in this case is by making your x equal to the
x-coordinate of your center and that's zero, so if
x is, if you, x is equal to the x squared of your center and this term becomes
zero, you can actually solve this equation. You can solve y squared
over nine equals one. So if x is equal to zero, so x is equal to the x-coordinate of its center, then this term goes away and you would get y squared over nine is equal to one. Or y squared is equal to nine. Or y is equal to plus or minus three. So you know that the coordinates, the x-coordinate of the
center, plus or minus three. That they're on the
hyperbola, that they're, and so you know that it's going to open upwards and downwards so you go to the center, the
x-coordinate of the center plus three and minus three
are on the hyperbola. Notice over here. Plus three and minus three
are not on this hyperbola. In fact, if plus three and minus three were on this hyperbola, you wouldn't be able to open up to
the right and the left. And so that's why,
whichever term is positive, that is the direction that
you open up or down with, or if, so if the x-term was
positive we would be opening to the left and the right,
for the exact same reason. And you can see, if we did
it the other way around, if we had the y equaling the
y-coordinate of the center, so this term, if the
y-term was zeroed out, you would end up with negative x squared over four is equal to one. Which is the same thing
as x squared over four is equal to negative
one, which is equal to x squared is equal to negative four. I just multiplied both
sides by negative one there. And then I multiplied both sides by four and this has no solution. And so that's why we
know that we're not going to intercept the, we're not
going to intercept the line. We're never going to
have a situation where y is equal to the
y-coordinate of the center. So that's, so y is never
going to be equal to zero. In this case, in cases B and D, y, there are points where y equals zero. So the thing to realize is whichever term is positive, that, and
whatever variable that is, so if it's the y-variable,
that's the direction we're going to open up in. And when I figured out what the actual vertices are, we
saw that the point zero, plus or minus three are on the graph, so A looks like a really good candidate. If we look at the other
choice, it opens up and down. It doesn't have zero, plus
or minus three on the graph. It has zero, plus or minus two. So we can feel pretty good
that, about, choice A.