Like what you saw?
Create FREE Account and:
Your video will begin after this quick intro to Brightstorm.

The Hyperbola - Problem 2

6,111 views
Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level 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're now going to find the equation for our hyperbola given the coordinates of the vertices and the co-vertices. So in the problem behind me what we are told is that we have our vertices at 0 plus or minus 4 and our co-vertices at +/- 3, 0.

The firth thing that always helps me visualize what's going on is to make a graph. Well this is a rough graph to sort of represent the general feel of what's happening. So our vertices are at 0, +/- 4 which tells me that I am up 4 and down 4. My co-vertices are at +/- 3 which tells us we are over 3 and over 3. Vertices refer to where the actual graph goes through, so my vertices are on the y axis which is telling me that my graph is going to be going up and down.

So I know that the graph has to be doing something like this and because we're going up and down that automatically that my y² term has to be first, my y² term has to be positive. So right now just looking at that little piece of information, I know that I'm dealing with y² minus x² is equal to 1 and I still just have to figure out my denominators.

The y² piece if it was an ellipse what I would call the x radius, what it really is in this case is half of the transverse squared, so we're looking for our transverse which in this case is 8, because we're going from +4 to -4, half of that is 4, so 4² is what goes underneath our y. 4² is easy enough to square which is just going to be 16.

X², this was an ellipse would be the minor axis, the shorter axis and in the hyperbola it's just called the conjugate because it's the opposite from the direction we're looking at for a hyperbola. Here we're looking at a conjugate of 6, half of our conjugate is 3, so 3² is what goes underneath our x term, 3² easy enough to do so we end up with 9, so easy enough to find our equation.

The other thing that I want to briefly touch on is not part of this problem that's on the board, but just some things that do come up is finding the focus for this graph. Remember the relationship to find the focus for hyperbola is a² plus b² is equal to c² so really all we're going to have to do is a is the square root of this so a² is 16, b² is 9 so 16 plus 9 is 25, so c would end up being 5.

What that tells us is the distance from the center to the focus, the focus is always going to lie within the curve, so that would tell me that my focus is going to be at 0,5 and 0,-5. Not part of the problem but I was good to touch base on everything that we can possibly do with a question.

Stuck on a Math Problem?

Ask Genie for a step-by-step solution