WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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So far in conic sections, we have discussed parabolas and circles.
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The next type of conic section we are going to cover is the ellipse.
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First of all, what is an ellipse? Well, an **ellipse** is formally defined as the set of points in a plane
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such that the sum of the distances from two fixed points is constant.
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Well, what does that mean? First of all, this is the general shape of an ellipse.
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And these two points here are the foci of the ellipse: we will call them f₁ and f₂.
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And looking back at this definition, it is the set of points in the plane...
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and if you pick any of these points (say right here) and measure the distance from this point to one focus
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(we will call that d₁), and then we measure the distance from that same point to f₂,
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to the other focus, we will call that distance d₂.
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This definition states that, if I add up these two distances, d₁ + d₂, they will equal a constant.
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I could pick any point; I could then pick this point and say, "OK, here is another distance," calling that, say, d₃, and this one d₄.
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And if I added up this distance and this distance, d₃ + d₄, I could get that same constant.
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And I could do that with any point on the ellipse.
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Continuing on with some properties of the ellipse: an ellipse actually has two axes of symmetry.
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One is called the major axis, and the other is the minor axis, and these intersect at the center of the ellipse.
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Here, we are going to look at an ellipse that is centered at the origin (it has a center at (0,0)).
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And again, I have foci f₁ and f₂.
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This is one of the vertices of the ellipse; here is a vertex; this is the second vertex of the ellipse.
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And the major axis runs from one vertex to the other vertex.
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And you can see that it passes through both of the foci; this is the major axis.
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And what we are looking at here is an ellipse that has a horizontal major axis.
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In a few minutes, we will look at ellipses that are oriented the other way--ellipses that are oriented with a vertical major axis.
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So, that does exist; but right now we will just focus on this for the general discussion.
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Here we have the major axis; and intersecting at the center is a second axis called the minor axis.
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Looking more closely at the relationships between the major axis, the minor axis, the foci, and the distances relating them:
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let's call the distance from one vertex to the center A: this distance is A.
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Therefore, the length of the major axis is 2A.
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And this is going to become important later on, when we are working with writing equations for ellipses,
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or taking an equation and then trying to graph the ellipse.
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That is my first distance: this is actually called the semimajor axis.
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The length of the semimajor axis is A; the length of the major axis is 2A.
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Now, looking at the minor axis: from this point to the center, this length is B.
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Therefore, the length of the minor axis is equal to 2B.
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Now, looking at the foci: the distance from one focus to the center is going to be C.
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Therefore, the distance from one focus to another, or the distance between the foci, is equal to 2C.
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There is also an equation relating these A, B, and C; and that is A² = B² + C².
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So, keep this equation in mind: again, it becomes important, because you might be given A and B, but not C;
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or you may be given a drawing, a sketch of the ellipse, and then asked to write an equation based on it.
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And sometimes you need to find this third component in order to write that equation.
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And you can do that by knowing that A² = B² + C².
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Again, A is the length of the semimajor axis; B is the distance from here to the center of the minor axis
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(2B is the length of the minor axis); and C is the distance between one focus and the center (2C gives the distance between the two foci).
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Those are important relationships to understand when working with the ellipse.
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Standard form: looking at what the standard form of the equation of an ellipse with a center at (0,0) and a horizontal major axis is:
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it is x²/A² + y²/B² = 1.
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This is the standard form; and again, we already discussed what A is and B is.
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And if you figure out those from the graph, and are given those, then you can go ahead and write your equation.
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Or, given the equation, you can graph the ellipse.
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Let's take a look at an example: A²/9 + y²/4 = 1.
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So again, this is just the equation for an ellipse with a horizontal major axis, so it is sketched out here that way.
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We can make this more precise by saying, "OK, A² = 9; therefore, A equals √9, or 3."
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That means that the distance from here to here is going to be 3.
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And since this is centered at the origin, this will actually be at the point (3,0).
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So, let's write that as a coordinate pair, (0,3).
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OK, that means that this length of A is 3; over here, the other vertex is going to be at (0,-3).
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So, I have one vertex at (0,3) and the other at (0,-3); and I have f₁ here and f₂ here.
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And the length of the major axis is actually 6: it is going to be equal to 2A.
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2A equals 6, and that is the length of the major axis.
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B² = 4: since B² = 4, B = √4, which equals 2.
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Therefore...actually, this needs to be written the other way; this is actually (3,0), and (-3,0).
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Now, up here, we have (0,2) and (0,-2).
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And 2, then, is the length; that is B--that is the length of half of the minor axis.
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2B = 4, and this is the length of the minor axis.
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You can get a lot of information just by looking at the equation of the ellipse in standard form.
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If I needed to, I could figure out the distance between the foci, and I could figure out what C is.
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And remember, C is this length; because recall that A² = B² + C².
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And I know that A² is 9; I know that B² is 4; so I am looking for C².
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9 - 4 gives me 5, so C² = 5; therefore, C = √5, and that is approximately equal to 2.2.
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So, this distance here is roughly 2.2; and the distance between these foci would be about 4.4.
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So, just by having this equation, I could graph the ellipse, and I could find this third component that was missing.
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OK, so that was standard form for an ellipse that has a horizontal major axis.
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For an ellipse that has a vertical major axis, you are going to see that the A² is associated with the y² term.
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In the horizontal major axis, we had x²/A².
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Now, if you were given an equation, how would you know what you are dealing with--
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if you were dealing with a vertical major axis or a horizontal major axis?
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Well, A² is the larger number; so let's say I was given something like y²/16 + x²/9 = 1.
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When I look at this, I see that the larger number is associated with y²; so that tells me that I have a vertical major axis.
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If it had been x² associated with 16, then I would have said, "OK, that is a horizontal major axis."
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Again, I can graph this ellipse by having this equation written in standard form.
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I know that A² equals 16; therefore, A equals the square root of 16; it equals 4.
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This time, I am going to go along here for the major axis; and that makes sense,
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because I have a focus here and another focus here, and the major axis passes through the two foci.
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OK, so (0,4) is going to give me one vertex; (0,-4) will give me the other vertex.
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And remember that 2A = 8, and that is going to be the length of the major axis.
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And again, right now, we are limiting our discussion to ellipses with a center at (0,0).
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Later on, we will expand the discussion to talk about the graphs of ellipses with centers in other areas of the coordinate plane.
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OK, so now I have B² = 9; therefore, √9 is going to equal B; this tells me that B equals 3.
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So, since B equals 3, the length of half of the minor axis is going to be 3.
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So, right here at (3,0) is going to be one point, and I am going to have the other point right here at (-3,0).
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And the length of the minor axis...2B = 6, and that is the length of the minor axis.
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Again, I can use the relationship A² = B² + C²
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to figure out what C is, and to figure out where the foci are located.
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I know that A² is 16; I know that B² is 9; and I am trying to figure out C².
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So, I take 16 - 9; that gives me 7 = C²; therefore, C = √7.
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And the square root of 7 is approximately equal to 2.6, so this is going to be up here at about (0,2.6).
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And f₂ is going to be down here at about (0,-2.6).
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So again, using standard form, you can graph the ellipse, and you can find where the foci are, based on the values of A² and B².
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We talked a little bit about graphing ellipses; but sometimes you are not given the equation in standard form.
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If the equation is not in standard form, you have to put it that way.
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And working with other conic sections, we have learned that you can put an equation into standard form for a conic section by completing the square.
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You can also use symmetry, just as we did when graphing parabolas or circles.
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So, let's say you are given an equation like this: 3x² + 4y² - 18x - 16y = -19; and you are asked to graph it.
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You are going to put it in standard form; but it is nice, first of all, to know what kind of shape you are working with.
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And you can tell that just by looking at this, even though it is not in standard form yet.
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And the reason is: I see that I have an x² term and a y² term
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on the same side of the equation, with the same sign, with different coefficients; that tells me that I am working with an ellipse.
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And we are going to go into more detail in the lecture on conic sections.
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We are going to review how to tell apart the equations for various conic sections.
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But just briefly now: recall that a parabola would have either an x² term or a y² term, not both.
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With a circle, the coefficients of x² and y² are the same; that is for a circle.
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For an ellipse, the coefficients of the x² and y² terms are different.
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Now, you see that these have the same sign; so with an ellipse,
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it is important to note that the x² and y² terms have the same sign.
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If they don't have the same sign, that is actually a different shape.
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They have the same sign; but this coefficient (the x² coefficient) is 3, and the y² coefficient is 4.
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That tells me I have an ellipse, not a circle.
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My next step is to complete the square and write this in standard form,
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so that, if I wanted to graph it, I would have all of the information that I need.
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The first thing to do is to group the x variable terms and the y variable terms.
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This gives me x² - 18x + 4y² - 16y = -19.
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Now, when I am looking at this, I remember that, to complete the square, I want to end up with a leading coefficient of 1.
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I am going to factor out this 3 to get 3(x² - 6x); and I am going to need to add
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something else over here to complete the square--a third term.
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Here, factor out the 4 to give me (y² - 4y) = -19.
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Recall that, to complete the square, you are going to add b²/4 to each set of terms.
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So, for this x group, we have b²/4 = -(6)²/4 = 36/4 = 9.
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So, I am going to add a 9 here; and very importantly, to the right side, I am going to add 9 times 3, which is 27.
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Add that to the right, because if I don't, the equation won't be balanced anymore.
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Now, the y variable terms: I need to add something here to complete the square.
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And I will work over here for this: b²/4 equals (-4)²/4 = 16/4 = 4.
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So, I am going to add 4 here; but to the right, what I need to add is 4 times 4; so I need to add 16.
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Now that I have this, my next step is to write this in standard form: 3...and this is x - 3, the quantity squared;
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if I squared this, I would get this back; plus 4y - 2, the quantity squared, equals...
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-19 plus 16 gives me, let's see, -3; 27 minus 3 gives me 24 on the right.
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Well, recall that for standard form, I want a 1 over here on the right; I want this to equal 1.
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So, this gives me...I need to divide both sides by 24; this is going to give me 3(x - 3)²/24 + 4(y - 2)²/24 = 24/24.
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This finally leaves me with (x - 3)²...this 3 will cancel, and I am going to get 8 in the denominator;
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this 4 will cancel, and I will get 6 in the denominator; equals 1.
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Now, I have the bigger term associated with x; that tells me that I have a horizontal major axis.
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So, the major axis is horizontal; and let me go ahead and write that up here.
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x²/a² + y²/b² = 1.
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OK, now you can see that this looks slightly different; and we have these extra terms here.
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And what that tells us is something that we are going to discuss in a moment.
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And what we are going to discuss is situations where the center of the ellipse is not at (0,0).
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But we have the same information that we would have with the center being at (0,0).
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And that is that I have A², which equals 8; and I know I have a horizontal major axis; and I have B² = 6.
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Looking, now, at equations for ellipses with centers at (h,k), at somewhere other than 0:
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if you look at the situation where we have a horizontal major axis, versus a vertical major axis,
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you can again see that it is very similar to when the center is at (0,0).
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The x is associated with the A² term when the major axis is horizontal.
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The y is associated with the A² term when the major axis is vertical.
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The only difference is that we now have these terms telling us where the center is.
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And if the center was going to be 0, all that would happen is that you would have x - 0 and y - 0,
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which then gives you back the equation we worked with before, x²/A² + y²/B²,
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or y²/A² + x²/B², all equal to 1.
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So, let's talk about an example for this: (x - 4)² + (y + 6)² = 1.
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This is all over 100, and this is all divided by 25.
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What this tells me is that the center is at (h,k); so this is h (that is 4).
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You need to be careful here, because what you have is a positive; but the standard form is a negative.
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And it is perfectly fine to write this like this, but you need to keep in mind that this is really saying...
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if you think about it, y + 6 is the same as y - -6.
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So, when I look at it this way, I realize that, if it is in this form, k is actually going to be -6.
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And if you are not careful about that, you can end up putting your center in the wrong place.
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Let's let this be 2, 4, 6, -2, -4, -6; the center is at (4,-6), right here.
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A² = 100; my larger term is my A²term, and I have a horizontal major axis.
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Therefore, A = √100, which is 10.
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So, since A equals 10 and the length of the major axis is horizontal, then I am going to need to go 10 over from 4.
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It is going to be all the way out here at 14.
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So, this is where one vertex would be--this is going to be at 4 + 10 gives me (14,-6).
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That is one vertex: -2, 4, 6, 8, 10, 12, 14...so -14 is going to be about here.
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And I am going to have the other vertex here at (-14,-6).
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I have a minor axis: B² = 25; therefore B = 5.
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So, what that tells me is that I come here at -6; I add 5 to that; and that is going to bring me right here at -1; that is going to be right about here.
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And then, -8, -10, -12...-(6 + 5) is going to be down here at -11; so this ellipse is going to roughly look like this.
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OK, and standard form allowed me to determine that the center is at (4,-6);
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that I have a vertex; if I add 10, that is going to give me this vertex right over here
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(let's draw this more at the vertex); that I have another vertex here; that the major axis is going to pass through here;
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and then, I am going to have a minor axis passing through here.
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All right, in the first example, we are asked to find the equation of the ellipse that is shown.
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And I am going to go ahead and label some of these points.
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This is going to be f₁; and let's say we are given f₁ as being at (0,3).
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This is going to be 1, 2, 3: each mark is going to stand for one.
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Down here, I have f₂; that should actually be down here a little bit lower; so that is f₂ at (0,-3).
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Let's say we are also given a vertex at (0,5), and the other vertex at (0,-5).
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All right, so I am asked to find the equation of the ellipse shown.
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And the one thing I see is that there is a vertical major axis.
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Since there is a vertical major axis, it is going to be in the general form (y - k)²/A² + (x - h)²/B² = 1.
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Now, I notice that the center is at (0,0); so that tells me that h and k are (0,0).
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So, this is actually going to become the simpler form, y/A² + x/B² = 1.
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The next piece of information I have is the length of A.
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So, this line is going to give me A; and I know that, since the center is here at (0,0), the length of this is 5; therefore, A = 5.
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Since A equals 5, A² is going to equal 5², or 25.
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I don't know B; this is going to be B, but I don't know what it is.
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However, I do know C; if I look here, this is C--from -3 to the center is 3, so C = 3.
00:26:22.700 --> 00:26:26.900
Therefore, C² = 3², or 9.
00:26:26.900 --> 00:26:30.800
The other piece of information I have is the equation we have been working with,
00:26:30.800 --> 00:26:41.900
that states that A² = B² + C².
00:26:41.900 --> 00:26:46.600
Therefore, I can find B², which I need in order to write my equation.
00:26:46.600 --> 00:26:59.700
So, I know that A² is 25; I know that C² is 9; and I am looking for B².
00:26:59.700 --> 00:27:07.900
25 - 9 is 16; therefore, B² = 16, and B = 4.
00:27:07.900 --> 00:27:13.300
Now, I can go ahead and write my equation; I have all the information I need, so let's put it all together right here.
00:27:13.300 --> 00:27:26.800
y, divided by A²...I know that A² is 25...(that is actually y² right here--
00:27:26.800 --> 00:27:35.300
let's not forget the squares)...plus x² divided by B²...B² is 16...equals 1.
00:27:35.300 --> 00:27:40.300
So again, this had a center at (0,0); so remember that this is the general form of the equation.
00:27:40.300 --> 00:27:42.800
When you have a center at (0,0), it becomes this form.
00:27:42.800 --> 00:27:50.700
I had a vertical major axis, so I am using this form, where the A² is associated with the y² term.
00:27:50.700 --> 00:27:58.100
And by having A² and C², I was able to determine what B² is.
00:27:58.100 --> 00:28:10.100
Example 2: let's find the equation of the ellipse satisfying a major axis 10 units long and parallel to the y-axis; minor axis 4 units long; center at (4,2).
00:28:10.100 --> 00:28:19.300
Let's start with the center--the center is at (4,2)--the center is right here at (4,2).
00:28:19.300 --> 00:28:22.900
And it says that the major axis is parallel to the y-axis.
00:28:22.900 --> 00:28:31.000
What that tells me is that I have a vertical major axis.
00:28:31.000 --> 00:28:39.500
Since this is a vertical major axis and I know the center (I know (h,k)), I am going to be working with the general form
00:28:39.500 --> 00:28:48.600
(y - k)²/A² + (x - h)²/B² = 1.
00:28:48.600 --> 00:28:51.800
I have h; I have k; I need to find A² and B².
00:28:51.800 --> 00:28:58.000
The other piece of information I have is that the major axis is 10 units long; and I know that it is vertical.
00:28:58.000 --> 00:29:11.400
Since it is 10 units long, that means that the major axis length is equal to 2A, and I know that that length is 10;
00:29:11.400 --> 00:29:16.900
therefore, I just take 10 divided by 2, and I get that A = 5.
00:29:16.900 --> 00:29:35.400
So, I am starting here; and if I take 5 + 2, I am going to get that I am going to have the vertex up here at 7.
00:29:35.400 --> 00:29:52.900
5 + 2 is going to give me 7, right there; so I go to (4,7); that is where this is going to be.
00:29:52.900 --> 00:30:03.500
And I got that by keeping the x where it was, and then adding 2 where the center was and adding 5 to that, which is the length of A.
00:30:03.500 --> 00:30:10.300
Then, I am going to go down; again, it is going to be at 4.
00:30:10.300 --> 00:30:15.600
But then, if I take 2 and subtract 5 from it, I am going to get -3.
00:30:15.600 --> 00:30:22.900
So, right here at (4,-3) is the other vertex.
00:30:22.900 --> 00:30:29.200
Now, I was not asked to graph this; but I am graphing it so that I have an understanding of what each of these means,
00:30:29.200 --> 00:30:32.000
so that I can go ahead and write the equation.
00:30:32.000 --> 00:30:44.300
What I have now is...I know what A is, which means I can figure out A²; and I know the center.
00:30:44.300 --> 00:30:57.400
The last thing I know is that the minor axis length equals 2B: I am given that that is 4 units long; therefore, B = 2.
00:30:57.400 --> 00:31:07.600
So, if I started here at 4, and I added 2 to that, I am going to end up with (6,2).
00:31:07.600 --> 00:31:21.400
And I am going to end up with...if I start at 4 and I subtract 2, I am going to end up with (2,2).
00:31:21.400 --> 00:31:27.100
And this is B; this is A.
00:31:27.100 --> 00:31:31.100
I have a minor axis that has a length of 2B, which tells me that B = 2.
00:31:31.100 --> 00:31:36.900
From this information, I can go ahead and write this equation.
00:31:36.900 --> 00:31:43.900
I know A equals 5, so A² equals 5², which is 25.
00:31:43.900 --> 00:31:50.900
B = 2; B², therefore, equals 2², or 4.
00:31:50.900 --> 00:32:00.200
So, I have everything I need to write this: (y - k)...well, k is 2; the quantity squared, divided by A²;
00:32:00.200 --> 00:32:15.800
A² is 25; plus (x - h)...h is 4...the quantity squared, divided by B², which is 4, equals 1.
00:32:15.800 --> 00:32:20.700
So, this standard form describes the ellipse with the major and minor axis here.
00:32:20.700 --> 00:32:33.300
And you could finish that out by just connecting these points and drawing the ellipse if you wanted to finish graphing it.
00:32:33.300 --> 00:32:42.400
Find the equation of the ellipse satisfying endpoints of the major axis at (11,3) and (-7,3) and foci at (7,3) and (-3,3).
00:32:42.400 --> 00:32:58.800
All right, endpoints of the major axis: let's do 2, 4, 6, 8, 10, 12, and -2, 4, -6...OK.
00:32:58.800 --> 00:33:11.600
The endpoint is at (11,3): 11 is right here, and then we will have 3 be right here.
00:33:11.600 --> 00:33:19.300
The other endpoint is at -7, which is going to be here, 3.
00:33:19.300 --> 00:33:29.900
And that tells me the major axis: since the major axis is horizontal, we are going to be working with the general form
00:33:29.900 --> 00:33:37.900
(x - h)²/A² + (y - k)²/B² = 1.
00:33:37.900 --> 00:33:40.200
So, that is my first piece of information.
00:33:40.200 --> 00:33:43.800
I can also figure out the length of the major axis.
00:33:43.800 --> 00:34:05.000
So, since the major axis goes from -7 to 11, if I take -7 minus 11, I am going to get -18.
00:34:05.000 --> 00:34:13.600
And a length is going to be an absolute value, so I am just going to take 18; the length is going to be the absolute value of this difference.
00:34:13.600 --> 00:34:21.600
All right, I know that the major axis length is 18; the other thing I know is that the major axis length equals 2A, as we have discussed.
00:34:21.600 --> 00:34:36.500
So, 2A = 18; therefore, A = 9; so I know that the distance from the center to this endpoint is 9.
00:34:36.500 --> 00:34:47.600
Therefore, I could just say, "OK, 9 minus 11 gives me 2; and I know that the y-coordinate will be 3; so that is (2,3)."
00:34:47.600 --> 00:34:53.500
Another way to solve this, without using all this graphing, would have simply been to find the midpoint.
00:34:53.500 --> 00:35:00.500
I am going to go ahead and do that, as well, because the midpoint of the major axis is the center of the ellipse.
00:35:00.500 --> 00:35:05.000
Let's try that, as well--the center, using the midpoint formula.
00:35:05.000 --> 00:35:12.100
Recall the midpoint formula: we are going to take x₁ + x₂ (that is 11 + -7);
00:35:12.100 --> 00:35:17.100
and we are going to take the average of that (we will divide it by 2); that is going to give me the x-coordinate.
00:35:17.100 --> 00:35:26.200
For the y-coordinate, I am going to take 3 + 3, and I am going to divide that by 2.
00:35:26.200 --> 00:35:38.200
And this will give you a center at...11 - 7 gives you 4; 4 divided by 2 is 2.
00:35:38.200 --> 00:35:44.200
3 + 3 is 6, divided by 2 gives you 3; and that is exactly what I came up with using the graphing method.
00:35:44.200 --> 00:35:54.500
So, just to show you: you could have figured this out algebraically (where the center is); or you could have figured it out using graphing.
00:35:54.500 --> 00:36:02.400
This gives me (h,k), so I have (h,k), and I have A, which is 9, so I can get A².
00:36:02.400 --> 00:36:07.600
The next thing I need to do is figure out B, and they don't give me B; but what they do give me are the foci.
00:36:07.600 --> 00:36:27.400
There are foci at (7,3) (7 is here--focus at (7,3)--we will call this one f₁ at (7,3)); and this is the center right here.
00:36:27.400 --> 00:36:41.000
There is another focus at (-3,3): f₂ is going to be at (-3,3).
00:36:41.000 --> 00:36:50.600
Recall that the distance from one focus to the other is 2C; the distance from one focus to the center is C.
00:36:50.600 --> 00:37:01.100
Let's just work with this; this is C; 7 - 2 is 5; therefore, C = 5.
00:37:01.100 --> 00:37:08.800
So, I have A = 9; I have C = 5 (again, that is the distance from the center to a focus);
00:37:08.800 --> 00:37:13.600
or I could have figured out the distance from one focus to another--that is 2C--and then divided by 2;
00:37:13.600 --> 00:37:20.400
So, I have A, and I have C, and I know that A² = B² + C².
00:37:20.400 --> 00:37:33.900
So, this gives me 9² = B² + 5²: 9² is 81, equals B² + 25.
00:37:33.900 --> 00:37:39.300
If I take 81 - 25, I am going to get 56 = B².
00:37:39.300 --> 00:37:45.300
And I don't even need to take the square root of that, because to put this into standard form, I actually need B².
00:37:45.300 --> 00:37:53.000
So, I get (x - h); remember, the center is (h,k), so that is 2; the quantity squared, divided by A²...
00:37:53.000 --> 00:38:04.200
recall that A² is 9², so it is 81; plus (y - k)²...k is 3, the quantity squared;
00:38:04.200 --> 00:38:09.300
divided by B²...I determined that B² is 56; all of this equal to 1.
00:38:09.300 --> 00:38:15.200
Just by knowing the endpoints of the major axis and the location of the foci, I could figure out A²
00:38:15.200 --> 00:38:27.400
and B², as well as the center, and then write this equation for the ellipse in standard form.
00:38:27.400 --> 00:38:31.100
Finally, we are asked to graph an ellipse that is not given to us in standard form.
00:38:31.100 --> 00:38:38.500
We have some extra work to do: we are going to actually have to complete the square in order to even graph this.
00:38:38.500 --> 00:38:49.500
So, let's go ahead and start by grouping the x terms together and y terms together.
00:38:49.500 --> 00:38:55.100
Also note that, since this has an x² term and a y² term on the same side of the equation,
00:38:55.100 --> 00:39:00.800
with the same sign (they are both positive), but different coefficients, I know I have an ellipse.
00:39:00.800 --> 00:39:07.400
It is not a circle, because for a circle, these coefficients would be the same.
00:39:07.400 --> 00:39:27.700
Grouping the terms together gives me 14x² - 56x + 6y² - 24y = -38.
00:39:27.700 --> 00:39:31.700
Looking at this, I see that I have a common factor of 2.
00:39:31.700 --> 00:39:38.000
If I divide both sides by 2, I can simplify this equation; so the numbers I will work with will be smaller.
00:39:38.000 --> 00:39:51.600
So, let's divide both sides by 2 to get 7x² - 28x + 3y² - 12y = -19.
00:39:51.600 --> 00:39:57.100
The next thing to do is to factor out the leading coefficient, since it is not 1.
00:39:57.100 --> 00:40:02.500
I am going to factor out a 7 to get x² - 4x, plus...
00:40:02.500 --> 00:40:11.000
over here, I am going to factor out a 3; that gives me y² - 4y, all equals -19.
00:40:11.000 --> 00:40:15.600
I need to complete the square, so I need to get b²/4.
00:40:15.600 --> 00:40:22.700
In this case, that is going to give me 4²/4 = 16/4 = 4.
00:40:22.700 --> 00:40:40.500
So, over here, I am going to add a 4; very important--on the right, I have -19, and I am adding to the left 7 times 4; that is 28.
00:40:40.500 --> 00:40:46.000
So, I need to add 28 to the right.
00:40:46.000 --> 00:41:01.700
All right, over here, for the y terms: b²/4: again, we have a b term that is 4, so I am going to end up with the same thing, 4.
00:41:01.700 --> 00:41:10.000
Now, here I am actually adding 4 times 3 (is 12), so I need to add that to the right, as well, to keep the equation balanced.
00:41:10.000 --> 00:41:13.900
This is the easiest step to mess up on: you are focused here on completing the square,
00:41:13.900 --> 00:41:18.900
and then you sometimes don't remember that you have to add the same thing to both sides.
00:41:18.900 --> 00:41:33.200
Now, working on writing this in standard form: this is going to give me (x - 2)² + 3(y - 2)² = -19 + 12...
00:41:33.200 --> 00:41:44.100
that is going to leave me with 28; -19 + 12 is going to be -7; 28 minus 7 equals 21.
00:41:44.100 --> 00:41:50.000
To get this into standard form, I need to have a 1 on the right, so I am going to divide both sides by 21.
00:41:50.000 --> 00:42:07.900
This is going to give me 7(x - 2)²/21 + 3(y - 2)²/21 = 21/21.
00:42:07.900 --> 00:42:27.400
This cancels to (x - 2)²/3; this becomes (y - 2)²/7; and this just becomes 1.
00:42:27.400 --> 00:42:30.400
Now, I have standard form; I can do some graphing.
00:42:30.400 --> 00:42:57.500
I have a center at, let's see, (2,2); that is right here; the center is at (2,2), so that is h and k.
00:42:57.500 --> 00:43:07.600
And I notice, actually, that the larger term is under the y; it is associated with the y.
00:43:07.600 --> 00:43:20.600
That tells me that this has a vertical major axis; and therefore, I am going to keep that in mind--that it is going to be oriented this way.
00:43:20.600 --> 00:43:24.200
The ellipse is going to end up like this, instead of like this.
00:43:24.200 --> 00:43:40.000
I have my center at (2,2); therefore, A² = 7; A = √7.
00:43:40.000 --> 00:43:46.600
The square root of 7 is about 2.6, so it gets messy, as always, when you are working with radicals.
00:43:46.600 --> 00:44:03.400
But it is about 2.6; so what I have to do is say, "OK, my vertex up here is going to be at x = 2, and then y is going to equal 2 + 2.6, which is 4.6."
00:44:03.400 --> 00:44:15.300
So, (2,4.6): and again, I got that by saying the length of A (the length from the vertex to the center) is 2.6.
00:44:15.300 --> 00:44:18.500
So, 2 + 2.6 gives me 4.6.
00:44:18.500 --> 00:44:33.900
Over here, I am going to take 2, and I am going to subtract 2.6 from it; so x is still going to be 2; now y is going to be about here, which is (2, -.6).
00:44:33.900 --> 00:44:45.000
Again, 2 is here; the length of A is 2.6, so it is going to be 2 - 2.6; this gives me my major axis.
00:44:45.000 --> 00:44:49.800
I know, from vertex to vertex, where this ellipse is going to land.
00:44:49.800 --> 00:45:00.500
Now, the minor axis, B: I know that B² = 3; therefore, B = √3, which is approximately equal to 1.7.
00:45:00.500 --> 00:45:06.600
This is not to do the same thing, but going along the x direction, the horizontal direction.
00:45:06.600 --> 00:45:15.600
So again, this is A; OK, now to get B, I am going to have 2, and I am going to add 1.7 to it.
00:45:15.600 --> 00:45:21.800
That is going to give me 3.7; it is going to land about here.
00:45:21.800 --> 00:45:38.800
In this direction, I am going to subtract; I am going to say that I have 2 - 1.7, so that is going to land here, at about .3.
00:45:38.800 --> 00:45:42.600
And in the y direction (in the vertical direction) it is still going to be up at 2.
00:45:42.600 --> 00:45:53.200
So again, to get this, I said 2 + 1.7; that brought me to (3.7,2)--that is this point.
00:45:53.200 --> 00:46:01.300
This point is at 2 minus 1.7, so that is (.3,2).
00:46:01.300 --> 00:46:22.500
So, this gives me the general shape of this ellipse, like this; so I can get a good sketch, based on this equation.
00:46:22.500 --> 00:46:27.300
I took this equation, and I recognized that it was an ellipse.
00:46:27.300 --> 00:46:34.800
I completed the square to get it in standard form, and saw that it had a vertical major axis and that it had a center at (2,2).
00:46:34.800 --> 00:46:45.900
I then found A to determine where the vertices would be, and then B to determine the width of the ellipse here; and then I could get a good sketch.
00:46:45.900 --> 00:46:51.000
That concludes this session of Educator.com on ellipses; thanks for visiting!