Like what you saw?
Create FREE Account and:
 Watch all FREE content in 21 subjects(388 videos for 23 hours)
 FREE advice on how to get better grades at school from an expert
 Attend and watch FREE live webinar on useful topics
The Hyperbola  Problem 1
FREECarl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
We now want to take a look at graphing a hyperbola. So what we do is approach this very much like we would an ellipse. The first think I look at is I'm looking at y² over 25 minus x² over something. When there's nothing there we know that this is actually just going to be over 1.
So I know that my y radius half of my major axis because the y term is bigger, is going to end up being 5 so that tells me from my center which is just 0,0 I go up 5 ad down 5 one, two, three, four, five two, four, five. So those are going to be my vertices because by looking at this I know that I'm going to have y intercepts, I'm not going to have x intercepts because the y coefficient is first, this tells me my hyperbola rather is going to be going to be going up and down, it's going vertically.
So I have my two points here, we need to figure out the same thing for the half of the minor axis in this case which is going to be the x axis, square root of 1 is 1, so that tells me I'm going out one to either side, I'm going to have these 4 points.
The next thing we wanted to do is draw our rectangle, fundamental box, formed at right angle silly box whatever you want to call it, this is the box that's going to dictate our asymptotes. So you connect where these two points would meet, so we have the point 1, 0 and the point 0, 5 you put those together and we get the point 1, 5, same thing for each of these points over here and hopefully you see that these four points create a rectangle.
We then want to connect the opposing corners of this rectangle. I'll try my best I don't have a meter stick or anything in front of me, so I will just approximate hopefully we'll get it roughly right not that great, pretty bad, but there you go. We have two different diagonals connecting these dots.
So now all we have to do is if we fill this in, we would actually get our rectangle. In general you don't have to actually make the physical box, you just want those diagonals connecting those points. If you want to draw the rectangle go right ahead.
So right no we know what our key point is, our vertices, we also know what our asymptotes are which are these. So I know that the graph is going to be sandwiched between these asymptotes touching this point. So coming down like this and then up like that and on the flip side something like that. So a very rough sketch of this hyperbola.
The things we're asked then what is the transverse. Transverse is the distance between the vertices, the main points that are on the graph, so I went up 5 and down 5 on the origin so the distance for the transverse is just going to be 10. The conjugate is the length of the minor axis if it was an ellipse which is just going to be side to side, so 1 over 1 over, this is just going to be 2.
And the last thing we are asked about is the focus, the foci where are the foci? And for this we need to go to our relationship for a hyperbola which is telling us a² plus b² is equal to c². For this one it doesn't really matter what a and b are because we're adding them so in an ellipse a has to be greater, hyperbola it doesn't really matter because we're adding.
So a was 5 because a² is 25, so we end up with 25 plus 1 is equal to c², so c is equal to the square root of 26. So our foci lie inside these two curves that we drew, so we know that one of them is going to be somewhere up here, the other one is going to be somewhere down here. That c is a distance from the center, so we start at 0,0 we go up square root of 26, we end up at the .0, root 26. We go down we end up at the point 0, negative root 26.
So graphing a hyperbola, taking what you know about an ellipse just adding a couple of little twists, a couple of different terminologies, but it's pretty much the exact same approach.
One last thing I want to talk about. If our xs and this term and this term is switched, so it was actually x² minus y² over 25, out fundamental rectangle actually box would be exactly the same because the 25 is still associated with the y², the 1 is still associated with the x². The only difference would be that instead of going up and down, the graph would have been going side to side.
Focus obviously would have been different, but we would have been able to figure out the exact same box by going through those exact same numbers.
Please enter your name.
Are you sure you want to delete this comment?
Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”
BRIGHTSTORM IS A REVOLUTION !!!”
because of you i ve got a 100/100 in my test thanks”
Get Peer Support on User Forum
Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.
Concept (1)
Sample Problems (5)
Need help with a problem?
Watch expert teachers solve similar problems.

The Hyperbola
Problem 1 7,767 viewsGraph:
y² − x² = 1 25 transverse?Foci?conjugate? 
The Hyperbola
Problem 2 6,527 viewsGive the equation for a hyperbola with vertices at (0,±4) and covertices at (±3,0).

The Hyperbola
Problem 3 1,070 views 
The Hyperbola
Problem 4 893 views 
The Hyperbola
Problem 5 949 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete