WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to be discussing the last type of conic section, which is hyperbolas.
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So far, we have covered parabolas, circles, and ellipses.
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As you can see, hyperbolas are a bit different in shape than the other conic sections we have worked with.
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And one thing that makes them unique is that there are two sections referred to as branches; there are two branches in this hyperbola.
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The formal definition is that a **hyperbola** is a set of points in the plane,
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such that the absolute value of the differences of the distances from two fixed points is constant.
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What does that mean? First, let's look at the foci.
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These two fixed points are the **foci**; and here is a focus, f₁, and here is the other, f₂.
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If I take a point on the hyperbola, and I measure the distance to f₁,
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and then the distance to f₂, that is going to give me d₁ and d₂.
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Recall that, with ellipses, we said that the distance from a point on the ellipse--if you measured the distance to one focus,
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and then the other focus, and then added those, that the sum would be a constant.
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Here, we are talking about the difference: the absolute value of this distance, d₁, minus (the difference) d₂, equals a constant.
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That is the formal definition of a hyperbola.
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Again, I could take some other point: I could take a point up here on this other branch.
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And I could find a distance, say, d₃, and then the distance to f₁ could be something--say d₄.
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Again, the absolute value of those differences would be equal to that same constant.
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All right, properties of hyperbolas: A hyperbola, like an ellipse, has two axes of symmetry.
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But these have different names: here you have a transverse axis and a conjugate axis, and they intersect at the center.
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We are looking here at a hyperbola with the center at (0,0).
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One thing to note is that you can also have a hyperbola that is oriented as such.
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But right now, we are looking at this one, with a more horizontal orientation.
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But just to note: this does exist, and we will be covering it.
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All right, first discussing the transverse axis: the transverse axis is going to go right through here--it is going to pass through the center.
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And this is the vertex, and this is the vertex of the other branch, and this is the transverse axis.
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The distance from one vertex to the center along this transverse axis is going to be A.
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Again, a lot of this is going to be similar from when we worked with ellipses; but there are some important differences, as well.
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So, the length of the transverse axis equals 2A; from here to here would be 2A.
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The foci: if you look at foci, say f₁, f₂...let's look at f₂...
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it would be the same over on f₁: if I looked at the distance from one focus to the center, that is going to be C.
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The distance between the foci is therefore 2C; if I measured from here to here, that length is going to be 2C.
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There is a second axis called the conjugate axis; and the two axes intersect here at the center.
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The length of half...if you take half the length of this conjugate axis, it is going to be equal to B.
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The transverse axis lies along here; the conjugate axis, in this case, is actually along the y-axis (and the transverse axis is along the x-axis).
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The length of the conjugate axis is 2B.
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As with the ellipse, there is an equation that relates A, B, and C; but it is a slightly different equation.
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Here, A, B, and C are related by C² = A² + B².
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This relationship will help us to look at the equation for a hyperbola and graph the hyperbola,
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or look at the graph, and then go back and write the equation.
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So again, there are two axes: transverse, which goes from vertex to vertex; and conjugate, which intersects the transverse axis
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at the center of the hyperbola and has a length of 2B (the transverse axis has a length of 2A).
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The distance from focus to focus (between the two foci) is 2C.
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The standard form of the hyperbola is also going to look somewhat familiar, because it is similar to an ellipse, but with a very important difference.
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Here we are talking about a difference instead of a sum.
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So, if you have a hyperbola with a center at (0,0) and a horizontal transverse axis, the equation is x²/A² - y²/B².
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And here, we again have that the center is at the origin, (0,0).
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Although the center certainly does not have to be at the origin, right now we are going to start out
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working with hyperbolas with a center at the origin, just to keep things simple.
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And again, by being given an equation in standard form, you can look at it and get a lot of information about what the hyperbola looks like.
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Therefore, the vertex is going to be at (A,0); the other vertex will be at (-A,0).
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You are going to have a point up here that is going to be (B,0); and this length, B, gives the length of half of the conjugate axis.
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And then, you are going to have another point...actually, that is (0,B), because it is along the y-axis...another point, (0,-B).
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This distance is B, from this point to the center; this distance is A.
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And then here, I have f₁ and f₂; and the distance from one of those to the center is C.
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As I mentioned, you can have a hyperbola that is oriented vertically.
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So, if the transverse axis is vertical, and the center is at (0,0), the standard form is such that y² is associated with the A² term.
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And here, it is positive: so you are taking y²/A² - x²/B² = 1.
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In this case, what you are going to have is a vertex right here at (0,A), the other vertex is here at (0,-A); here is the transverse axis.
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And then, you are going to have the conjugate axis; the length of half of that is going to be B; the length of the entire thing is 2B.
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So, this is going to be some point, (B,0); and B² is given here, so you could easily find B by taking the square root.
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And then, over here is (-B,0).
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So again, there are two different standard forms, depending on if you are working with a hyperbola
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that has a horizontal transverse axis or one that has a vertical transverse axis.
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Something new that we didn't talk about with ellipses is asymptotes.
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Recall that an asymptote is a line that a curve on a graph approaches, but it never actually reaches.
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And asymptotes are very useful when you are trying to graph a hyperbola.
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The equations are given here: let's go ahead and draw these first.
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Now, recall that this vertex is at a point (A,0); this vertex is at (-A,0).
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If I measure the length...this is the transverse axis, and it is horizontal...let's say that it turns out that B is right up here, (0,B).
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And B is going to be the length from this point to the center; 2B will be the length of the conjugate axis.
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Then, I am going to have another point down here, (0,-B).
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What I can do is form a box, a rectangle; and the rectangle is going to have vertices...I am going to go straight up here and across here.
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Therefore, this is going to be given by (-A,B); that is going to be one vertex.
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I can go over here and do the same thing; I am going to go straight up from this vertex, and straight across from this point.
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And that is going to give me the point (A,B).
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I'll do the same thing here: I go down directly and draw a line across here; this point is going to be (-A,-B).
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One final vertex is right here: and this is given by (A,-B).
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OK, now you draw a box using these A and B points; and then you take that rectangle and draw the diagonals.
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If you continue those diagonals out, you will have the two asymptotes for the hyperbola.
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OK, so each of these lines is an asymptote.
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And notice that the hyperbola is going to approach this, but it is not actually going to reach it.
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So, it is going to continue on and approach, but not reach, it; it is going to approach like that.
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All right, now this is one way to just graph out the asymptotes.
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You can also find the equation; and for right here, we are working with a hyperbola with a horizontal transverse axis.
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So, we are going to look at this equation.
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If I was working with a vertical one, I would look at this equation.
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Now, what does this mean? Well, y equals ±(B/A)x.
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What this actually is: this B/A gives the slope of the asymptote.
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Recall that y = mx + b; since the center is at (0,0), the y-intercept is 0; so here, b = 0, so I am going to have y = mx.
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The slope, m, is B/A; B/A for this line is increasing to the right (m = B/A);
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and the slope here equals -B/A, where the line is decreasing as we go towards the right.
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So again, there are two ways to figure out these asymptotes.
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You can just sketch it out by drawing this rectangle with vertices at (-A,B), (A,B), (-A,-B)...that is actually (A,B), positive (A,B)...or (-A,-B).
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Draw that rectangle and extend the diagonals.
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Or you can use the formula, which will give you the slope for these two asymptotes.
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If you just started out knowing the A's and B's and drew these, then you could easily sketch the hyperbola,
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because you know that it is going to approach these asymptotes.
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OK, so we have talked a lot about graphing.
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And just to bring it all together: you are going to begin by writing the equation in standard form.
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And then, for hyperbolas, you are going to graph the two asymptotes, as I just showed.
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So, let's start out with an example: let's make this x²/9 - y²/4 = 1.
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Since I have this in the form x²/A² (this x² term is positive here),
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divided by y²/B² = 1, what I have is a horizontal transverse axis.
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So, this tells me that there is a horizontal transverse axis.
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So, that is how this is just roughly sketched out already, showing the transverse axis along here.
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Since it is in this form, I know that A² = 9; therefore, A = 3.
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This has a center right here at (0,0).
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And this point here is going to be A, which is 3, 0.
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Right here, I am going to have -A, or -3, 0.
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So, my goal is to make that rectangle extend out the diagonals.
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And then, I would be able to graph this correctly.
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OK, B² = 4; therefore, B = 2; so right up here at (I'll put that right there) (0,2)...that is going to be B.
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And then, right down here at (0,-2)...
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Now, all I have to do is extend the line up here and here; and these are going to meet at (2,3).
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Extend a line out here; I am going to have a vertex right here at (-3,2).
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I am going to have another vertex here at (-3,-2), and then finally, one over here at (3,-2).
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Now, this was already sketched on here for me; but assuming it was not there, I would have started out by drawing this box,
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and then, drawing these lines extending out--the asymptotes.
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And what is going to happen is that this hyperbola is actually going to approach, but it is never going to intersect with, the asymptote.
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So again, write the equation in standard form, which might require completing the square.
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I gave it to you in standard form already; use that to figure out this rectangle.
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And you are going to need to know A and B to figure out this rectangle.
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Draw the asymptotes, and draw then the hyperbola approaching (but not reaching) those asymptotes.
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So far, we have been talking about hyperbolas with a center at the origin (0,0).
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However, that is not going to always be the case.
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If the center is at another point, (h,k), that is not (0,0), then standard form looks like this.
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It is very similar to what we saw with the origin of the center, except instead of just x²/A², we now have an h and a k.
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For a horizontal transverse axis, you are going to have (x - h)²/A² - (y - k)²/B².
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For a vertical transverse axis, this term is going to be first; it will be positive.
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And then, you are going to subtract (x - h)²/B².
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But the k stays associated with the y term.
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For example, given this equation, (y - 3)² - (x - 2)²...and we are going to divide that by 16,
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and divide this by 9, and set it all equal to 1: what this is telling me is that the center is at (2,3), because this is h;
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that A² = 16, so A = 4; and that B² = 9, so B = 3.
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From that, I can graph out this hyperbola.
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And this has a vertical transverse axis; something else to be careful of--let's say I had something like this:
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(y + 5)/10, the quantity squared, plus (x + 4), the quantity squared, divided by 12, equals 1.
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The center is actually at (that actually should be a negative right here--this is a difference) (-4,-5).
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And the reason for that is that this is the same as (y - -5)², and then (x - -4)².
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A negative and a negative is a positive.
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So, you need to be careful: even though it is acceptable to write it like this, it is good practice,
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if you are trying to figure out what the center is, to maybe write it out like this,
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so that you have a negative here, so that whatever is in here is already k, or already h.
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You don't have to say, "Oh, I need to make that a negative; I need to change the sign."
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So, that is just something to be careful of.
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Here, I already had negative signs in here; they are completely in standard form--I have h and k here; h and k is (-4,-5).
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All right, to get some practice, we are going to first find the equation of a hyperbola that I am going to give you some information on.
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I will give you that one of the vertices is at (0,2); the other vertex is at (0,-2).
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The other piece of information is that you have a focus at (0,4), and a focus called f₂ at (0,-4).
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So, looking at this, I can see that this is the transverse axis, and then the center is right there.
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So, I have a horizontal transverse axis.
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I can also see that the midpoint right here, the center, is at the origin; so the center equals (0,0).
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So, this is actually vertical--correction--a vertical transverse axis, going up and down: a *vertical* transverse axis.
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Since this is actually a vertical transverse axis with a center at (0,0), I am working with this standard form:
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y²/A² - x²/B² = 1.
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So, the A² term is with the y² term, since this is a vertical transverse axis.
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All right, in order to find the equation, I need to find A².
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This distance, from 0 to the vertex, is 2, because this is at (0,2).
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Therefore, A equals 2; since A = 2, A² = 2², or 4.
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I have A²; I need to find B²; I am not given that.
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But what I am given is an additional piece of information, and that is that there is a focus here and a focus here.
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This allows me to find C: the distance from the center to either focus (let's look at this one)--from the center, 0, down to -4--
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the absolute value of that is 4; therefore, C = 4; the distance is 4.
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C², therefore, equals 4², or 16.
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Recall the relationship: C² = A² + B² for a hyperbola.
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So, I have C², which is 16, equals A², which is 4, plus B².
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16 - 4 is 12; 12 = B²; therefore, B = √12, which is about 3.5.
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If you wanted to draw B, then you could, because that is right here at (3.5,0).
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But what we are just asked to do is write the equation; and we have enough information to do that,
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because I have that y² divided by A²; I determined that A² is 4;
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minus x²/B²; I determined that that is 12; equals 1.
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So, this is the equation for this hyperbola, with a vertical transverse axis and a center at (0,0) in standard form.
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The next example: Find the equation of the hyperbola satisfying vertices at (-5,0) and (5,0) and a conjugate axis that has a length of 12.
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Just sketching this out to get a general idea of what we are looking at--just a rough sketch--vertices are at (-5,0) and (5,0).
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That means that the center is going to be right here at (0,0).
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So, the center is at the origin; since the vertices are here and here, then I have a horizontal transverse axis;
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this is going to go through like this, and then like this.
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So, my second piece of information is that I have a horizontal transverse axis.
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Since I have a horizontal transverse axis, then I am going to have an equation in the form x²/A² - y²/B² = 1.
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The center is at the origin; it has a horizontal transverse axis; this is a standard form that I am working with.
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I need to find A: well, I know that the center is here, and that A is this length; so from this point to the center,
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or from this point (the vertex) to the center, is 5: A = 5.
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Since A = 5, A² = 5²; it equals 25.
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The other information I have is that the conjugate axis has a length of 12.
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So, the length of the conjugate axis, recall, is 2B; here they are telling me that that length is 12.
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Therefore, 12/2 gives me B; B = 6; so, that would be up here and here: (0,6) and then (0,-6).
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This would be the conjugate axis; so this is B = 6.
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Since B equals 6, I want B² that equals 6², which equals 36.
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Now, I can write this equation: I have (this is my final one) x²/A², which is 25,
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minus y²/B², and I determined that that is 36, equals 1.
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So, this is a hyperbola with a center at the origin.
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And A² is 25; B² is 36; and it has a horizontal transverse axis.
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We are asked to graph this equation; and it is not in standard form.
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But when I look at it, I see that I have a y² term and an x² term, and they have opposite signs.
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So, I am working with the difference between a y² term
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and an x² term, which tells me that this is the equation for a hyperbola.
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If they were a sum, this would have been an ellipse, since they have different coefficients.
00:26:28.800 --> 00:26:31.300
But it is a difference, so it is a graph of a hyperbola.
00:26:31.300 --> 00:26:37.200
What I need to do is complete the square to get this in standard form.
00:26:37.200 --> 00:26:57.400
OK, so first I am grouping y terms and x terms: y² + 12y - 6x² + 12x - 36 = 0.
00:26:57.400 --> 00:27:07.200
What I am going to do is move this 36 to the other side and get that out of the way for a moment by adding 36 to both sides.
00:27:07.200 --> 00:27:14.300
The next thing I need to do with completing the square is factor out the leading coefficient, since it is something other than 1.
00:27:14.300 --> 00:27:20.200
So, from the y terms, I will factor out a 2; that is going to leave me with y² + 6y.
00:27:20.200 --> 00:27:29.500
You have to be careful here, because you are factoring out a -6, so I need to make sure that I worry about the signs.
00:27:29.500 --> 00:27:36.100
And that is going to leave behind an x² here; here, it is going to leave behind, actually, -2x.
00:27:36.100 --> 00:27:41.200
So, checking that, -6 times x² is -6x²--I got that back.
00:27:41.200 --> 00:27:48.900
-6 times -2x is + 12x; equals 36.
00:27:48.900 --> 00:28:05.700
Now, to complete the square, I have to add b²/4 in here, which equals...b is 6; 6²/4 is 36/4; that is 9.
00:28:05.700 --> 00:28:08.500
I need to be careful to keep this equation balanced.
00:28:08.500 --> 00:28:17.400
Now, this is really 9 times 2 that I am adding; 9(2) = 18--I need to add that to the right.
00:28:17.400 --> 00:28:34.300
Working with the x terms: b²/4 = 2²/4, which is 4/4; that is 1, so I am going to add 1 here.
00:28:34.300 --> 00:28:46.100
-6 times 1 needs to be added to the other side; so I am going to actually subtract 6 from the right to keep it balanced.
00:28:46.100 --> 00:29:07.400
Now, I am rewriting this as (y + 3)² - 6(x - 1)² =...18 - 6 is 12; 36 + 12 gives me 48.
00:29:07.400 --> 00:29:13.000
The next step, because standard form would have a 1 on this side, is: I need to set all this equal to 1.
00:29:13.000 --> 00:29:29.400
I need to divide both sides of the equation by 48.
00:29:29.400 --> 00:29:40.400
This cancels, so it becomes y + 3²; the 2 is gone; this becomes a 24; minus...6 cancels out,
00:29:40.400 --> 00:29:48.500
and that leaves me with (x - 1)²; 6 goes into 48 eight times; and this is a 1.
00:29:48.500 --> 00:29:52.500
OK, so it is a lot of work just to get this to the point where it is in standard form.
00:29:52.500 --> 00:29:59.600
But once it is in standard form, we can do the graph, because now I know the center; I know A² and B².
00:29:59.600 --> 00:30:03.000
We have this in standard form; so now we are going to go ahead and graph it.
00:30:03.000 --> 00:30:15.300
I will rewrite the standard form that we came up with, (y + 3)²/24 - (x - 1)²/8 = 1.
00:30:15.300 --> 00:30:23.700
Looking at this; since this is positive, I see that I have a vertical transverse axis.
00:30:23.700 --> 00:30:35.000
The other thing to note is this plus here: recall that, if you have (y + 3)², this is the same as (y - -3)².
00:30:35.000 --> 00:30:38.100
And when we look at standard form, we actually have a negative here.
00:30:38.100 --> 00:30:45.800
So, you need to be careful to realize that the center is at (1,-3), not at (1,3).
00:30:45.800 --> 00:30:56.400
Let's make this 2, 4, 6, 8, -2, -4, -6, -8; the center, then, is going to be at (1,-3).
00:30:56.400 --> 00:31:11.200
The next piece of information: A² = 24; therefore, A = √24, which is approximately 4.9.
00:31:11.200 --> 00:31:23.200
B² = 8; therefore, B = √8; therefore, if you figure that out on your calculator, that is approximately 2.8.
00:31:23.200 --> 00:31:32.500
Since I have the center, and I have A and B, I can draw the rectangle that will allow me to extend diagonals out to form the asymptotes.
00:31:32.500 --> 00:31:42.200
The goal is to write this in standard form, find A and B, find the center, make the rectangle, and make the asymptotes;
00:31:42.200 --> 00:31:46.000
and then, you can finally draw both branches of the hyperbola.
00:31:46.000 --> 00:31:51.700
All right, so if the center is here at (1,3), then I am going to have 2 vertices.
00:31:51.700 --> 00:32:06.200
And what is going to happen, since this is a vertical transverse axis, is: one vertex is going to be up here; the other is going to be down here.
00:32:06.200 --> 00:32:13.400
The center is at (1,3); that means I am going to have a vertex at 1, and then it is going to start at the center,
00:32:13.400 --> 00:32:23.900
and then it is going to be 4.9 directly above that center; so that means this is going to be at -3 + 4.9.
00:32:23.900 --> 00:32:30.400
The y-coordinate will be at -3 + 4.9, which equals (1,1.9).
00:32:30.400 --> 00:32:38.600
Therefore, at (1,1.9) (that is right there)--that is where there is going to be one vertex.
00:32:38.600 --> 00:32:42.800
And this is A--this is the length of A.
00:32:42.800 --> 00:32:59.000
The second vertex is going to be at (1,-3); that is the center; and then I am going to go down 4.9--that is the length, again, of A.
00:32:59.000 --> 00:33:19.200
That is -3 - 4.9 (or + -4.9; you can look at it that way) = (1,-7.9), down here.
00:33:19.200 --> 00:33:33.800
OK, vertices are at (1,-7.9) and (1,1.9).
00:33:33.800 --> 00:33:44.200
Now, I need to find where B is--where that endpoint over here is, horizontally--so that I can make this rectangle.
00:33:44.200 --> 00:33:59.900
I know that B equals approximately 2.8; that means that I am going to have a point over here at 1 + 2.8...-3.
00:33:59.900 --> 00:34:13.700
Well, 1 + 2.8 is (3.8,-3); so (3.8,-3) is right there.
00:34:13.700 --> 00:34:27.700
I can reflect across; and I am going to have a point at 1 - 2.8, -3, which is going to give me...1 - 2.8 is (-1.8,-3).
00:34:27.700 --> 00:34:36.100
That is going to be...this is 2...-2 is right here; so that is going to be right about there.
00:34:36.100 --> 00:34:43.200
I now have these points; and recall that I can then extend out to make a box.
00:34:43.200 --> 00:34:49.700
There is going to be a vertex here; I am going to extend across; there is going to be a vertex here.
00:34:49.700 --> 00:34:56.000
Bring this directly down; there is a vertex here, and then another vertex right here.
00:34:56.000 --> 00:35:09.200
Again, I got these points by knowing where the center is, knowing where the vertices of the hyperbola are, and then knowing the length of B.
00:35:09.200 --> 00:35:13.100
This is B; then this length is A.
00:35:13.100 --> 00:35:30.300
Once I have this rectangle, I can go ahead and draw the asymptotes by extending diagonals out.
00:35:30.300 --> 00:35:40.200
Another way to approach this, recall, would have been to use the formula for the slope that we discussed, for the slope of the asymptotes.
00:35:40.200 --> 00:35:43.400
Either method works.
00:35:43.400 --> 00:35:55.700
I know that I am going to have a hyperbola branch up here; the vertex is right here, and it is going to approach, but never reach, this asymptote.
00:35:55.700 --> 00:36:10.000
It is going to do the same thing with the other branch: a vertex is here; it is going to approach, but never reach, the asymptote.
00:36:10.000 --> 00:36:14.700
OK, so this was a difficult problem; we were given an equation in this form.
00:36:14.700 --> 00:36:17.700
We had to do a lot of work just to get it in standard form.
00:36:17.700 --> 00:36:29.500
And then, once we did, we were able to find the center and form this rectangle, draw the asymptotes, and then (at last) graph the hyperbola.
00:36:29.500 --> 00:36:31.900
Example 4: We don't need to do graphing on this one.
00:36:31.900 --> 00:36:40.300
We are just finding the equation of a hyperbola with the center here, (0,0), and a horizontal transverse axis.
00:36:40.300 --> 00:36:46.400
I am going to stop right there and think, "OK, I have a center at (0,0) and a horizontal transverse axis."
00:36:46.400 --> 00:36:55.300
So, the standard form is going to be x²/A² - y²/B² = 1.
00:36:55.300 --> 00:36:59.000
Since the center is at (0,0), I don't have to worry about h and k.
00:36:59.000 --> 00:37:10.700
The horizontal transverse axis has a length of 12; well, the transverse axis length, recall, is equal to 2A.
00:37:10.700 --> 00:37:20.300
I am given that that length is 12; if I take 12/2, that is going to give me A = 6.
00:37:20.300 --> 00:37:37.000
The conjugate axis--recall that the length of the conjugate axis is equal to 2B, which is 6: B = 3.
00:37:37.000 --> 00:37:43.900
Now, I need to find A², which is 6², or 36, to put in here.
00:37:43.900 --> 00:37:47.400
B² is 3², which is 9.
00:37:47.400 --> 00:37:58.700
Now, x²/36 - y²/B² (which is 9) = 1.
00:37:58.700 --> 00:38:10.400
So, this is the equation for a hyperbola with the center at the origin, a horizontal transverse axis, and a conjugate axis with a length of 6.
00:38:10.400 --> 00:38:15.000
That concludes this lesson on hyperbolas; thanks for visiting Educator.com!