Vertices & direction of a hyperbola | Precalculus | High School Math | Khan Academy

Which of the following graphs can represent the hyperbola ( \frac{y^2}{9} - \frac{x^2}{4} = 1 )?

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 can see, D here opens to the left and the right. In the ones that open up to the left and right, or up and down, 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.

All right, so there's a bunch of ways to think about it. One thing you might say is, what's the center of this hyperbola? Since in our equation, we just have a simple ( y^2 ) or a simple ( x^2 ), we know that the center is going to be at ( (0, 0) ). If the center was anywhere else, if the center was at the point ( (h, k) ), then this equation would be ( \frac{(y - k)^2}{9} - \frac{(x - h)^2}{4} = 1 ).

In this case, this is just the case where ( k ) and ( h ) are both equal to 0. So we just get, you could view this as ( (y - 0)^2 ) and ( (x - 0)^2 ). So the center in this case is going to be at ( (0, 0) ), and you see that for all of them.

Now, the next question you might ask is, well, is this going to be opening up and down, or is it going to be opening left and right? The key thing to realize is, you just need to look at whichever term, when it's written in standard form like this. When you have ( \frac{(y - k)^2}{something^2} - \frac{(x - h)^2}{something^2} = 1 ) 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 a 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 that's never too satisfying. I always want to know, why does that work?

The key thing to realize is, if the ( y ) term is positive, then you could set the other term equal to zero. The way that 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 0. So, if ( x ) is equal to the ( x ) coordinate of your center, and this term becomes 0, you can actually solve this equation.

You can solve ( \frac{y^2}{9} = 1 ). So if ( x ) is equal to 0, the ( x ) coordinate of its center, then this term goes away, and you would get ( \frac{y^2}{9} = 1 ) or ( y^2 = 9 ) or ( y = \pm 3 ).

So, you know that the coordinates ( (0, \pm 3) ) are on the hyperbola. And so you know that's going to open upwards and downwards. You go to the center; the ( x ) coordinate of the center plus 3 and minus 3 are on the hyperbola. Notice over here ( (0, \pm 3) ) are not on this hyperbola. In fact, if ( \pm 3 ) were on this hyperbola, you wouldn't be able to open up to the right and the left.

That's why whichever term is positive, that is the direction that you open up or down with. If the ( x ) term was positive, we would be opening to the left and the right for the exact same reason. You could see, if we did 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 ( -\frac{x^2}{4} = 1 ), which is the same thing as ( \frac{x^2}{4} = -1 ), which is equal to ( x^2 = -4 ). I just multiplied both sides by -1 there, and then I multiplied both sides by 4. This has no solution, and so that's why we know that 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 what, so ( y ) is never going to be equal to 0 in this case, in cases B and D, why there are points where ( y = 0 ).

So the thing to realize is, whichever term is positive, that and whatever variable that is—so if it's for the ( y ) variable, that's the direction that we're going to open up in. When I figured out what the actual vertices are, we saw that the point ( (0, \pm 3) ) are on the graph.

So A looks like a really good candidate. If we look at the other choice that opens up and down, it doesn't have ( (0, \pm 3) ) on the graph; it has ( (0, \pm 2) ). So we can feel pretty good about choice A.