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 2 answersLast reply by: julius mogyorossyFri Aug 15, 2014 2:23 PMPost by Kenneth Montfort on March 6, 2013So, 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?

### 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 c2 = a2 + b2.
• Understand the standard formula for the equation of a hyperbola.
• Know how to put an equation in standard form by completing the square.

### 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 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

### 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, f1, and here is the other, f2.0041

If I take a point on the hyperbola, and I measure the distance to f1,0048

and then the distance to f2, that is going to give me d1 and d2.0056

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, d1, minus (the difference) d2, 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, d3, and then the distance to f1 could be something--say d4.0100

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 f1, f2...let's look at f2...0212

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.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 C2 = A2 + B2.0303

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 axis0321

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

So, if you have a hyperbola with a center at (0,0) and a horizontal transverse axis, the equation is x2/A2 - y2/B2.0355

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 out0373

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 f1 and f2; 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 y2 is associated with the A2 term.0451

And here, it is positive: so you are taking y2/A2 - x2/B2 = 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 B2 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 hyperbola0505

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 x2/9 - y2/4 = 1.0790

Since I have this in the form x2/A2 (this x2 term is positive here),0801

divided by y2/B2 = 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 A2 = 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, B2 = 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 x2/A2, we now have an h and a k.1008

For a horizontal transverse axis, you are going to have (x - h)2/A2 - (y - k)2/B2.1015

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/B2.1033

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 A2 = 16, so A = 4; and that B2 = 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 f2 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

y2/A2 - x2/B2 = 1.1234

So, the A2 term is with the y2 term, since this is a vertical transverse axis.1240

All right, in order to find the equation, I need to find A2.1247

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

Therefore, A equals 2; since A = 2, A2 = 22, or 4.1258

I have A2; I need to find B2; 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

C2, therefore, equals 42, or 16.1298

Recall the relationship: C2 = A2 + B2 for a hyperbola.1303

So, I have C2, which is 16, equals A2, which is 4, plus B2.1309

16 - 4 is 12; 12 = B2; 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 y2 divided by A2; I determined that A2 is 4;1342

minus x2/B2; 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 x2/A2 - y2/B2 = 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, A2 = 52; 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 B2 that equals 62, which equals 36.1523

Now, I can write this equation: I have (this is my final one) x2/A2, which is 25,1533

minus y2/B2, and I determined that that is 36, equals 1.1545

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

And A2 is 25; B2 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 y2 term and an x2 term, and they have opposite signs.1571

So, I am working with the difference between a y2 term1577

and an x2 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: y2 + 12y - 6x2 + 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 y2 + 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 x2 here; here, it is going to leave behind, actually, -2x.1649

So, checking that, -6 times x2 is -6x2--I got that back.1656

-6 times -2x is + 12x; equals 36.1661

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 9.1669

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: b2/4 = 22/4, 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 + 32; 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 A2 and B2.1792

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: A2 = 24; therefore, A = √24, which is approximately 4.9.1856

B2 = 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 x2/A2 - y2/B2 = 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 A2, which is 62, or 36, to put in here.2257

B2 is 32, which is 9.2264

Now, x2/36 - y2/B2 (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