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Conic section from expanded equation: ellipse

Sal manipulates the equation 9x^2+4y^2+54x-8y+49=0 in order to find that it represents an ellipse. Created by Sal Khan.

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Video transcript

The standard question you often get in your algebra class is they will give you this equation and it'll say identify the conic section and graph it if you can. And the equation they give you won't be in the standard form, because if it was you could just kind of pattern match with what I showed in some of the previous videos and you'd be able to get it. So let's do a question like and let's see if we can figure it out. So what I have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0. And once again, I mean who knows what this is it's just not in the standard form. And actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are. If there's only an x squared term and then there's just a y and not a y squared term, then you're probably dealing with a parabola, and we'll go into that more later. Or if it's the other way around, if it's just an x term and a y squared term, it's probably a parabola. But assuming that we're dealing with a circle, an ellipse, or a hyperbola, there will be an x squared term and a y squared term. If they both kind of have the same number in front of them, that's a pretty good clue that we're going to be dealing with a circle. If they both have different numbers, but they're both positive in front of them, that's a pretty good clue we're probably going to be dealing with an ellipse. If one of them has a negative number in front of them and the other one has a positive number, that tells you that we're probably going to be dealing with a hyperbola. But with that said, I mean that might help you identify things very quickly at this level, but it doesn't help you graph it or get into the standard form. So let's get it in the standard form. And the key to getting it in the standard form is really just completing the square. And I encourage you to re-watch the completing the square video, because that's all we're going to do right here to get it into the standard form. So the first thing I like to do to complete the square, and you're going to have to do it for the x variables and for the y terms, is group the x and y terms. Let's see. The x terms are 9x squared plus 54x. And let's do the y terms in magenta. So then you have plus 4y squared minus 8y and then you have-- let me do this in a different color-- plus 49 is equal to 0. And so the easy thing to do when you complete the square, the thing I like to do is, it's very clear we can factor out a 9 out of both of these numbers, and we can factor out a 4 out of both of those. Let's do that, because that will help us complete the square. So this is the same thing is 9 times x squared plus 9 times 6 is 54, 6x. I'm going to add something else here, but I'll leave it blank for now. Plus 4 times y squared minus 2y I'm probably going to add something here too, so I'll leave it blank for now. Plus 49 is equal to 0. So what are we going to add here? We're going to complete the square. We want to add some number here so that this whole three term expression becomes a perfect square. Likewise, we're going to add some number here, so this three term number expression becomes a perfect square. And of course whatever we add on the side, we're going to have to multiply it by 9, because we're really adding nine times that. And add it on to that side. Whatever we add here, we're going to have to multiply it times 4 and add it on that side. If I put a 1 here, it's really like as if I had a 4 here, because 1 times 4 is 4 and if I had a 1 here it's 1 times 9. So 9 there. Let's do that. When we complete the square, we just take half of this coefficient. This coefficient is 6, we take half of it is 3, we square it, we get a 9. Remember it's an equation, so what you do to one side, you have to do to the other. So if we added a 9 here, we're actually adding 9 times 9 to the left-hand side of the equation, so we have to add 81 to the right-hand side to make the equation still hold. And you could kind of view it if we go back up here. This is the same thing, just to make that clear as if I added plus 81 right here. Of course I would have had to add plus 81 up here. Now let's go to the y terms. You take half of this coefficient is minus 2, half of that is minus 1. You square it, you get plus 1. 1 times 4, so we're really adding 4 to the left-hand side of the equation. And just so you understand what I did here. This is equivalent as if I just added a 4 here, and then I later just factored out this 4. And so what does this become? This expression is 9 times what? This is the square of-- you could factor this, but we did it on purpose-- it's x plus 3 squared and then we have plus 4 times-- What is this right here? That's y minus 1 squared. You might want to review factoring of polynomial or completing the square if you found that step a little daunting. And then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85. All right, so now we have 9 times plus 3 squared plus 4 times y minus 1 squared. And let's subtract 49 from both sides. That is equal to-- let's see if I subtract 50 from 85 I get 35, so if I subtract 49, I get 36. And now we are getting close to the standard form of something, but remember all the standard forms we did except for the circle-- we had a y-- and we know this isn't a circle, because we have these weird coefficients, well not weird but different coefficients in front of these terms. So to get the 1 on the right-hand side let's divide everything by 36. If you divide everything by 36, this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4, and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1. And there you go. We have it in the standard form, and you can see our intuition at the beginning the problem was correct. This is indeed an ellipse, and now we can actually graph it. So first of all, actually good place to start, where's the center of the this ellipse going to be? It's going to be x is equal to negative 3. What x value makes this whole terms 0? So it's going to be x is equal to minus 3, and y is going to be equal to 1. What y value makes this term 0? y is equal to 1. That's our center. So let's graph that, and then we can draw the ellipse. It's going to be in the negative quadrant. This is our x-axis and this is our y-axis. And then the center of our ellipse is at minus 3 and positive 1, so that's the center. And then, what is the radius in the x direction? We just take the square root of this, so it's 2. So in the x direction we go two to the right. We go two to the left. And in the y direction, what do we do? Well we go up three and down three. The square root of this. Let me do that. Remember you have to take the square root of both of those. The vertical axis is actually the major radius or the semi-major axis is 3, because that's the longer one. And then the 2 is the minor radius, because that's the shorter one. And now we're ready to draw this ellipse. I'll draw it in brown. Let me see if I can do this properly. I have a shaky hand. All right, it looks something like that. And there you go. We took this kind of crazy looking thing, and all we did is algebraically manipulate it. We just completed the squares with the x's and the y terms. And then we divided both sides by this number right here and we got it into the standard form. We said oh this is an ellipse. We have both of these terms, they're both positive, we're adding we're not subtracting, they have different coefficients underneath here. So we're ready to go over the ellipse, and we realized that the center was at minus 3,1, and then we just drew the major radius, or the major axis and the minor axis. See you in the next video.