Hey, everyone, and welcome back. So up to this point, we've been talking about the various shapes and conic sections, and we've specifically been looking at the ellipse in recent videos. Now, all of the ellipses we've dealt with so far have been centered at the origin, where the shape is right in the center of the graph. What would happen if we had our ellipse shifted to some new location, which we'll call (h, k)? How could we deal with these types of ellipses? That's what we're going to be looking at in this video. It turns out the equation and graph do change for this kind of ellipse, but don't sweat it because we're actually going to find that not only is a lot of this stuff straightforward, but it's also pretty familiar to what we've already learned with function transformations. So let's get right into this.
If you have an ellipse that is not at the origin, it is going to be shifted by a certain amount (h, k), where h is the horizontal position and k is the vertical position. We've seen this equation in previous videos, which deals with ellipses at the origin. The equation for an ellipse not at the origin looks like that for this kind of ellipse. And notice that it's actually very similar to the other equation. The only difference is we have this h and k. But just like we get when doing a function transformation where we have a shift, the h and k represent the horizontal and vertical shifts respectively.
To understand this a bit better, let's take a look at an example where we have to graph an ellipse that's not at the origin. In this example, we are given this equation and we're asked to graph the following ellipse. Our first step should be to determine the major and minor axes. To do this, I can look at the equation that we have and see what these are. Recall that the major axis, \(a^2\), is associated with the larger number in the denominator. Since I can see that the larger number is 64, \(a^2 = 64\). This means that \(a\) would be the square root of 64, which is 8. We can also see here that \(b^2\) is going to be associated with the smaller number; that's the minor axis. So \(b^2 = 9\). If \(b^2 = 9\), then \(b\) is going to be the square root of 9, which is 3. So that's our \(a\) and \(b\), which is our major and minor axes.
Our second step asks us to find the orientation of the ellipse, whether it's vertical or horizontal. I can do this by looking for the larger number, which is the major axis squared. I can see that it's 64, and 64 is underneath the \(y\). So because the larger number is underneath the \(y\), that means we're going to be stretched on the \(y\) axis, meaning we're dealing with a vertical ellipse.
The third step asks us to find the center of our ellipse, and to do this, I need to look for \(h\) and \(k\). Based on the equation that we have, \(h\) is going to be what's subtracted from the \(x\), and I can see here that we have \(x-4\). So our \(h\) is going to be 4. And our \(k\) value is going to be associated with what's subtracted from the \(y\) which is 2. So our \(k\) is 2. This means the center of our ellipse is going to be at the horizontal position of 4, and the vertical position of 2.
The fourth step asks us to find the vertices of our ellipse. Since we're dealing with a vertical ellipse, we're going to use these coordinates right here to find the vertices. This tells us we need to add and subtract our major axis from the \(k\) value, which is the vertical position. So, basically, what this is saying is that we need to go up and down rather than left and right to find our vertices. Since I see that our \(a\) value is 8, that means we need to go up 8 units, which would put us right here at \(4,10\). We're going to have a point at \(4, 10\) and then you need to go down 8 units, which would put us down here at \(4, -6\).
The next step asks us to find the \(b\) points. The \(b\) points are associated with vertical oriented ellipses, as we have here. Notice how this asks us to add and subtract from the \(h\) value. That means we need to go left and right. To find the \(b\) points, if I start here at the center, I can take the horizontal position we have \(h\), which is 4, and I can subtract off 3, and that will put us back here because \(4-3\) is positive 1. So that puts us at that point, and then what I need to do is add 3 to 4, and that would put us over here between 6-8 which would be at 7. So the \(b\) points are going to be here at \(1,2\), and over there at \(7,2\).
Our last step asks us to connect these points with a smooth curve. To find the ellipse, what we need to do is take these outside points and connect them with a smooth curve. This will give us the graph of our ellipse.
One more thing that I want to mention is that it's also important to recognize how we could find the foci of an ellipse that's not at the origin. It's actually pretty straightforward because when dealing with a vertical ellipse, you want to use these coordinates for the foci. And when dealing with a horizontal ellipse, you want to use those coordinates for the foci. In this example, we had a vertical ellipse, and all this is saying is that the \(c\) value we talked about calculating in the previous video, you just need to add and subtract that from the vertical position. So if you wanted to find your foci, they're going to be somewhere up here and somewhere down there. That's how you can find the foci of an ellipse. I hope you found this video helpful. Thanks for watching.
