 ###### 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!

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# Transformations of a Hyperbola - Problem 1

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!

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Finding the equation for a hyperbola when we are given information about the vertices and foci. So what we want to do is in this problem we are given some information and the first thing I want to do is take this information and put it on a graph so I can sort of see what is going on.

So what we are told is that our vertices are at 0,1 and 4,1 so I’m going to plot those points 0,1 and 4,1 and our foci are located at negative 1,1 and 5,1. And what we what to do is from these points find the equation for this hyperbola.

So the first thing I see is that my foci are horizontal from my vertices and also these are referred to as vertices telling us that they are the major points on this graph. So we know that the graph is then going to be facing horizontal it’s going to be facing that way because our foci have to be inside this curve.

So what we need to do is first find this center point. And the center point is actually pretty easy to find because it’s just going to be the middle of our vertices. Take our vertices find the midpoint there’s our center. Midpoint is just the average of the x values, average of the y values so our x value has to be 2 and our y value has to be 1.

So our midpoint is our center which is going to be at the point 2,1. So from there I can already get a pretty good chunk of my equation written. I know that I’m looking at a horizontal hyperbola so therefore my x term goes first. And we do x minus our x coordinate to the center 2 a hyperbola is minus then we do y minus the y coordinate of the center, all these are squared obviously and then this is going to be 2,1, let’s erase a little bit of our graph.

So the only thing we have left to find is our denominators. The x denominator is actually not that bad to find either because we are given the distance between the vertices we are actually given the transverse axis. The distance between these two points is 4 half if that is 2, 2² is what the half of the transverse is what goes into our fraction so 2², 4 is our denominator on x’s.

We still have yet to find our y denominator and for that we just need to go to our relationship between our distance between the vertices co-vertice and the foci from the center, the relationship a² plus b² is equal to c², where a b and c are the distance of those three things from the center. So we know that the distance from the center to our vertices is 2, so that’s just going to be 4 and because we are adding we can interchange our co-vertices and vertices depth distances.

B² is the distance from the center to our co-vertices, we don’t know what that is so we have to leave it as b² and c² is the distance from the center to our foci. Our center is at 2,1 our foci is at 5,1 so that distance is going to be 3² or 9. So what we end up with is 9 minus 4, b² is equal to 5. So the distance from our center to our co-vertice is b square root of x² to 5 but what we put into our equation is just the distance squared so all we have to put in for this is 5.

So by making a picture sorting out our information and then using the relationship a² plus b² equals c² you are able to find the equation for this hyperbola.