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### Hyperbolas

- Understand the concepts of vertices, transverse axis, and conjugate axis.
- Understand the role of the asymptotes in graphing a hyperbola. Know their equations.
- Understand the fundamental equation c
^{2}= a^{2}+ b^{2}. - Use symmetry to help you graph a hyperbola.
- Understand the standard formula for the equation of a hyperbola.
- Know how to put an equation in standard form by completing the square.

### Hyperbolas

^{2}− y

^{2}− 40x − 16y + 20 = 0

- Group the x's with the x's and the y's with the y's, move the constant term to the right of the equation
- (4x
^{2}− 40x) + ( − y^{2}− 16y) = − 20 - Factor out a 4 from the x's and negative from the y's
- 4(x
^{2}− 10x) − (y^{2}+ 16y) = − 20 - Complete the square by adding [(b
^{2})/4] - 4(x
^{2}− 10x + [(b^{2})/4]) − (y^{2}+ 16y + [(b^{2})/4]) = − 20 + 4( [(b^{2})/4] ) − ( [(b^{2})/4] ) - 4(x
^{2}− 10x + [(( − 10)^{2})/4]) − (y^{2}+ 16y + [(16^{2})/4]) = − 20 + 4( [(( − 10)^{2})/4] ) − ( [(16^{2})/4] ) - 4(x
^{2}− 10x + [100/4]) − (y^{2}+ 16y + [256/4]) = − 20 + 4( [100/4] ) − ( [256/4] ) - 4(x
^{2}− 10x + 25) − (y^{2}+ 16y + 64) = − 20 + 4( 25 ) − ( 64 ) - 4(x − 5)
^{2}− (y + 8)^{2}= 16 - Divide left and right side of equation by 16. Simplify
- [(4(x − 5)
^{2})/16] − [((y + 8)^{2})/16] = [16/16]

^{2})/4] − [((y + 8)

^{2})/16] = 1

^{2}− y

^{2}− 40x − 16y + 20 = 0

- Write in Standard Form equation of the hyperbola
- Notice how this problem is the same as problem # 1 Proceed with the the solution [((x − 5)
^{2})/4] − [((y + 8)^{2})/16] = 1 - Find the center of the hyperbola and graph it.
- Center = (h,k) = (5, − 8)
- Find the lenght of a and b to write the rectangle in order to draw the asymptotes
- a
^{2}= 4 - a = √4 = 2
- b
^{2}= 16 - b = √{16} = 4
- Starting from the center, move 2 units to the right, label it A, then starting from the center move 2 units to the left, label it B
- Starting from the center, move 4 units up, label it C; starting from the center, move 4 units down, label it D.
- Draw a rectangle that passes through the points A, C, B, D.
- Draw your asymptotes
- Draw your asymptotes, these must pass throgh the corners of the rectangle formed by points A, B, C, D.
- Draw the hyperbola
- Points B and A are your vertex. Sketch the hyperbola. The hyperbola opens to the left and to the right

^{2}− 25y

^{2}+ 108x + 200y − 301 = 0

- Group the x's with the x's and the y's with the y's, move the constant term to the right of the equation
- (9x
^{2}+ 108x) + ( − 25y^{2}+ 200y) = 301 - Factor out a 9 from the x's and − 25 from the y's
- 9(x
^{2}+ 12x) − 25(y^{2}− 8y) = 301 - Complete the square by adding [(b
^{2})/4] - 9(x
^{2}+ 12x + [(b^{2})/4]) − 25(y^{2}− 8y + [(b^{2})/4]) = 301 + 9( [(b^{2})/4] ) − 25( [(b^{2})/4] ) - 9(x
^{2}+ 12x + [(12^{2})/4]) − 25(y^{2}− 8y + [( − 8^{2})/4]) = 301 + 9( [(12^{2})/4] ) − 25( [( − 8^{2})/4] ) - 9(x
^{2}+ 12x + [144/4]) − 25(y^{2}− 8y + [64/4]) = 301 + 9( [144/4] ) − 25( [64/4] ) - 9(x
^{2}+ 12x + 36) − 25(y^{2}− 8y + 16) = 301 + 9( 36 ) − 25( 16 ) - 9(x + 6)
^{2}− 25(y − 4)^{2}= 225 - Divide left and right side of equation by 225. Simplify
- [(9(x + 6)
^{2})/225] − [(25(y − 4)^{2})/225] = [225/255]

^{2})/25] − [((y − 4)

^{2})/9] = 1

^{2}− 25y

^{2}+ 108x + 200y − 301 = 0

- Write in Standard Form equation of the hyperbola
- Notice how this problem is the same as problem # 3 Proceed with the the solution [((x + 6)
^{2})/25] − [((y − 4)^{2})/9] = 1 - Find the center of the hyperbola and graph it.
- Center = (h,k) = ( − 6,4)
- Find the lenght of a and b to write the rectangle in order to draw the asymptotes
- a
^{2}= 25 - a = √{25} = 5
- b
^{2}= 9 - b = √9 = 3
- Starting from the center, move 5 units to the right, label it A, then starting from the center move 5 units to the left, label it B.
- These are your vertices.
- Starting from the center, move 3 units up, label it C; starting from the center, move 3 units down, label it D.
- Draw a rectangle that passes through the points A, C, B, D.
- Draw your asymptotes
- Draw your asymptotes, these must pass throgh the corners of the rectangle formed by points A, B, C, D.
- Draw the hyperbola
- Points B and A are your vertex. Sketch the hyperbola. The hyperbola opens to the left and to the right

Vertices at (11,0) and ( − 11,0)

Conjugate Axis has lenght − 16

- Identify type of hyperbola
- Looking at the information provided, you can see that the hyperbola will be in the format [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - which means it will open to the left and to the right
- Find a and b
- Since a is the length from the center (0,0) to one of the vertices, a = 11
- Since the lenght of the conjugate axis equals 2b
- 2b = 16
- b = 8
- Write the formula given a and b
- [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - [(x
^{2})/(11^{2})] − [(y^{2})/(8^{2})] = 1

^{2})/121] − [(y

^{2})/64] = 1

Vertices at (6,0) and ( − 6,0)

Conjugate Axis has lenght of 18

- Identify type of hyperbola
- Looking at the information provided, you can see that the hyperbola will be in the format [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - which means it will open to the left and to the right
- Find a and b
- Since a is the length from the center (0,0) to one of the vertices, a = 6
- Since the lenght of the conjugate axis equals 2b
- 2b = 18
- b = 9
- Write the formula given a and b
- [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - [(x
^{2})/(6^{2})] − [(y^{2})/(9^{2})] = 1

^{2})/36] − [(y

^{2})/81] = 1

Vertices at (14,0) and ( − 14,0)

Conjugate Axis has lenght of 22

- Identify type of hyperbola
- Looking at the information provided, you can see that the hyperbola will be in the format [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - which means it will open to the left and to the right
- Find a and b
- Since a is the length from the center (0,0) to one of the vertices, a = 14
- Since the lenght of the conjugate axis equals 2b
- 2b = 22
- b = 11
- Write the formula given a and b
- [(x
^{2})/(a^{2})] − [(y^{2})/(b^{2})] = 1 - [(x
^{2})/(14^{2})] − [(y^{2})/(11^{2})] = 1

^{2})/196] − [(y

^{2})/121] = 1

Center(2,4)

Vertical Transverse axis of length 16

Conjugate axis of length 26

- Given that the transverse axis is vertical, the equation of the hyperbola will be in the format [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - Find a and b
- 2a = transverse axis
- 2a = 16
- a = 8
- 2b = conjugate axis
- 2b = 26
- b = 13
- Write equation with the ceenter and a and b
- [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - [((y − 4)
^{2})/(8^{2})] − [((x − 2)^{2})/(13^{2})] = 1

^{2})/64] − [((x − 2)

^{2})/169] = 1

Center(10,6)

Vertical Transverse axis of length 20

Conjugate axis of length 8

- Given that the transverse axis is vertical, the equation of the hyperbola will be in the format [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - Find a and b
- 2a = transverse axis
- 2a = 20
- a = 10
- 2b = conjugate axis
- 2b = 8
- b = 4
- Write equation with the ceenter and a and b
- [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - [((y − 6)
^{2})/(10^{2})] − [((x − 10)^{2})/(4^{2})] = 1

^{2})/100] − [((x − 10)

^{2})/16] = 1

Center( − 7,10)

Vertical Transverse axis of length 2

Conjugate axis of length 6

- Given that the transverse axis is vertical, the equation of the hyperbola will be in the format [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - Find a and b
- 2a = transverse axis
- 2a = 2
- a = 1
- 2b = conjugate axis
- 2b = 6
- b = 3
- Write equation with the ceenter and a and b
- [((y − k)
^{2})/(a^{2})] − [((x − h)^{2})/(b^{2})] = 1 - [((y − 10)
^{2})/(1^{2})] − [((x + 7)^{2})/(3^{2})] = 1

^{2})/1] − [((x + 7)

^{2})/9] = 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Hyperbolas

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- What are Hyperbolas?
- Properties
- Transverse Axis and Conjugate Axis
- Vertices
- Length of Transverse Axis
- Distance Between Foci
- Length of Conjugate Axis
- Standard Form
- Vertical Transverse Axis
- Asymptotes
- Graphing Hyperbolas
- Equation with Center at (h, k)
- Example 1: Equation of Hyperbola
- Example 2: Equation of Hyperbola
- Example 3: Graph Hyperbola
- Example 4: Equation of Hyperbola

- Intro 0:00
- What are Hyperbolas? 0:12
- Two Branches
- Foci
- Properties 2:00
- Transverse Axis and Conjugate Axis
- Vertices
- Length of Transverse Axis
- Distance Between Foci
- Length of Conjugate Axis
- Standard Form 5:45
- Vertex Location
- Known Points
- Vertical Transverse Axis 7:26
- Vertex Location
- Asymptotes 8:36
- Vertex Location
- Rectangle
- Diagonals
- Graphing Hyperbolas 12:58
- Example: Hyperbola
- Equation with Center at (h, k) 16:32
- Example: Center at (h, k)
- Example 1: Equation of Hyperbola 19:20
- Example 2: Equation of Hyperbola 22:48
- Example 3: Graph Hyperbola 26:05
- Example 4: Equation of Hyperbola 36:29

### Algebra 2

### Transcription: Hyperbolas

*Welcome to Educator.com.*0000

*Today, we are going to be discussing the last type of conic section, which is hyperbolas.*0002

*So far, we have covered parabolas, circles, and ellipses.*0006

*As you can see, hyperbolas are a bit different in shape than the other conic sections we have worked with.*0012

*And one thing that makes them unique is that there are two sections referred to as branches; there are two branches in this hyperbola.*0018

*The formal definition is that a hyperbola is a set of points in the plane,*0025

*such that the absolute value of the differences of the distances from two fixed points is constant.*0029

*What does that mean? First, let's look at the foci.*0037

*These two fixed points are the foci; and here is a focus, f_{1}, and here is the other, f_{2}.*0041

*If I take a point on the hyperbola, and I measure the distance to f _{1},*0048

*and then the distance to f _{2}, that is going to give me d_{1} and d_{2}.*0056

*Recall that, with ellipses, we said that the distance from a point on the ellipse--if you measured the distance to one focus,*0067

*and then the other focus, and then added those, that the sum would be a constant.*0074

*Here, we are talking about the difference: the absolute value of this distance, d _{1}, minus (the difference) d_{2}, equals a constant.*0078

*That is the formal definition of a hyperbola.*0092

*Again, I could take some other point: I could take a point up here on this other branch.*0094

*And I could find a distance, say, d _{3}, and then the distance to f_{1} could be something--say d_{4}.*0100

*Again, the absolute value of those differences would be equal to that same constant.*0107

*All right, properties of hyperbolas: A hyperbola, like an ellipse, has two axes of symmetry.*0117

*But these have different names: here you have a transverse axis and a conjugate axis, and they intersect at the center.*0124

*We are looking here at a hyperbola with the center at (0,0).*0133

*One thing to note is that you can also have a hyperbola that is oriented as such.*0139

*But right now, we are looking at this one, with a more horizontal orientation.*0147

*But just to note: this does exist, and we will be covering it.*0153

*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.*0156

*And this is the vertex, and this is the vertex of the other branch, and this is the transverse axis.*0166

*The distance from one vertex to the center along this transverse axis is going to be A.*0179

*Again, a lot of this is going to be similar from when we worked with ellipses; but there are some important differences, as well.*0186

*So, the length of the transverse axis equals 2A; from here to here would be 2A.*0195

*The foci: if you look at foci, say f _{1}, f_{2}...let's look at f_{2}...*0212

*it would be the same over on f _{1}: if I looked at the distance from one focus to the center, that is going to be C.*0218

*The distance between the foci is therefore 2C; if I measured from here to here, that length is going to be 2C.*0229

*There is a second axis called the conjugate axis; and the two axes intersect here at the center.*0246

*The length of half...if you take half the length of this conjugate axis, it is going to be equal to B.*0260

*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).*0273

*The length of the conjugate axis is 2B.*0288

*As with the ellipse, there is an equation that relates A, B, and C; but it is a slightly different equation.*0297

*Here, A, B, and C are related by C ^{2} = A^{2} + B^{2}.*0303

*This relationship will help us to look at the equation for a hyperbola and graph the hyperbola,*0311

*or look at the graph, and then go back and write the equation.*0317

*So again, there are two axes: transverse, which goes from vertex to vertex; and conjugate, which intersects the transverse axis*0321

*at the center of the hyperbola and has a length of 2B (the transverse axis has a length of 2A).*0331

*The distance from focus to focus (between the two foci) is 2C.*0338

*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.*0346

*Here we are talking about a difference instead of a sum.*0352

*So, if you have a hyperbola with a center at (0,0) and a horizontal transverse axis, the equation is x ^{2}/A^{2} - y^{2}/B^{2}.*0355

* And here, we again have that the center is at the origin, (0,0).*0368

*Although the center certainly does not have to be at the origin, right now we are going to start out*0373

*working with hyperbolas with a center at the origin, just to keep things simple.*0377

*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.*0384

*Therefore, the vertex is going to be at (A,0); the other vertex will be at (-A,0).*0396

*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.*0408

*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).*0420

*This distance is B, from this point to the center; this distance is A.*0429

*And then here, I have f _{1} and f_{2}; and the distance from one of those to the center is C.*0436

*As I mentioned, you can have a hyperbola that is oriented vertically.*0447

*So, if the transverse axis is vertical, and the center is at (0,0), the standard form is such that y ^{2} is associated with the A^{2} term.*0451

*And here, it is positive: so you are taking y ^{2}/A^{2} - x^{2}/B^{2} = 1.*0463

*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.*0470

*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.*0482

*So, this is going to be some point, (B,0); and B ^{2} is given here, so you could easily find B by taking the square root.*0495

*And then, over here is (-B,0).*0502

*So again, there are two different standard forms, depending on if you are working with a hyperbola*0505

*that has a horizontal transverse axis or one that has a vertical transverse axis.*0508

*Something new that we didn't talk about with ellipses is asymptotes.*0516

*Recall that an asymptote is a line that a curve on a graph approaches, but it never actually reaches.*0520

*And asymptotes are very useful when you are trying to graph a hyperbola.*0528

*The equations are given here: let's go ahead and draw these first.*0534

*Now, recall that this vertex is at a point (A,0); this vertex is at (-A,0).*0536

*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).*0545

*And B is going to be the length from this point to the center; 2B will be the length of the conjugate axis.*0555

*Then, I am going to have another point down here, (0,-B).*0564

*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.*0568

*Therefore, this is going to be given by (-A,B); that is going to be one vertex.*0578

*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.*0586

*And that is going to give me the point (A,B).*0592

*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).*0597

*One final vertex is right here: and this is given by (A,-B).*0608

*OK, now you draw a box using these A and B points; and then you take that rectangle and draw the diagonals.*0618

*If you continue those diagonals out, you will have the two asymptotes for the hyperbola.*0630

*OK, so each of these lines is an asymptote.*0650

*And notice that the hyperbola is going to approach this, but it is not actually going to reach it.*0654

*So, it is going to continue on and approach, but not reach, it; it is going to approach like that.*0663

*All right, now this is one way to just graph out the asymptotes.*0671

*You can also find the equation; and for right here, we are working with a hyperbola with a horizontal transverse axis.*0676

*So, we are going to look at this equation.*0686

*If I was working with a vertical one, I would look at this equation.*0688

*Now, what does this mean? Well, y equals ±(B/A)x.*0691

*What this actually is: this B/A gives the slope of the asymptote.*0697

*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.*0703

*The slope, m, is B/A; B/A for this line is increasing to the right (m = B/A);*0714

*and the slope here equals -B/A, where the line is decreasing as we go towards the right.*0724

*So again, there are two ways to figure out these asymptotes.*0734

*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).*0737

*Draw that rectangle and extend the diagonals.*0754

*Or you can use the formula, which will give you the slope for these two asymptotes.*0758

*If you just started out knowing the A's and B's and drew these, then you could easily sketch the hyperbola,*0765

*because you know that it is going to approach these asymptotes.*0772

*OK, so we have talked a lot about graphing.*0778

*And just to bring it all together: you are going to begin by writing the equation in standard form.*0780

*And then, for hyperbolas, you are going to graph the two asymptotes, as I just showed.*0786

*So, let's start out with an example: let's make this x ^{2}/9 - y^{2}/4 = 1.*0790

*Since I have this in the form x ^{2}/A^{2} (this x^{2} term is positive here),*0801

*divided by y ^{2}/B^{2} = 1, what I have is a horizontal transverse axis.*0809

*So, this tells me that there is a horizontal transverse axis.*0819

*So, that is how this is just roughly sketched out already, showing the transverse axis along here.*0830

*Since it is in this form, I know that A ^{2} = 9; therefore, A = 3.*0841

*This has a center right here at (0,0).*0853

*And this point here is going to be A, which is 3, 0.*0860

*Right here, I am going to have -A, or -3, 0.*0867

*So, my goal is to make that rectangle extend out the diagonals.*0872

*And then, I would be able to graph this correctly.*0876

*OK, B ^{2} = 4; therefore, B = 2; so right up here at (I'll put that right there) (0,2)...that is going to be B.*0882

*And then, right down here at (0,-2)...*0903

*Now, all I have to do is extend the line up here and here; and these are going to meet at (2,3).*0907

*Extend a line out here; I am going to have a vertex right here at (-3,2).*0916

*I am going to have another vertex here at (-3,-2), and then finally, one over here at (3,-2).*0924

*Now, this was already sketched on here for me; but assuming it was not there, I would have started out by drawing this box,*0936

*and then, drawing these lines extending out--the asymptotes.*0944

*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.*0960

*So again, write the equation in standard form, which might require completing the square.*0972

*I gave it to you in standard form already; use that to figure out this rectangle.*0977

*And you are going to need to know A and B to figure out this rectangle.*0982

*Draw the asymptotes, and draw then the hyperbola approaching (but not reaching) those asymptotes.*0985

*So far, we have been talking about hyperbolas with a center at the origin (0,0).*0993

*However, that is not going to always be the case.*1000

*If the center is at another point, (h,k), that is not (0,0), then standard form looks like this.*1002

*It is very similar to what we saw with the origin of the center, except instead of just x ^{2}/A^{2}, we now have an h and a k.*1008

*For a horizontal transverse axis, you are going to have (x - h) ^{2}/A^{2} - (y - k)^{2}/B^{2}.*1015

*For a vertical transverse axis, this term is going to be first; it will be positive.*1027

*And then, you are going to subtract (x - h) ^{2}/B^{2}.*1033

*But the k stays associated with the y term.*1037

*For example, given this equation, (y - 3) ^{2} - (x - 2)^{2}...and we are going to divide that by 16,*1041

*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;*1053

*that A ^{2} = 16, so A = 4; and that B^{2} = 9, so B = 3.*1065

*From that, I can graph out this hyperbola.*1072

*And this has a vertical transverse axis; something else to be careful of--let's say I had something like this:*1076

*(y + 5)/10, the quantity squared, plus (x + 4), the quantity squared, divided by 12, equals 1.*1087

*The center is actually at (that actually should be a negative right here--this is a difference) (-4,-5).*1101

*And the reason for that is that this is the same as (y - -5) ^{2}, and then (x - -4)^{2}.*1112

*A negative and a negative is a positive.*1125

*So, you need to be careful: even though it is acceptable to write it like this, it is good practice,*1128

*if you are trying to figure out what the center is, to maybe write it out like this,*1134

*so that you have a negative here, so that whatever is in here is already k, or already h.*1139

*You don't have to say, "Oh, I need to make that a negative; I need to change the sign."*1145

*So, that is just something to be careful of.*1148

*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).*1151

*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.*1160

*I will give you that one of the vertices is at (0,2); the other vertex is at (0,-2).*1168

*The other piece of information is that you have a focus at (0,4), and a focus called f _{2} at (0,-4).*1178

*So, looking at this, I can see that this is the transverse axis, and then the center is right there.*1188

*So, I have a horizontal transverse axis.*1196

*I can also see that the midpoint right here, the center, is at the origin; so the center equals (0,0).*1207

*So, this is actually vertical--correction--a vertical transverse axis, going up and down: a vertical transverse axis.*1217

*Since this is actually a vertical transverse axis with a center at (0,0), I am working with this standard form:*1227

*y ^{2}/A^{2} - x^{2}/B^{2} = 1.*1234

*So, the A ^{2} term is with the y^{2} term, since this is a vertical transverse axis.*1240

*All right, in order to find the equation, I need to find A ^{2}.*1247

*This distance, from 0 to the vertex, is 2, because this is at (0,2).*1251

*Therefore, A equals 2; since A = 2, A ^{2} = 2^{2}, or 4.*1258

*I have A ^{2}; I need to find B^{2}; I am not given that.*1267

*But what I am given is an additional piece of information, and that is that there is a focus here and a focus here.*1270

*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--*1280

*the absolute value of that is 4; therefore, C = 4; the distance is 4.*1291

*C ^{2}, therefore, equals 4^{2}, or 16.*1298

*Recall the relationship: C ^{2} = A^{2} + B^{2} for a hyperbola.*1303

*So, I have C ^{2}, which is 16, equals A^{2}, which is 4, plus B^{2}.*1309

*16 - 4 is 12; 12 = B ^{2}; therefore, B = √12, which is about 3.5.*1315

*If you wanted to draw B, then you could, because that is right here at (3.5,0).*1330

*But what we are just asked to do is write the equation; and we have enough information to do that,*1338

*because I have that y ^{2} divided by A^{2}; I determined that A^{2} is 4;*1342

*minus x ^{2}/B^{2}; I determined that that is 12; equals 1.*1348

*So, this is the equation for this hyperbola, with a vertical transverse axis and a center at (0,0) in standard form.*1355

*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.*1368

*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).*1380

*That means that the center is going to be right here at (0,0).*1400

*So, the center is at the origin; since the vertices are here and here, then I have a horizontal transverse axis;*1406

*this is going to go through like this, and then like this.*1424

*So, my second piece of information is that I have a horizontal transverse axis.*1432

*Since I have a horizontal transverse axis, then I am going to have an equation in the form x ^{2}/A^{2} - y^{2}/B^{2} = 1.*1440

*The center is at the origin; it has a horizontal transverse axis; this is a standard form that I am working with.*1451

*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,*1457

*or from this point (the vertex) to the center, is 5: A = 5.*1466

*Since A = 5, A ^{2} = 5^{2}; it equals 25.*1473

*The other information I have is that the conjugate axis has a length of 12.*1486

*So, the length of the conjugate axis, recall, is 2B; here they are telling me that that length is 12.*1490

*Therefore, 12/2 gives me B; B = 6; so, that would be up here and here: (0,6) and then (0,-6).*1503

*This would be the conjugate axis; so this is B = 6.*1517

*Since B equals 6, I want B ^{2} that equals 6^{2}, which equals 36.*1523

*Now, I can write this equation: I have (this is my final one) x ^{2}/A^{2}, which is 25,*1533

*minus y ^{2}/B^{2}, and I determined that that is 36, equals 1.*1545

*So, this is a hyperbola with a center at the origin.*1551

*And A ^{2} is 25; B^{2} is 36; and it has a horizontal transverse axis.*1554

*We are asked to graph this equation; and it is not in standard form.*1566

*But when I look at it, I see that I have a y ^{2} term and an x^{2} term, and they have opposite signs.*1571

*So, I am working with the difference between a y ^{2} term*1577

*and an x ^{2} term, which tells me that this is the equation for a hyperbola.*1579

*If they were a sum, this would have been an ellipse, since they have different coefficients.*1585

*But it is a difference, so it is a graph of a hyperbola.*1589

*What I need to do is complete the square to get this in standard form.*1591

*OK, so first I am grouping y terms and x terms: y ^{2} + 12y - 6x^{2} + 12x - 36 = 0.*1597

*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.*1617

*The next thing I need to do with completing the square is factor out the leading coefficient, since it is something other than 1.*1627

*So, from the y terms, I will factor out a 2; that is going to leave me with y ^{2} + 6y.*1634

*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.*1640

*And that is going to leave behind an x ^{2} here; here, it is going to leave behind, actually, -2x.*1649

*So, checking that, -6 times x ^{2} is -6x^{2}--I got that back.*1656

*-6 times -2x is + 12x; equals 36.*1661

*Now, to complete the square, I have to add b ^{2}/4 in here, which equals...b is 6; 6^{2}/4 is 36/4; that is 9.*1669

*I need to be careful to keep this equation balanced.*1686

*Now, this is really 9 times 2 that I am adding; 9(2) = 18--I need to add that to the right.*1688

*Working with the x terms: b ^{2}/4 = 2^{2}/4, which is 4/4; that is 1, so I am going to add 1 here.*1697

*-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.*1714

*Now, I am rewriting this as (y + 3) ^{2} - 6(x - 1)^{2} =...18 - 6 is 12; 36 + 12 gives me 48.*1726

*The next step, because standard form would have a 1 on this side, is: I need to set all this equal to 1.*1747

*I need to divide both sides of the equation by 48.*1753

*This cancels, so it becomes y + 3 ^{2}; the 2 is gone; this becomes a 24; minus...6 cancels out,*1769

*and that leaves me with (x - 1) ^{2}; 6 goes into 48 eight times; and this is a 1.*1780

*OK, so it is a lot of work just to get this to the point where it is in standard form.*1788

*But once it is in standard form, we can do the graph, because now I know the center; I know A ^{2} and B^{2}.*1792

*We have this in standard form; so now we are going to go ahead and graph it.*1799

*I will rewrite the standard form that we came up with, (y + 3) ^{2}/24 - (x - 1)^{2}/8 = 1.*1803

*Looking at this; since this is positive, I see that I have a vertical transverse axis.*1815

*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}.*1824

*And when we look at standard form, we actually have a negative here.*1835

*So, you need to be careful to realize that the center is at (1,-3), not at (1,3).*1838

*Let's make this 2, 4, 6, 8, -2, -4, -6, -8; the center, then, is going to be at (1,-3).*1846

*The next piece of information: A ^{2} = 24; therefore, A = √24, which is approximately 4.9.*1856

*B ^{2} = 8; therefore, B = √8; therefore, if you figure that out on your calculator, that is approximately 2.8.*1871

*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.*1883

*The goal is to write this in standard form, find A and B, find the center, make the rectangle, and make the asymptotes;*1892

*and then, you can finally draw both branches of the hyperbola.*1902

*All right, so if the center is here at (1,3), then I am going to have 2 vertices.*1906

*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.*1912

*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,*1926

*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.*1933

*The y-coordinate will be at -3 + 4.9, which equals (1,1.9).*1944

*Therefore, at (1,1.9) (that is right there)--that is where there is going to be one vertex.*1950

*And this is A--this is the length of A.*1958

*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.*1963

*That is -3 - 4.9 (or + -4.9; you can look at it that way) = (1,-7.9), down here.*1979

*OK, vertices are at (1,-7.9) and (1,1.9).*1999

*Now, I need to find where B is--where that endpoint over here is, horizontally--so that I can make this rectangle.*2014

*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.*2024

*Well, 1 + 2.8 is (3.8,-3); so (3.8,-3) is right there.*2040

*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).*2054

*That is going to be...this is 2...-2 is right here; so that is going to be right about there.*2068

*I now have these points; and recall that I can then extend out to make a box.*2076

*There is going to be a vertex here; I am going to extend across; there is going to be a vertex here.*2083

*Bring this directly down; there is a vertex here, and then another vertex right here.*2090

*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.*2096

*This is B; then this length is A.*2109

*Once I have this rectangle, I can go ahead and draw the asymptotes by extending diagonals out.*2113

*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.*2130

*Either method works.*2140

*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.*2143

*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.*2156

*OK, so this was a difficult problem; we were given an equation in this form.*2170

*We had to do a lot of work just to get it in standard form.*2175

*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.*2178

*Example 4: We don't need to do graphing on this one.*2189

*We are just finding the equation of a hyperbola with the center here, (0,0), and a horizontal transverse axis.*2192

*I am going to stop right there and think, "OK, I have a center at (0,0) and a horizontal transverse axis."*2200

*So, the standard form is going to be x ^{2}/A^{2} - y^{2}/B^{2} = 1.*2206

*Since the center is at (0,0), I don't have to worry about h and k.*2215

*The horizontal transverse axis has a length of 12; well, the transverse axis length, recall, is equal to 2A.*2219

*I am given that that length is 12; if I take 12/2, that is going to give me A = 6.*2231

*The conjugate axis--recall that the length of the conjugate axis is equal to 2B, which is 6: B = 3.*2240

*Now, I need to find A ^{2}, which is 6^{2}, or 36, to put in here.*2257

*B ^{2} is 3^{2}, which is 9.*2264

*Now, x ^{2}/36 - y^{2}/B^{2} (which is 9) = 1.*2267

*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.*2279

*That concludes this lesson on hyperbolas; thanks for visiting Educator.com!*2290

0 answers

Post by Shiden Yemane on March 21 at 12:52:14 PM

In example 3, when you were completing the square for the x terms, shouldn't the the (b^2) over 4 be (-2)^2 over 4, instead of 2^2 over 4? Let me know if I'm wrong.

2 answers

Last reply by: julius mogyorossy

Fri Aug 15, 2014 2:23 PM

Post by Kenneth Montfort on March 6, 2013

So, you said in the lecture on ellipses, that a^2 was the larger term, it seems here that b^2 is the larger term...is that another clue about how to tell which formula you are using?