Subscribe! http://www.freemathvideos.com Welcome ladies and gentlemen. So what I'd like to do is show you how to determine the center, the foci, the vertices, and the covertices of an ellipse when it's not written in our standard form up here and up here. You can see that it's all expanded out. So basically to find the center, if you remember-- I have all the information written up there. The center is your h and k, which is clearly displayed in these two formulas, in these two forms, but they're not clearly displayed in these two forms. a represents the distance from your vertices to your center. b represents the distance from your covertices from your center, which are clearly represented here, which are not clearly represented here.
So basically what we need to do to determine our center, foci, our vertices, covertices, and eventually our foci is to transform these equations into these forms. And the way that we can do that is one thing we kind of notice, and something that we've done in quadratics as well as did it hyperbolas and with circles, is we notice that we have binomial squares. And we can create a binomial square by factoring out down a perfect square trinomial. So basically what we need to do is create perfect square trinomials.
Now you can see that my x and my y are separated. So basically what I'm going to do to create a perfect square trinomial is I'm going to separate my x terms and my y terms. You can see here I have a 9x squared and a negative 54x. I'm going to group those together. And I have a 4y squared and a 40y. I'm going to group those together.
Now let me just make sure-- I believe that's equal to 0. And then they're always equal to 1. So I'm going to set that over to the other side.
All right, so first step, 9x squared minus 54x plus 4y squared plus 40y equals negative 37. So I subtracted the 37 to the other side, all right. Now basically what I'm going to do is I'm going to use some parentheses to group the x terms and group the y terms.
Now by grouping these terms, basically what I'm going to do is-- you can see that these are not perfect trinomials. These are two binomials, right? And what I'm going to do is I need to create a perfect square trinomial. And to do that, we're going to do completing the square.
Now I erased that over here, but just remember-- remember for completing the square though, we cannot have a-- let me just kind of write that out-- ax squared plus bx plus c. So that's the standard form of a quadratic. And basically you can see that now inside these parentheses, I have two quadratics. They're just not trinomials, they're binomials. So I'm going to create a perfect square trinomial.
To do that though, we cannot have a be equal to any other number besides 1. So the first thing I'm going to do is I'm going to factor out a 9. I'm going to use blue on this on. All right, so I'm going to factor a 9 in this case. And that's going to leave me with a x squared minus 6x. And then over here I'm going to factor out a 4. And that's going to leave me with a y squared plus 10y equals negative 37. OK, follow me? Good.
So now that we have a is equal to 1, you can see here now we can complete the square. And the process of completing the square is we're going to take the value b of our now simplified trinomial, we're going to take the value b and divide it by 2 and then square it. So it basically looks like this, b divided by 2 squared.
So for our first trinomial, I'm going to take a negative 6, divide it by 2, and square it. Negative 6 divided by 2 is negative 3. Negative 3 squared is equal to 9. And then I do the same thing over here. I do 10 divided by 2, because again, that's my b. It's the coefficient of my linear term. 10 divided by 2 is 5. 5 squared is 25.
Now I'm going to take those values and I'm going to insert them into each parenthesis to create a perfect square trinomial. So now I got a lot of colors going on. So I'm going to have 9 times x squared minus 6x plus 9 plus 4 times y squared plus 10y plus 25, end parentheses, equals negative 37.
So here's where it gets interesting, because what I did is I took these values and I added them to the left side of the equation. But if you remember when you're first solving equations, whatever you add to one side, you have to add to the other side. Whatever you multiply on one side, you have to multiply to the other side.
So that's exactly what we need to do in this case. However, we need to understand that based on the distributed property, I'm not really adding the 9. I'm adding the 9 that's being multiplied by 9. I'm not really adding the 25, I'm adding the 25 that's being multiplied by a 4. So I'm still going to add my 25, but I need to multiply that by 4. I'm still adding the 9, but that 9 is being multiplied by 9.
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