 So let's consider the unit circle centered at the origin, which has equation x squared plus y squared equals 1. Now if we apply a horizontal stretch by a factor of a, we obtain a curve with equation x divided by a squared plus y squared equal to 1. If we then apply a vertical stretch by a factor of b, we obtain a curve with equation x divided by a squared plus y divided by b squared equals 1. But this is the equation for a familiar curve, and so we have this connection between circles and ellipses. Namely that the ellipse is the unit circle centered at the origin stretched horizontally by a factor of a and vertically by a factor of b. Now in a circle the distance from the center to the circle itself is the same. But if we stretch the circle to form an ellipse, the greatest distance from the center to the ellipse forms the semi-major axis, and the least distance from the center to the ellipse forms the semi-minor axis. And in the equation, since a and b represent the horizontal and vertical stretches, these semi-major and semi-minor axes will correspond to the greater and lesser of the values a and b. It may be helpful to remember that the semi-major and semi-minor axes are analogous to the radius of a circle, and the major and minor axes, no semi, are analogous to the diameter of a circle. The ellipse also has two special points corresponding to the points nearest and farthest from the directrix. These are called the vertices, and they correspond to the endpoints of the major axis, and the formula for finding the vertices is Well, remember, don't memorize formulas. Understand concepts. If we know where the endpoints of the major axis are, we know where the vertices are. Now, if we move our ellipse to the origin, we'll have it in this form, and let's talk symmetry. This equation gives us a graph that is symmetric about the x-axis, the y-axis, and the origin. And what this means is that all features of the ellipse will also be symmetric about the x-axis, the y-axis, and the origin. But all features means all features, and this includes the focus. And so we could say that an ellipse has two focuses. Except we don't. Focus is actually a Latin word, and the plural of focus in Latin is spelled this way and usually pronounced foci. And so we can say that an ellipse has two. Oh, wait, let's fix this. An ellipse has two foci. Now, given the equation of the ellipse in standard form, we can find the focus as follows. If we know the length of the semi-major axis and the length of the semi-minor axis, the foci are located c units away from the center along the major axis, where c squared is a squared minus b squared, and c is called the focal distance. Meanwhile, the eccentricity is this ratio c divided by a, and that's the ratio of the focal distance to the distance between the center and the vertex. So, for example, let's sketch the ellipse with this equation. We'll find the foci, the vertices, and the eccentricity. Now, let's think about this as a transformation. So we could begin with the unit circle, x squared plus y squared equals one, then stretch horizontally by a factor of five, stretch vertically by a factor of three, and these two get us the ellipse, although it's not in the right place. To move it to the right place, we need to shift horizontally by three units and then shift vertically by two units. And the important thing to remember is that in all of this, the center undergoes the same transformations. So the unit circle we started with was centered at the origin, zero, zero, and the horizontal and vertical stretches don't affect that location, but the horizontal and vertical translations do. And so our origin moves from zero, zero, horizontally three units, vertically two units, two, three, two. Since five is greater than three, in other words, we stretched along the horizontal axis more than we stretched along the vertical axis, then our semi-major axis runs horizontally and has length five. This means the vertices will be located five units away from the center at the end points of the major axis, which will be, so to find the foci, remember the foci of an ellipse with semi-major axis a, and semi-minor axis b will be located c unit away from the center along the major axis, where c squared equals a squared minus b squared. And just in case you think I wasn't paying attention, sometimes you do memorize formulas. This one is worth memorizing because there's no easy way of recovering this from the information about the ellipse. So we have a equals five, b equals three, and c squared will be the difference between the squares. And remember c is the focal length, and so the foci will be located four units away from the center along the major axis, and since that major axis runs horizontally, that's four units to the right and four units to the left. And so they'll be located at seven, two, and negative one, two. Finally, the eccentricity is the ratio of the focal distance to the semi-major axis. So that's going to be this quantity for fifths.