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Contents

- 1 Hyperbolas
- 2 Graphing Hyperbolas 2
- 3 Graphing Hyperbolas 2
- 4 Graphing Hyperbolas 3
- 5 Graphing Hyperbolas 3
- 6 Equations for Hyperbolas
- 7 Graphing Hyperbolas 4
- 8 Graphing Hyperbolas 4
- 9 Graphing Hyperbolas 5
- 10 Graphing Hyperbolas 5
- 11 Graphing Hyperbolas 6
- 12 Graphing Hyperbolas 6
- 13 Graphing Hyperbolas 7
- 14 Graphing Hyperbolas 7
- 15 Graphing Hyperbolas 8
- 16 Graphing Hyperbolas 8

We know this equation produces an ellipse. We start at the center at the origin move three units in the y direction, up and down. This gives us four main points on our ellipse so that way we can graph it easily. We're going to use this knowledge of ellipses to help us graph a hyperbola. A hyperbola is another type of conic section and its equation looks more like this. Notice that the difference between a hyperbola equation and an ellipse equation is this subtraction sign here. For ellipses, we have addition between the x squared term and the y squared term. But for hyperbolas, we have subtraction between the terms. One is positive and one is negative. To graph this hyperbola, we're going to start by graphing it as an ellipse. So we'll start with this shape that we already have. Now, those first four points were helpful for graphing the start of our elipse and they're also going to be helpful for graphing our hyperbola. We'll draw a dashed rectangle around our ellipse so that way we can start graphing our hyperbola. Once we have the rectangle, we'll draw dash lines across the diagonals. Now, this minus sign is like a knife. It's going to cut open our ellipse. And then we're going to pivot these two halves outward. These two points will become the vertices of two branches that will face this way and this way. So notice, I'm going to take off my ellipse now and graph my hyperbola. Notice how the branches of this hyperbola approach the diagonals from the rectangle. The same is true for this branch. The ins of the graph approach the dash line, and this will always be true for our hyperbola. The ins will always stretch towards the diagonals that we created from before. Now one other important piece of information is that the x square term comes first here. This means that our parabola opens to the left and to the right. And you might be wondering, how can I get a parabola to open up and open down? Well it turns out we just have the y squared term conversed. For example if I make this y squared term positive here and we subtract off x squared divded by 4 here we'll have a different type of hyperbola. Since the y squared term comes first this hyperbola will open up. And it will open down. I'm going to take this hyperbola off for now, and then we'll start over with our basic ellipse. Now, I could switch the order of the y squared and the x squared terms, since addition is commutative. So, I'm just going to move it this term to the right side of the y squared divided by 9. I notice when I make this change, I still have the same exact ellipse as from before. We started with our center at the origin and we went two units left and right from that center and three units vertically up and down. This time to get this hyperbola, we'll do one thing different. We'll still draw our box around our ellipse and we'll have the diagonals to that box. Now we'll use these two points as the vertices for the two hyperbola branches. They'll face this way. The subtraction sign is going to be the knife that cuts our ellipse in half, and then the top and bottom of the ellipse will fold outward to create our hyperbola. So taking away our ellipse, we get this as our hyperbola. And again notice that we have the y squared term coming first, which means the parabola opens up and it opens down. Now that you've seen a couple of hyperbolas, try creating some on your own. Try changing the values of the denominators and see how the hyperbolas change. Have some fun and then continue on with the practice.

Try graphing this hyperbola. What do you think is its center? What do you think's the a value and what do you think is the b value? Remember to draw the graph on your own, and then compare your result in the solution video.

This hyperbola is centered at 0,0. It has an a value of 3 and a b value of positive 5. Nice work if you got these correct. Now, let's check your graph. We'll start with an ellipse. So, this ellipse will be centered at the origin and will go 3 units in the x direction. And 5 units in the y direction. We'll draw a dashed box around our ellipse, so that way we can start to draw a hyperbola. Next, we'll draw in our two diagonals across our rectangle. We want to extend these along the graph, so that way we can draw in our hyperbola branches. So now we're ready to cut the ellipse open. We turn each of the pieces out. And then we can see we have our hyperbola. And notice again, that we can get our rectangle, just by looking at the a value and the b value. We start with the origin and we move 3 units in the x direction for a and 5 units in the y direction for b. This hyperbola opens to the left and to the right since the x squared term comes first. Whichever term is positive, the x squared term or the y squared term. That will determine which way our hyperbola opens. Since the x squared term here is positive, then we know that it opens left and right.

Now, just as with other conic sections, hyperbolas are not necessarily centered at the origin. What do you think would be the center for this hyperbola based on what you know about circles and ellipses. I think you can figure this out.

This Hyperbolas is centered at three negative 2, has an a value of two and a b value of 3. Great thinking if you got all three correct. We know the center is at three negative two, since we have the opposite sign in the parenthesis. So this is our h and this is our k. H will be positive 3, the opposite sign of negative 3, and k will be negative 2, the opposite sign of positive 2. This is our center for our hyperbola. Finding a and b are the same as just before. We just take the square root of this to get a, and the square of this to get b. Now we're ready to graph our hyperbola. We start by graphing an ellipse with a center of positive 3 and negative 2. So, positive 3 on the x, negative 2 on the y, here's our new center. We go two units in the x direction and three units in the y direction to get the remaining points. But we're not really graphing an ellipse, we're graphing a hyperbola. So we draw a dashed rectangle around our ellipse and then we draw in the 2 diagonals. We know our hyperbola is going to open to the left and to the right, since the x squared term is positive. So we split our ellipse in half, and then we pivot the two sides outward and extend them to the diagonals. This gives us our hyperbola.

For a hyperbola that is not centered at the origin, we can use h comma k as its center. This actually gives us two equations for hyperbolas. This equation has the x squared term coming first. So the parabola will open to the left and to the right. H and k will be our center. And the value of a will be how many units we move in the x directions. And the value of b will be how many units we move in the y direction. This allows us to graph our ellipse, and then draw the box around it for our rectangle. Other hyperbolas however open up and down. These hyperbolas start with the y squared term. This term will be positive instead of being negative. But notice that h and k have been switched here. H is still with x and k is still with y. But k comes first in this equation and h comes second. So, you want to keep this mind when finding the center if the hyperbola opens up and down. The h value will be this value and the k value will be this one. This might be confusing but you can always remember that anything associated with the x direction will be with the variable. For example, the x coordinate of the center will be h. That's with the x. And the way we move in the x direction, a will be here. Underneath the x squared term. Likewise, anything having to do with the y or the y direction will be next to the y variable. The y coordinate of the center k is here, and the way we move vertically up and down, the value of b is here. These equations aren't that different. And really the things with a squared and b squared work exactly the same as before. The only difference is that this hyperbola opens left and right since the x squared comes first and this parabola opens up and down since the y squared term comes first and is positive. The other subtle difference is hk. Here they're in order, h and then k. Here we need to be careful and get the h next to the x. And then get the k that's with the y.

Let's try graphing some more hyperbolas. I want you to graph this equation on your own paper and then tell me what's the center and what's a and what's b. Just like before, we'll check to see if your graph matches this one and the solution.

This hyporbola is centered at 1,2 and has an a value of 4 and a b value of 3. Good work if you got all three. We'll start by graphing the ellipse. The ellipse will be centered at 1,2. So we go 1 unit right and 2 units up. Next we move 4 units left and right in the x direction from the value of a, then we move 3 units up and down and the y direction for the value of b. Now we're ready to start constructing our hyperbola. We draw a dashed rectangle around our ellipse and then we connect the diagonals. We'll cut open our ellipse and then we'll flip the two halves outward, and this will give us our hyperbola. We want to make sure that the ends of the hyperbola are stretching out to the diagonals. This curve will always approach this dash diagonal and this dash diagonal. It's one of the special properties for hyperbolas and you should look into it more if you're interested.

Try graphing this hyperbola. What's the center? What's a? And what's b?

The center will be at negative 1 negative 2 the opposite signs of these and a value is 3 and the b value is 2. Nice thinking if you got these right. Now let's check out your graph. We'll start by graphing an ellipse and the center is at negative 1 negative 2. We'll go one unit to the left, and two units down. Now we use our a and b values to get the other points. Three units in the x direction for a, and two units in the y direction for b. We carefully construct our rectangle around our ellipse. This is going to start our hyperbola graph. Next, we draw on our two diagonals across the rectangle. And now we're ready to draw on our hyperbola. Our hyperbola will open left and right, since the x squared term comes first. So we'll cut open our ellipse, and we'll split the two halves outward. We'll stretch the ends of the hyperbola to our diagonals, which gives us our final graph. Keep in mind, that the reason this parabola opened left and right was because the x squared term came first. It was positive in this case. We use the opposite signs for h and k to get our center, negative 1 and negative 2. These are some of the most common pit falls students make. So you want to make sure you pay close attention to signs and which term comes first for hyperbolas.

Try drawing a graph for this hyperbola. Draw your graph first and then indicate the center, the value of a and the value of b, just like before.

The center is at negative 2,3, since we take the opposite signs in the parentheses. The opposite of positive 2 is negative 2, and the opposite of negative 3 is positive 3. This is our h and our k. The value of a is the square root of 25, which equals 5, and the value of b is the square root of 36, which equals 6. Great work if you got these three correct. Now, what's even more important is your graph. Let's see if yours will match mine. We'll start with an ellipse, as usual and it will be centered at negative 2, 3. Negative 2 in the x direction, positive 3 in the y direction. We move 5 units left and right from the center, and 6 units up and down from that center to form our ellipse. Now, we draw in our dashed rectangle around our ellipse, and then we draw in the two diagonals. Now we're ready to create our hyperbola. The x squared term comes first. So we know our hyperbola will open left and right. We'll split our ellipse in half and we'll flip the two ends out, so that way we get the graph of our hyperbola. Nice work if you got this graph.

Now how about here. Try giving me a sketch of this graph, then tell me it's center, the value of a and the value of b.

This hyperbola was centered at 4,2, has an a value of 2, and a b value of 4. Nice thinking if you got these correct. Now, the key is that y squared comes first, so this means our hyperbola will open up and down. Remember, too, that k is this value, and h is this value. H is always with x, and k is always with y. So, our center is really at positive 4 and positive 2. We take the opposite signs of the terms in the parentheses. To get the value of a, we just take the square root of 4, since it's underneath the x squared term. The square foot of equals 4. We start by graphing the ellipse. The ellipse is centered at 4, 2. So, we go 4 units right, and 2 units up vertically. Now we need to go a and b, we need to find the four other points on the ellipse. From the center we move 2 units in the x direction, and we move 4 units in the y direction. 2 is for a, and 4 is for b. Remember that a is always associated with the x squared term, this square root of 4 here. And b is always associated with the y squared term. In this case, the square root of 16. Now, we're ready to draw in our dash rectangle and the two diagonals, so we can draw our hyperbola. We need to keep in mind that our ellipse is going to open up and down, since y comes first. It's the positive term. So, we're going to cut our ellipse and them flip these two ends outward. When we do that, we get our entire hyperbola. Now, let's make sure this whole picture makes sense. Our hyperbola is centered at 4,2 and then we moved 2 units horizontally for our rectangle, an 4 units vertically.

Here's your last type of hyperbolic graph. What do you think would be this center and the values of a and b. Keep in mind that you want to set this equation equal to 1. What do you think you need to do here.

This hyperbola will be centered at 0, 5 give, it will have an a value of 8, and a b value of 4. Good thinking if you got these three right. We'll start by dividing each of these terms by 64. When we divide this term by 64, we'll simplify it by 1 16th. This term divided by 64 turns into negative x squared divided by 64. And finally, 64 divided by 64, equals 1. Our y squared term comes first, which means that our hyperbola needs to open up, an it needs to open down. There's no number next to the x variable, so that means the center, or the x coordinate of the center, is 0. The y coordinate of the center would be positive 5. This is the k it is always next to y. The a is the number that we move in the x direction so that's underneath the the x squared term. It's the square root of this number. Since the square root of 64 equals 8, we know the value of a, it must be 8. For the value of b, we simply take the square root of 16 which equals 4. We start by graphing the ellipse with a center at 0, the y direction for two more points. Next we draw our dash rectangle around our ellipse. Then we draw in the two diagonals. And since we know the y squared term is first and is positive, we know our hyperbola opens up and it opens down. So we cut our ellipse horizontally. And we flip over these two pieces. Ultimately, this produces our hyperbola. As a quick check, we can make sure that these two points are four units vertically away from the center. Our center is at 0, 5, so yes, this point is four units away vertically, and this point is four units away in the other direction.