Ellipses: Standard Form - Video Tutorials & Practice Problems

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 and the ellipse forms when you take a three dimensional cone and you slice it with a two dimensional plane at a slight angle. So if you slightly tilt the plane into the cone, this is going to give you this ellipse shape, which we're gonna talk about in this video. Now, 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. And 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 look at a circle already and we figured out that the circle depends on its radius and this radius tells us the distance from the center of the circle to any point on it. So we can see from this graph here that the circle has a radius of 123 units. And that means that any distance you travel to any point on the circle is always going to be a distance of three from the center. Now, this equation for a circle looks like this where we have X squared plus Y squared equals the radius squared. Now, I want you to imagine that we take this circle and we stretch it horizontally. If we were to stretch this circle, we would end up getting a shape which looks something like that. But notice that this shape is a bit different than the circle and that it doesn't have the same symmetry all the way around because the distance from the center of the ellipse to this point is not the same as the distance from the center to that point. So in order to define this new shape, there are two distances that we're going to need to keep track of. And this shape is what we call the ellipse because there are going to be two distances which are the semi major axis and the semi minor axis that define the shape. Now the semi major axis is just going to be the larger of the two axes. So if I look at a here, I can see the A is 1234 units long. And I can see that B here is 123 units long. And since A is bigger than B, that means A is going to be the semi major axis whereas B is going to be the semi minor axis. So that's the main idea of the horizontal ellipse. But what do you think would happen if we took the same circle and now stretched it vertically instead of horizontally? Well, if we vertically stretch the circle, we would end up getting a vertical ellipse. And the way that the vertical ellipse works is the major axis or where the ellipse is stretched is now going to be on the Y rather than the X. Now, if we actually take a look at the major and minor axis, we can see that this A value is still going to be the same, it's going to be 1234 units. But this time, the A is going to be up and down rather than left and right. And if we look at our B value, well, we can see that B is still 123 units long, but now the B is left and right rather than being up and down like before. So we can see that A is still the semi major axis and B is still the semi minor axis. But now it's just the directions that are switched. So this is what would happen if you had a vertically stretched ellipse. Now, it's also important to know how the equations are going to work. So when it comes to the horizontal ellipse, this is what the equation is going to look like. Notice how we have the semi major axis squared, which is underneath the X when we are stretched on the X axis and we have the semi minor axis square underneath the Y. Now, when it comes to the vertical ellipse, the way that this equation is going to be different is we're going to have a squared and B squared switch. So we'll have B squared underneath the X and we'll have a squared underneath the Y. And this is because we're now stretched on the Y axis. So A, the major axis shows up under the Y, whereas B shows up under the X. Now, in both of these equations, it's only going to be the A and the B that you need to plug in or replace the variables with. So if we go over to the horizontal, our equation is going to be X squared over A squared and A squared is four squared or 16 plus Y squared over three squared, which is nine and that's going to be equal to one. And for our vertical ellipse, we're going to have X squared over B squared, which is X squared over nine plus Y squared over A squared, which is 16 is equal to one. So this would be the equation for our horizontal ellipse. And that would be the equation for the vertical ellipse. And it's important to know that whenever you see the variable A a always tells you the longest distance from the center of the ellipse to the outside. Now, one more thing I want to mention as just kind of a way to remember these equations is you may notice that the ellipse equation versus the circle equation are a bit different. But if you actually take the circle equation and rewrite it by dividing R squared by every term which you can do. If you do it to the both sides of the equal sign, you can cancel the R squared giving you X squared over R squared plus Y squared over R squared is equal to one. And notice how this equation looks very similar to the other two equations for the ellipses. The only main difference is this distance is the same. It's the radius because we have symmetry all the way around the circle. Whereas these two equations have the two different distances A and B because we don't have the same symmetry all the way around the shape. So this is the main idea and concept behind the graphs of ellipses at the origin where the center is right in the middle of the graph. So 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 this 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 foi. Now when it comes to vertices and foi, there are two vertices and two foi 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 foi 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. So 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 farthest from the center of the ellipses. And it's gonna be along the major axis which the ellipse is stretch. So these are the vertices. Now you'll also need to know how to find the foi and the foi are going to be the points that tell you the general symmetry of the ellipse. So the foresite could be right here, for example. And what these tell you is that the sum of any distance from the foresite to a single point are 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, to a point on the and this distance is one. And let's say we have a distance right there, which is 33 plus one is equal to four. And what the fite tell us is that any point we look at the, at the ellipse, the distance there will always sum to four. 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 four. So notice how they show you this general symmetry all the way around the ellipse. Now, in order to calculate the vertices, you are 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 two vertex points. And in order to find the foi 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 to the A scored and the B squared together. So you get that C scored is equal to A squared minus B squared. And that will allow you to calculate the distance to the 24 side. 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 foi 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 five. 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 five. So we need to go 12345 units to the right and 12345 units to the left. And this will tell us two vertices which I can see are at 50 and negative 50. Now, we're also asked to find the foi and to find the foi, 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 nine. And if I take the square root on both sides of this equation, the square root of nine will give us three. So our B value is three. Now, from here, I can calculate the C value which will tell us the distance to the two F side. So I can see here that C squared is A squared minus B squared and A squared is going to be five squared which is 25 minus B squared, which is three squared or nine. Now 25 minus nine, that's equal to 16. And 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 four. So C value is four, meaning if we start at the center of our ellipse, we can go 1234 units to the right and 1234 units to the left and this will get us to the two pho which we can see are at 40 and negative 40. So this is how you can calculate the vertices and foi 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 fite 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 zero and negative A zero which ended up landing us right here and right there. Now we also saw that the two foci were also on the major axis of the ellipse and they ended up being a distant C and a distant C would be somewhere like here and there. So basically the vertices and foresite end up on the X axis when looking at a horizontal ellipse. Now, if we instead had a vertical ellipse, the vertices and foresite 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 you F I are going to end up somewhere around here and there, which are also on the Y axis at zero C zero negative C. So for a vertical ellipse, the vertices and foi are on the Y axis and for a horizontal ellipse, the vertices and foresite are on the X axis. So that's the basic idea of finding the vertices and foresite 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. Now, what would happen if we had our ellipse that was shifted to some new location which we'll call hk how could we deal with these types of ellipses? Well, that's what we're gonna be looking at in this video and it turns out the equation and graph does 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. So if you have an ellipse that is not at the origin, it is going to be shifted by a certain amount HK where H is the horizontal position and K is the vertical position. Now, 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. 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 shift respectively. So 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. So in this example, we are given this equation and we're asked to graph the following ellipse. Now our first step should be to determine the major and minor axis. And to do this, what I can do is look at the equation that we have and see what these are. Now recall that the major axis A squared is associated with the larger number in the denominator. Since I can see that the larger number is 64 we can say that A squared is 64. This means that A would be the square root of 64 which is eight. Now we can also see here that B squared is going to be associated with the smaller number. That's the minor axis. So B squared would be equal to nine. So if B squared is equal to nine, then B is going to be the square root of nine, which is three. So that's our A and B which is our major and minor axis. Now, our second step asks us to find the orientation of the ellipse, whether it's vertical or horizontal. Now, the way I can do this is by looking for the larger number, which is the major axis square. And I can see that that's 64 and 64 is underneath the Y. So because the larger numbers underneath the Y, that means we're going to be stretched on the Y axis, meaning we're dealing with a vertical ellipse. Now, the third step asks us to find the center of our lips and to do this, I need to look for H and K. Well, 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 four. So our H is going to be four and our K value U is going to be associated with what's subtracted from the Y, which is two. So our K is two. So that means the center of our ellipse is going to be at the horizontal position of four, which is right there and the vertical position of two. So this is the center. Now, our fourth step asks us to find the vertices of our ellipse. And since we're dealing with a vertical ellipse, we're going to use these coordinates right here to find the vertices and this tells us we need to add and subtract our major axis from the K value, which is the 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. So since I see that our a value is eight, that means that we need to go up eight units which would put us right here at 410. So we're gonna have a point at four comma 10. And then I need to go down eight units which would put us down here at four, negative six. So this is how you can find the vertices. Now, our next step asks us to find the B points and the B points are going to be the vertical oriented elliptic because that's what we have and notice how this asks us to add and subtract from the H value. That means we need to go left and right. So to find the B points. Well, if I start here at the center, I can take the, the horizontal position, we have H which is four and I can subtract off three. And that will put us back here because four minus three is positive one. So that put us at that point. And then what I need to do is add four to add 3 to 4. And that would put us O over here between six and eight, which would be at seven. So the B points are going to be here at 12 and there at 72. Now, our last step asks us to connect to these points with a smooth curve to define the ellipse. What we need to do is take these outside points and connect them with a smooth curve. And this will give us the graph of our ellipse. Now, one more thing that I want to mention is that it's also important to recognize how we could find the folk side of an ellipse that's not of the origin. And it's actually pretty straightforward because when dealing with a vertical ellipse, what you want to do is use these coordinates for the foi and when dealing with a horizontal ellipse, you want to use those coordinates for the fori. And in this example, we had a vertical ellipse and all this is saying is that the C value we talked, 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 folk side, they're going to be somewhere up here and somewhere down there. And that is how you can find the folk side of an ellipse. So 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: $\left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right)$