 So, fair warning, my ordinarily malifluous voice is going to be rather badly affected by the spraying allergy season. So let's talk more about conic sections. Remember if the eccentricity is greater than 1, then the point on the conic is farther from the focus than from the directrix, and this produces a hyperbola. Now since the focus is farther from the point than the directrix is, there are actually points on the hyperbola on the opposite side of the directrix. And the important thing here is the two curves do not connect, and we say the hyperbola has two branches. And there are two special points, a point directly between the focus and the directrix, and a point on the far side of the directrix on the same line, and these are known as the vertices of the hyperbola. And the center will be the midpoint of the vertices. Finally the hyperbola itself will run through the vertices, and it will separate the center from the focus. And again with some effort and a lot of algebra, we can find the following. The hyperbola centered at 0, 0 with vertical directrix has the equation x squared divided by a squared minus y squared divided by b squared equals 1, and our vertices will be located at a0, negative a0, and foci at c0, negative c0, where a squared plus b squared equals c squared. And similarly if our directrix is horizontal, we'll switch the x and y coordinates. And as with ellipses, the distance from the center to the focus is the focal distance, and the eccentricity is the ratio of the focal distance to the distance of the vertex, c divided by a. And again, as with ellipses, the hyperbola is symmetric about the x and y axes, so there will be two foci. For example, let's consider this hyperbola. So we can start with hyperbola y squared divided by 5 squared minus x squared divided by 4 squared, and this is a hyperbola with a horizontal directrix, and the vertices at 0a, the denominator of the y squared term, 0 minus a, and foci at 0c, 0 minus c, where a squared plus b squared equals c squared. So our hyperbola will have horizontal directrix, center 0, 0, vertices at 0, 5, and 0, negative 5. So we'll need to calculate that focal distance, and so c squared is equal to a squared plus b squared, and so we find. And so our foci are going to be located at 0 root 41 and 0 negative root 41. Finally our eccentricity is the ratio of the focal distance to the semi-major axis, and so that's square root 41 divided by 5. Now let's go ahead and graph these points. Our center at 0, 0, 2 vertices at 0, 5, and 0 negative 5, and the foci at 0 root 41, 0 negative root 41. And since we know where the vertices and the foci are, we can draw the two branches of the hyperbola. Well that's great, we now know everything there is to know about the hyperbola y squared divided by 5 squared minus x squared divided by 4 squared, but this isn't that hyperbola. So we need to add a few translations. So we'll translate this hyperbola one unit to the left, then three units upward. And what's important to recognize is that all features of the graph undergo these same transformations, and so our center moves one unit to the left, and three units up to negative 1 and 3. The vertices move from 0, 5, and 0 negative 5, one unit left, and three units upward. So they move to, and our foci move from 0 root 41 and 0 negative root 41, again one unit left, and three units upward. And the eccentricity is unaffected by this translation.