 ###### 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 2

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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Whenever we are looking for information about a conic section we really need to have to things in a form that we recognize. So what we are looking at right now is just this big long string of information and we are asked about the center, vertices, co-vertices and focus.

As this question doesn’t tell us anything about what kind of conic section we are dealing with. Both ellipses and hyperbolas have centers, vertices, co-vertices and foci. So what we really need to do is figure out what this equation is. And the way we do that is by completing the square.

The first thing we want to do is isolate our x’s and y’s and our constant term the other side, so we want to bring our x’s together 4x² minus 8x minus y² minus 54 y is equal to 113. So the next thing we want to do is to complete the square and in order to complete the square our coefficients and our square terms have to be 1.

So what that tells us I we have to factor out the 4 from the x terms and the -9 from y terms. The -9 because we have a negative coefficient so it need to come out altogether. So pick out our 4, x² minus 2x, I’m going to leave a little space for our completing the square, minus 9. Y² when we take out negative 9 from 54 this turns into a plus 6y and then this is just equal to 113.

Hopefully by now completing the square is becoming easier for you, so I’m going to skip a couple step what we end up with here is x minus 1 quantities squared. I’ve added 1 inside my parenthesis so what I really added to this side is 4. To keep it balanced we have to add 4 to the other side as well, minus 9 oops I forgot my 4 of my coefficient, minus 9 here we end up with y plus 3 quantity squared. So we ended up adding 9 inside of our parenthesis and then we have -9 times 9 so we really subtracted 81 from this side. So we have to subtract 81 from the other side to keep it balanced. So what we end up with is the statement is equal to, let’s quickly calculate what we ended up with over there, 113 plus 4 minus 81 so that’s 36. This is equal to 36.

So right now I know that I have a hyperbola because we are minusing. But we are used to seeing a hyperbola is being compared to the number 1. So what we have to do is divide by 36 and to get this term 1 and to get in a form that we recognize. So what we end up with then is x minus 1² over 9, minus y plus 3² over 4 is equal to 1.

So we were able to complete the square to get to our standard form that we are used to looking at it. Now we are just asked to find some information about this. So we know that our center is being transformed an its being moved 1 unit forward and 3 units down so our center is then going to be, let’s use a different color in here, 1,-3.

Our vertices are where the graph goes. So looking at this equation I know my x is first so this is telling me my hyperbola is going to be side to side. So my vertices then are going to be moved over and the y value is going to be the same. So I do know that my y coordinate of my vertices is going to be -3 as well and we are just left with finding what the x values are. The x value is basically just going to be the square root of this in either direction from my center.

So our center is 1, so for one vertice we go over to 3 over to 4, and the other one we go back 3. So 1 back to -2. So we found our center and our vertices. Our co-vertices are exactly the same idea but instead of dealing with our transverse axis our main axis we are dealing with the other one. So we are just going to go up and down 2 from our center to find our co-vertices.

So for this let’s go over here I’m running out of space and what we end up with is our co-vertices are, in this case our x values are going to stay the same, so our x values are 1 and then we are going up and down 2 from here, so up 2 will take us to -1 and down 2 will take us to -5.

The last thing we have to do is find our focus and our focus the main way we find that is by dealing with our relationship which for a hyperbola is a² plus b² is equal to c². a² is 9, b² is 4, so what we end up with is 9 plus 4 is equal to c², 13 is equal to c², c is equal to the square root of 13.

So what this is these points are on that transverse axis inside of the curve and so we are going to add that to basically the center going on the x direction. So what we end up with and for our focus it's getting a little bit messy in here so we’ll just draw it in over here. We had our center at 1,-3 we know that it is side to side so then our x value is what’s changing for our focus because it’s going to have to go into the curves. So what we end up with is for our focus 1 plus the square root of 13, -3 and also 1 minus the square root of 13, -3.

So just remembering that on that transverse axis that main axis is where our focus also lie using the relationship a² plus b² is equals c². So by completing the square we were able to take an equation that we didn’t recognize it all, get it down to a formula we did recognize and then just extracting the key information from that equation.