Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A hyperbola is the set of points in a plane the difference of whose distance from 2 fixed points, called the foci, is constant. So when we look at a hyperbola, we're going to have the vertex is going to be the closest point to the center, the focus. The hyperbola is going to curve around the focus locations. So we're going to have a curve of going through the vertex going around each of the focus. The transverse axis is the axis that the vertex and the focus lie upon. The center is also on the transverse axis. So when we look at this, we're going to develop the formula and we know that it's going to be the distance. The difference is going to equal some constant. This constant's going to be two A. So the distance between these, the distance of the first two points is going to be actually let's write it smaller. Let's do X - -C ^2 + y - 0 ^2. So this is PF 1 -, X -, C ^2 + y - 0 ^2. This is P to F2, and that's going to equal to A. So we're going to have the absolute value of the distance, but in a minute we're going to square each side. So the absolute value is not going to matter when we square each side, we're going to get a positive anyway. So we're going to have sqrt X ^2 + 2 CX plus C ^2 + y ^2 equaling 2A plus the square root X ^2 - 2 CX plus C ^2 + y ^2 if we foil the left side. No, If we square each side, we have to square a square root on the left side, but we have to foil the right side. So the left side just ends up being X ^2 + 2 CX plus C ^2 + y ^2. But the right side we have to foil. So we're going to get four a ^2 + 4 a square roots, X ^2 -, 2 CX plus C ^2 + y ^2, and then the last term squared. So the square roots are going to cancel X ^2 -, 2 CX plus C ^2 + y ^2. So now we're going to have things that cancel. When we look, we have an X ^2 on one side and an X ^2 on the other, AC squared on one side and AC squared on the other, AY squared on one side and AY squared on the other. So I need to get everything to one side that's not the square root. So we're going to get 4 CX -4 A squared equaling 4A square roots X ^2 -, 2 CX plus C ^2 + y ^2. Because I like keeping my numbers small, I'm going to divide everything through by 4 also. So I'm going to get CX minus a ^2 equaling a square roots X ^2 -, 2 CX plus C ^2 + y ^2, squaring each side again. We're going to get C ^2 X ^2 -, 2 A squared CX plus A to the 4th, equaling a ^2 times the quantity X ^2 -, 2 CX plus C ^2 + y ^2. So C ^2 X ^2 - 2 A squared CX plus A to the 4th, distributing that a ^2 a ^2 X ^2 - 2 A squared CX plus a ^2 C ^2 + a ^2 y ^2. So here we're going to have things that cancel again. Let's see this -2 A squared CX and -2 A squared CX. I'm going to take all of the X&Y terms to one side and everything without an X or Y term to the other. So we're going to have A to the fourth minus a ^2, C ^2 equaling a ^2 -, C ^2 times the X ^2 plus a ^2, y ^2. So I took all of the X ^2 terms together and factored out the X ^2. So I got a ^2 -, C ^2, a ^2, y ^2, and then I took everything to the other side. On this left side, we're going to factor out an A ^2, so we get a ^2 -, C ^2. So when we divide, we get one equaling X ^2 / a ^2 + y ^2 / a ^2 -, C ^2, now a ^2 -, C ^2. If we think about our hyperbola, we know that the furthest distance out now is our C because it's the distance between the center to the foci. The distance between the center to the vertex was A. So our relationship here has got to be C ^2 equal a ^2 + b ^2. The furthest away has to equal the two smaller ones, so C ^2 equal a ^2 + b ^2. So if I wanted to solve for b ^2 here, I'd get b ^2 equaling C ^2 -, a ^2. Well, in this equation we have a ^2 -, C ^2. So what we're going to do is we're going to factor out a -1 to get C ^2 -, a ^2. When I factor out that -1, what happens is I get X ^2 / a ^2. The plus and negative is really the same thing as A -, y ^2 and C ^2 -. A ^2 is just b ^2. So this is the formula for a hyperbola that's opening side to side. No longer does A have to be the bigger one like in an ellipse, it just has to be the positive. So 1 equal y ^2 / a ^2 -, X ^2 / b ^2 would be hyperbolas that we're opening up and down. So the generic formula if we're not at the center anymore being the origin, if we have a center shift, we'd have HK over a ^2 and b ^2 equaling 1. So the centers HK, we're parallel to the X axis for our transverse axis, our vertices is going to be H -, A and H + a comma K. We also, if we were going with our transverse act axis parallel to the Y axis or vertical, our center would be HK and now we'd have HK minus A and HK plus a for our vertices. If we look at some examples here, we have a few graphs here. The center is going to be at 00 for this one and we're opening left and right. So the X term is going to be positive. If we count how far out we went, 123, we went three to the right or three to the left. So our a is going to be 3 and we're going to have 3 ^2. We're going to go up 12345 S underneath the Y is going to be 5 ^2. So X ^2 / 9 -, y ^2 / 25 equal 1. If we look at this next one, we're opening up and down this time. So our y ^2 is going to be the positive term. Our center is still at the origin, so we went up to. So underneath the Y we're going to have two squared to the right. We went 123 to get the edge of that box. The edge of that box is going to form our asymptotes. Our asymptotes are the lines that the graph gets closer and closer and closer to this next one. We're opening left and right, so the X is going to be the positive term. Our center is still at the origin. We went over 123, so we're going to have 3 ^2. We went up to 2 ^2 from the center, so X ^2 / 9 -, y ^2 / 4 equaling 1. The next one, the Y, it opens up and down, so the Y term is positive. Our center is still at the origin. We went up five and we went to the right 3. So we're going to have 5 ^2 and 3 ^2. So y ^2 / 25 -, X ^2 / 9 equaling 1. So let's do some other types of examples. Let's do an example. If we're given our foci being zero -3 and 03 and our vertices being zero -1 and 01, If we have our foci, we know that our C distance is going to be 3. If we have our vertices, we know that our A distance is 1. The fact that the Y is the value that's changing tells us we know that the Y is going to be the positive term. We also know that the center because the symmetry has got to be halfway in between the two vertices or the two foci. So we know that the center here is at the origin 00. So underneath the Y is always going to be the A term or the a ^2 term. And in this one we know that the foci is further away than everything else. So we have our relationship of the furthest away. C ^2 has to equal a ^2 + b ^2, so 3 ^2 equal 1 ^2 + b ^2 or b ^2 = 8. Remember, we no longer care about which one is bigger, we care about which one is opening in the positive direction. What if we had one that gave us the endpoints of the transverse axis zero -6 and 06 the endpoints of transverse axis and it gave us an asymptote of Y equal to X? Well, if we have the end points of the transverse, we really really have the a value or the vertices. So we're going to have y ^2 / 6 ^2 -, X ^2 over something equal 1. Now, if we thought about graphing this, we know we're going to have a box that has our asymptotes in it. So if we went up six, but our slope was 2, so we went up to and over 1, and up two more, and over 1 and up two more and over 1. So if we went up six and we want to figure out how far over we went, so 6 over some letter M has to equal our slope. Our slope being 2 / 1 so our M has got to be 3. So if we cross multiplied we get 2M equaling 6 so our M is 3. So this distance here would have be a three. So our B is going to be 3 ^2. So we'd get y ^2 / 36 -, X ^2 / 9 equaling 1. What if we had a problem that started out with some coefficients? Let's say we had nine y ^2 - 4 X squared minus eighteen y + 24 X -63 = 0. The first thing we're going to do is we're going to group together the Y terms and factor out the coefficient on the squared. Then we're going to group together the X terms and factor out the coefficient on the X, and we're going to take all the constants to the other side. Then we're going to complete the square. So half a -2 ^2 is 1 and half of -6 ^2 is 9. Now we didn't really add one. We really added the coefficient of 9 * 1. We didn't really add nine. We really added -4 * 9. So pay attention to the fact that this was nine times the one and this other one was -4 times the +9. So when we look at this, we get 9 times the quantity y - 1 ^2 - 4 times the quantity X - 3 ^2 equaling 63 + 9 - 36. So 63 + 9 is 72 - 36 is 36. If we then divide by 36, we'd have y -, 1 ^2 / 4 -, X - 3 ^2 / 9 equaling 1. So our center here is going to be 3 one. Our vertices are going to have be changing in the Y direction. So we're going to add 2 to the Y and we're going to subtract 2 from the Y. Our foci, we're going to have our C ^2. The furthest thing away is a ^2 + b ^2, so 4 + 9 = C ^2, and it's going to be moving in the same direction as our vertices. So we're going to combine this with our Y value. So we're going to have 1 ± sqrt 13 for the Y. If we were going to graph this, we would go to our center +3 positive one from that location. An easy way to do it is just to square root the numbers underneath each term. So sqrt 4 is 2, so I'm going to go up two and down 2. Sqrt 9 is 3, so I'm going to go right three and I'm going to go left three. Those are going to be where my box gets drawn in. My asymptotes always go through the corner of the boxes, always. So the equation here for the asymptotes, and I think it's easiest if we draw it to find them personally, we're going to have y -, 1 equaling the slope. Well, if we start from the center, how did we get to one of the corners? We went up to and to the right 3, so we went up two to the right 3A slope of 2/3 times the X -, 3. If we started at the center, how did we get to the other asymptote line? We went down to and to the right 3. So y - 1 equal positive 2/3 and negative 2/3 * X - 3. Thank you and have a wonderful day.