Precalculus (2018 edition)
Sal picks the graph of y²/9-x²/4=1 based on the hyperbola's center, direction, & vertices.
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- 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.(18 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(17 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.(3 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)
- Be careful, hyperbolas don't have major and minor axes---they have asymptotes (lines that the graph approaches but never touches). With ellipses, the values of a and b tell you the length of the major and minor radii, but here a and b are used to find the equations of the asymptotes (+/-(b/a)x for a hyperbola centered at the origin, where b is the square root of the number that y is being divided by, and a is the square root of the number that x is being divided by). I hope this helps!(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)
- [ 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.