Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Ellipse, the set of all points in a plane whose sum from 2 fixed points is a constant. The fixed points are called the foci. The singular is focus. So if we look at an ellipse, we have two fixed points F1 and F2. In this particular problem, we're going to let the center be at the origin. We have some point, any point pxy. So the distance between the center by definition and the focus is going to be represented by the variable C. The distance between the center and the vertex is going to be represented by the variable A. So if we thought about the distance from P to F1 plus the distance of P to F2, that's going to equal to A. And the reason is that if we look at this graph and we think about the distance from the focus one to our vertices and then the distance from the vertices to the focus 2. So if this is F2 to V2, that's going to be the same distance because of symmetry as V1 to F1. And if we know that the center to the end point is A, then the center to the other end point is also A or the center to the vertices. So we have a + A or two a. So when we look at this, we're going to put in the points using the distance formula X - -C ^2 + y - 0 ^2 + X - C ^2 + y - 0 ^2. When we square each of these, we're going to take the one term to the other side and square each side. When we square a square root, that'll cancel and we're going to get X ^2 + 2 CX plus C ^2 + y ^2. When we square the right side, we have to foil it out, so we get four a ^2 -, 4 a square root, X ^2 -, 2 CX plus C ^2 + y ^2 plus X ^2 - 2 CX plus C ^2 + y ^2. We have things on both sides that will cancel the X squareds, the C squareds, and the Y squareds. I'm going to take the square root to one side and everything else to the other again. And because I like small numbers, I'm going to also divide everything through by a four. Now to get rid of the square root, we're going to square each side again. So we get a squared times the quantity X ^2 -, 2 CX plus C ^2 + y ^2. On the right side, we need to foil A to the 4th -2 A squared CX plus C ^2 X ^2. Then we're going to distribute a ^2 X ^2 -, 2 A squared CX plus a ^2 C ^2 + a ^2 y ^2, equaling A to the 4th -2 A squared CX plus C ^2 X ^2. From there, the -2 A squared CX cancels on each side. I'm going to take all the XS to one side, everything else to the other, all the XS and YS to one side, everything else to the other. When we factor out the a ^2 from the right hand side, we're going to then divide all the terms by a ^2 times the quantity a ^2 -, C ^2. So we get X ^2 / a ^2 plus y ^2 / a ^2 -, C ^2 equaling one. Now, because the distance to any point of the sum is 2A, if we have the point up here, we know that the distance from here to here from the two focace is going to be two a. So we have A and we have a. So now I have a right triangle where I can think about the distance between the center and the focus was C The distance between the center and the end point of the minor axis was B. Our hypotenuse has to be A. So we have a relationship of b ^2 + C ^2 equal a ^2. Or we could think of that as a ^2 -, C ^2 equal b ^2. So instead of this a ^2 -, C ^2 here we're going to replace it with b ^2. So that's our equation of an ellipse if it's being elongated along the X axis. If it was being elongated along the Y axis, a, which is our bigger term, would be underneath our Y. So X -, H quantity squared over a ^2 + y -, K ^2 / b ^2 equal 1. This is if the center has shifted to HK. The vertices. Remember it in ellipse A is bigger than B, so the vertices are going to be going along the X direction. So H ± a comma K the focus or the foc IH plus or minus C, K, the endpoints of the minor axis H, K ± b That's if it was being elongated along the X axis. If it was being elongated in the other direction or along the Y axis, the center HK, the vertices H K ± a foci H, K ± C, the endpoints of the minor H ± b, K. Now actually when I say along the axis, I mean parallel to the axis because we're not going through the center being the origin anymore. We've shifted the centers to H&K. So this first example, if I have 40, nine X ^2 + 4 Y squared equal 196. The first thing we're going to do is we're going to divide everything through by 196. So we're going to get X ^2 / 49 + y ^2 / 49. Oops, sorry, X ^2 / 4 + y ^2 / 49 equaling one. Our center is going to be 00. Our vertices, the 49 is bigger, so it's going to be 0 ± 7. Our end points of the minor is going to be underneath the X, the smaller number, so positive -2, zero. And our focus, the way that I think about this is in an ellipse, the vertice is the furthest point out. So we know that the a ^2 has to equal b ^2 + C ^2. So if the a ^2 is 49 and the b ^2 is 4, we can find our C ^2 to be 45. So C is going to be positive or -3 square roots of five. It's going to be changing along the Y direction because the 49, the bigger number was underneath the Y. So 0, ±3 square roots of five. If we wanted to graph this, we'd start at 00. We'd move two to the right, two to the left, 7UP, and seven down. The closer the numbers and the denominator are together, the more like a circle it would be. If the numbers were exactly the same, it would be a circle because we'd go left and right, up and down the same quantity. If we look at this next one, we're going to start by completing the square we were. Actually, the original problem was nine X ^2 -, 72 X plus four y ^2 + 16 Y plus 124 = 0. So we're going to start by grouping the axis together and the YS together, pulling out the leading coefficient and completing the square. So half a -8 ^2 is 16, half of 4 ^2 is 4. Now we didn't really add 16, we really added the nine times the 16. So if I do it to one side, I have to do it to the other. We didn't really just add 4, we added 4 * 4. If I do it to one side, I have to do it to the other. So when we group our like terms in factor, we get 9 times the quantity X -, 4 ^2 + 4 times the quantity y + 2 ^2 equaling 36. Dividing everything through by a 36, we get X - 4 ^2 / 4 + 1 + 2 ^2 / 9 equaling 1. So the center here is going to be 4 negative 2. The vertice is going to be 4, 1:00 and 4:00, -5 because we look at the bigger of these two numbers, which was 9 sqrt 9 is 3 and we're going to change it in the Y direction. So if I add 3 and I subtract 3, the smaller one was the four. So I'm going to take sqrt 4 and I'm going to change it in the X direction. So I'm going to add 2 and I'm going to subtract 2. The focus a is going to be the furthest out, so a ^2 equal b ^2 + C ^2. The a was 9, the B was 4. So 9 -, 4, five square root, it positive -5 and it's going to be changing along the bigger direction or where the 9 is. So in the Y so -2 ± sqrt 5, if we wanted to graph it, we go to the center, we go up three and down 3. We go right two and left 2. If they were giving us pieces like we had our two focus and we had two vertices, if we know our focus, we know that our C has got to be two. If we know our vertices, we know our A is going to be 6. We also know due to symmetry, that the center's got to be right in the middle or at 00, so we have 00. the Y is the one that's changing for our focus and our vertices, so the bigger numbers got to be underneath the Y. So y ^2 / 6 ^2. The X ^2. We don't know what it's over yet, but we know it equals 1. So if we know A is 6 and C is 2, by the Pythagorean identity a ^2 has to equal b ^2 + C ^2. Remember the A is the furthest one out the vertices and then... The is further than the minor axis in points or the focus. So 36 -, 4 is going to give us 32 is our b ^2. So underneath the X ^2 is going to be our 32. So thank you and have a wonderful day.