Ellipses: Standard Form - Online Tutor, Practice Problems & Exam Prep

Ellipses are formed by slicing a cone with a plane at an angle, resulting in two axes: the semi-major axis (a) and the semi-minor axis (b). The equations for horizontal and vertical ellipses differ based on their orientation. For a horizontal ellipse, the equation is $(x^2/a^2+y^2/b^2)=1$. The vertices and foci are critical points that define the ellipse's shape and symmetry, calculated using $c=\sqrt{(a^2-b^2)}$.

Graph Ellipses at Origin

Welcome back, everyone. So up to this point, we've been talking about conic sections. And in the last video, we looked at the circle shape. In this video, we're going to take a look at the ellipse. The ellipse forms when you take a three-dimensional cone and slice it with a two-dimensional plane at a slight angle. If you slightly tilt the plane into the cone, this is going to give you the ellipse shape, which we're going to talk about in this video. The ellipse is a bit more complicated than the circle because there's more that you need to keep track of when it comes to your graph and equation. But don't sweat it because in this video, we're going to figure out that there are actually some similarities between the circle and the ellipse. Even though it's a bit more complicated, the problem should be pretty straightforward once you know how the equations and graph work. So let's get right into this.

Now we've looked at a circle already, and we figured out that the circle depends on its radius. This radius tells us the distance from the center of the circle to any point on it. From this graph here, the circle has a radius of 3 units. That means any distance you travel to any point on the circle is always going to be a distance of 3 from the center. The equation for a circle looks like this:

x2 + y2 = r  2

Imagine that we take this circle and stretch it horizontally. This new shape, different from the circle, does not have the same symmetry all around. In order to define this new shape, we need to keep track of two distances: the semi-major axis and the semi-minor axis. When a (4 units long) is larger than b (3 units long), a is the semi-major axis and b the semi-minor axis. That's the main idea of the horizontal ellipse. What if we stretched the same circle vertically instead of horizontally? The vertically stretched ellipse now has the major axis on the y rather than the x, switching the directions of a and b.

The equations are important to note: For the horizontal ellipse:

x2 a2 + y2 b2 = 1

And for the vertical ellipse:

x2 b2 + y2 a2 = 1

A simple way to remember these equations is by comparing them to the circle equation when divided by r squared:

x2 r2 + y2 r2 = 1

This looks very similar to the ellipse equations, with the only main difference being a and b, due to the lack of symmetry all the way around the ellipse. This is the main idea and concepts behind the graphs of ellipses at the origin where the center is right in the middle of the graph. I hope you found this video helpful. Thanks for watching, and let's move on.

Given the equation $\frac{x^2}{4}+\frac{y^2}{9}=1$, sketch a graph of the ellipse.

• A.
• B.
• C.
• D.

Given the ellipse equation $\frac{x^2}{16}+\frac{y^2}{4}=1$, determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

• A. $a=16$, $b=4$
• B. $a=4$, $b=16$
• C. $a=4$, $b=2$
• D. $a=2$, $b=4$

Foci and Vertices of an Ellipse

Hey, everyone, and welcome back. So in the last video, we got introduced to the idea of an ellipse, and how you can graph this shape. And in this video, we're going to be learning about new elements of the ellipse, specifically the vertices and foci. Now, when it comes to vertices and foci, there are 2 vertices and 2 foci for each ellipse. And sometimes this could be a bit tedious to calculate, because you're going to need to know different variables and equations to be able to find the vertices and foci. But don't worry about it because in this video, we're going to find out that the vertices and foci are just these various points that you can find along the major axis of the ellipse, basically, however, the ellipse is stretched. So without further ado, let's get right into this.

Now, when it comes to the vertices, these are going to be the points on the ellipse that are furthest away from the center. So if you had a horizontally stretched ellipse, the vertices are going to be the two points that are furthest from the center of the ellipses, and it's going to be along the major axis which the ellipses stretch. So these are the vertices. Now you'll also need to know how to find the foci, and the foci are going to be the points that tell you the general symmetry of the ellipse. So the foci could be right here, for example. And what these tell you is that the sum of any distance from the foci to a single point on the ellipse is always going to be constant.

Now if this sounds a little confusing, let me explain. Let's say we have this distance right here to a point on the ellipse, and this distance is 1. And let's say we have a distance right there, which is 3. 3 plus 1 is equal to 4, and what the foci tell us is that any point we look at on the ellipse, the distance there will always sum to 4. So, if we had a point right here on the ellipse, and we measured this point and that point, adding these two distances together would also give us 4. So notice how they show you this general symmetry all the way around the ellipse.

Now, in order to calculate the vertices, you're going to need the distance a, which is the distance of the semi-major axis. And that makes sense, because if you travel the semi-major axis this direction or that direction, it will get you to the 2 vertex points. And in order to find the foci, you will need a distance c, which is a distance we're going to talk about right now.

Now recall the equation for an ellipse, which looks something like this. And we mentioned how a squared and b squared are in the denominators of x and y depending on how your ellipse is stretched. Well, in order to calculate c, you're going to need this equation, which relates the a squared and the b squared together. So you get that c2=a2-b2, and that will allow you to calculate the distance to the two foci.

Now to make sense of this equation and how this relates to the ellipse, let's actually take a look at an example where we are asked to find the vertices and foci of an ellipse. Now a squared is the largest number that you see in the denominator for the ellipse equation. And I can see that that largest number is 25. So what that means is that a squared is equal to 25. So to find a, I just need to take the square root on both sides of this equation. The square root of 25 will give us 5, and that is our distance for the semi major axis. Now since we have a calculated, this will allow us to calculate our vertices. So if I go over to this ellipse, I can start at the center here, and our a value is 5, so we need to go 1, 2, 3, 4, 5 units to the right, and 1, 2, 3, 4, 5 units to the left. And this will tell us our 2 vertices, which I can see are at 50 and negative 50.

Now we're also asked to find the foci. And to find the foci, I first need to identify b. Well, I can see here that b or b squared is going to be the smaller number. So that means that b squared is equal to 9. And if we take the square root on both sides of this equation, the square root of 9 will give us 3. So our b value is 3. Now from here, I can calculate the c value, which will tell us the distance to the 2 foci. So I can see here that c squared is a squared minus b squared, and a squared is going to be 5 squared, which is 25, minus b squared, which is 3 squared or 9. Now 25 minus 9, that's equal to 16. If I take the square root on both sides of this equation, we're going to get that c is equal to the square root of 16, which is 4. So our C value is 4, meaning if we start at the center of our ellipse, we can go 1, 2, 3, 4 units to the right, and 1, 2, 3, 4 units to the left. And this will get us to the 2 foci, which we can see are at 40 and negative 40.

So this is how you can calculate the vertices and foci for an ellipse that's centered at the origin like we have here. Now, something else I want to mention is that if you're dealing with a vertical ellipse rather than a horizontal ellipse, the coordinates for the vertices and foci are going to be a bit different. And to understand this better, well, we saw that for the horizontal ellipse that our vertices ended up being at the distance of the semi-major axis. So it was a 0 and negative a 0, which ended up landing us right here and right there. Now we also saw that the 2 foci were also on the major axis of the ellipse, and they ended up being a distance c, and a distance c would be somewhere like here and there. So basically the vertices and foci end up on the x-axis when looking at a horizontal ellipse. Now if we instead had a vertical ellipse, the vertices and foci would actually end up on the y-axis. So what you would end up seeing is that the vertices are going to end up here and there, where a is the y coordinate this time, and then you see that your foci are going to end up somewhere around here and there, which are also on the y axis at 0 c, 0 negative c. So for a vertical ellipse, the vertices and foci are on the y-axis, and for a horizontal ellipse, the vertices and foci are on the x-axis. So that's the basic idea of finding the vertices and foci of an ellipse. Hope you found this video helpful, and thanks for watching.

Determine the vertices and foci of the following ellipse: $\frac{x^2}{49}+\frac{y^2}{36}=1$.

• A. Vertices: $\left(7,0\right),\left(-7,0\right)$Foci: $\left(6,0\right),\left(-6,0\right)$
• B. Vertices: $\left(6,0\right),\left(-6,0\right)$Foci: $\left(7,0\right),\left(-7,0\right)$
• C. Vertices: $\left(7,0\right),\left(-7,0\right)$Foci: $\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)$
• D. Vertices: $\left(0,7\right),\left(0,-7\right)$Foci: $\left(0,\sqrt{13}\right),\left(0,-\sqrt{13}\right)$

Determine the vertices and foci of the following ellipse: $\frac{x^2}{9}+\frac{y^2}{16}=1$.

• A. Vertices: $\left(4,0\right),\left(-4,0\right)$Foci: $\left(\sqrt7,0\right),\left(-\sqrt7,0\right)$
• B. Vertices: $\left(0,4\right),\left(0,-4\right)$Foci: $\left(0,\sqrt7\right),\left(0,-\sqrt7\right)$
• C. Vertices: $\left(4,0\right),\left(-4,0\right)$Foci: $\left(3,0\right),\left(-3,0\right)$
• D. Vertices: $\left(0,4\right),\left(0,-4\right)$Foci: $\left(0,3\right),\left(0,-3\right)$

Find the standard form of the equation for an ellipse with the following conditions.

Foci = $\left(-5,0\right),\left(5,0\right)$

Vertices = $\left(-8,0\right),\left(8,0\right)$

• A. $\frac{x^2}{64}+\frac{y^2}{25}=1$
• B. $\frac{x^2}{25}+\frac{y^2}{64}=1$
• C. $\frac{x^2}{8}+\frac{y^2}{5}=1$
• D. $\frac{x^2}{64}+\frac{y^2}{39}=1$

Graph Ellipses NOT at Origin

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 shift to a new location, which we'll call h and k? How could we deal with these types of ellipses? Well, 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 figure out 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 \(h\) and \(k\), 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. And when it comes to an ellipse not at the origin, the equation looks like that for this kind of ellipse. Notice that it's actually very similar to the other equation. The only difference is we have \(h\) and \(k\). But just like when doing a function transformation where we have a shift, the \(h\) and \(k\) represent the horizontal and vertical shift 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 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, we can conclude that \(a^2\) is 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, which is the minor axis. So \(b^2\) would be equal to 9. So if \(b^2\) is equal to 9, then \(b\) is going to be the square root of 9, which is 3. That's our \(a\) and \(b\), which are our major and minor axes.

The second step asks us to find the orientation of the ellipse, whether it's vertical or horizontal. The way I can do this is by looking for the larger number, which is the major axis square. I can see that it's 64, and 64 is underneath the \(y\). So because the larger number is underneath the \(y\), it 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 \(k\). I can see based on the equation that we have that \(h\) is going to be what's subtracted from the \(x\), and I can see here that we have \(x\) minus 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. That 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. Since our \(a\) value is 8, that means that we need to go up 8 units, which would put us right here at (4, 10). So 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). This is how you can find the vertices.

The next step asks us to find the \(b\) points. The \(b\) points are going to be for vertical-oriented ellipses, and 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 minus 3 is 1. So that put 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 and 8, which would be at 7. So the \(b\) points are going to be here at (1, 2) and there at (7, 2).

The 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 pretty straightforward because when dealing with a vertical ellipse, you want to use these coordinates for the foci. And 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. And that's how you can find the foci of an ellipse. I hope you found this video helpful. Thanks for watching.

Graph the ellipse $\frac{\left(x-1\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$.

• A.
• B.
• C.
• D.

Determine the vertices and foci of the ellipse $\left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1$.

• A. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right)$
• B. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-2,2\right),\left(0,2\right)$
• C. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(1,2+\sqrt3\right),\left(1,2-\sqrt3\right)$
• D. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right)$