# Relationship between foci and vertices

### How do you find the equation of an ellipse with vertices (0,+-8) and foci (0, +-4)? | Socratic

The distance of every point on parabola curve from its focus point and from its directrix is always same. Remember the two patterns for an ellipse: Each ellipse has two foci (plural of focus) as shown in the picture here: As you can see, c is the distance from the. You can click on the link if you'd like to see the development, but the short version is that Move p units along the axis of symmetry from the vertex to the focus.

And if I were to measure the distance from this point to this focus, let's call that point d3, and then measure the distance from this point to that focus -- let's call that point d4. If I were to sum up these two points, it's still going to be equal to 2a. Let me write that down. And, actually, this is often used as the definition for an ellipse, where they say that the ellipse is the set of all points, or sometimes they'll use the word locus, which is kind of the graphical representation of the set of all points, that where the sum of the distances to each of these focuses is equal to a constant.

And we'll play with that a little bit, and we'll figure out, how do you figure out the focuses of an ellipse. But the first thing to do is just to feel satisfied that the distance, if this is true, that it is equal to 2a.

And the easiest way to figure that out is to pick these, I guess you could call them, the extreme points along the x-axis here and here. We're already making the claim that the distance from here to here, let me draw that in another color. That this distance plus this distance over here, is going to be equal to some constant number. And using this extreme point, I'm going to show you that that constant number is equal to 2a, So let's figure out how to do that.

So one thing to realize is that these two focus points are symmetric around the origin. So, whatever distance this is, right here, it's going to be the same as this distance. Because these two points are symmetric around the origin. So this, right here, is the same distance as that.

### Equation of an Ellipse in standard form and how it relates to the graph of the Ellipse.

And, of course, we have -- what we want to do is figure out the sum of this distance and this longer distance right there. Well, what's the sum of this plus this green distance? Well, this right here is the same as that. So this plus the green -- let me write that down. So let me write down these, let me call this distance g, just to say, let's call that g, and let's call this h. Now, if this is g and this is h, we also know that this is g because everything's symmetric.

So what's g plus h? Well, that's the same thing as g plus h. Which is the entire major diameter of this ellipse.

• Equation of an Ellipse
• Focus of Ellipse
• What is the relationship between a parabola's curve, directrix, and focus point?

Well, we know the minor radius is a, so this length right here is also a. So the distance, or the sum of the distance from this point on the ellipse to this focus, plus this point on the ellipse to that focus, is equal to g plus h, or this big green part, which is the same thing as the major diameter of this ellipse, which is the same thing as 2a.

Hopefully that that is good enough for you. Now, the next thing, now that we've realized that, is how do we figure out where these foci stand. Or, if we have this equation, how can we figure out what these two points are?

Let's figure that out. So, the first thing we realize, all of a sudden is that no matter where we go, it was easy to do it with these points. But even if we take this point right here and we say, OK, what's this distance, and then sum it to that distance, that should also be equal to 2a. And we could use that information to actually figure out where the foci lie.

So, let's say I have -- let me draw another one.

## Finding the Foci of an Ellipse

So that's my ellipse. And then we want to draw the axes. Let me write down the equation again. Just so we don't lose it. Let's take this point right here. These extreme points are always useful when you're trying to prove something. Or they can be, I don't want to say always. Now, we said that we have these two foci that are symmetric around the center of the ellipse.

This is f1, this is f2. And we've already said that an ellipse is the locus of all points, or the set of all points, that if you take each of these points' distance from each of the focuses, and add them up, you get a constant number. And we've figured out that that constant number is 2a. So we've figured out that if you take this distance right here and add it to this distance right here, it'll be equal to 2a.

So we could say that if we call this d, d1, this is d2. We know that d1 plus d2 is equal to 2a. And an interesting thing here is that this is all symmetric, right? This length is going to be the same, d1 is is going to be the same, as d2, because everything we're doing is symmetric. These two focal lengths are symmetric. This distance is the same distance as this distance right there. So, d1 and d2 have to be the same. There's no way that you could -- this is the exact center point the ellipse.

### What is the relationship between a parabola's curve, directrix, and focus point? | Socratic

The ellipse is symmetric around the y-axis. So if d1 is equal to d2, and that equals 2a, then we know that this has to be equal to a. And this has to be equal to a. I think we're making progress. And the other thing to think about, and we already did that in the previous drawing of the ellipse is, what is this distance? This distance is the semi-minor radius. Which we already learned is b. And this of course is the focal length that we're trying to figure out. This should already pop into your brain as a Pythagorean theorem problem.

So we have the focal length. And we could do it on this triangle or this triangle. I'll do it on this right one here. This focal length is f. Let's call that f.

Which is equal to a squared.

And now we have a nice equation in terms of b and a. We know what b and a are, from the equation we were given for this ellipse. So let's solve for the focal length. The focal length, f squared, is equal to a squared minus b squared. So, f, the focal length, is going to be equal to the square root of a squared minus b squared. Pretty neat and clean, and a pretty intuitive way to think about something. So you just literally take the difference of these two numbers, whichever is larger, or whichever is smaller you subtract from the other one.

You take the square root, and that's the focal distance. Now, let's see if we can use that to apply it to some some real problems where they might ask you, hey, find the focal length. Or find the coordinates of the focuses. So let's add the equation x minus 1 squared over 9 plus y plus 2 squared over 4 is equal to 1. So let's just graph this first of all. That constant will be 2a. You can use that definition to derive the equation of an ellipsebut I'll give you the short form below.

The ellipse is a stretched circle. Begin with the unit circle circle with a radius of 1 centered at the origin. What you have done is multiplied every x by a and multiplied every y by b. In translation form, you represent that by x divided by a and y divided by b. The center is the starting point at h,k. The major axis contains the foci and the vertices. This is also the constant that the sum of the distances must add to be. The foci are within the curve. Since the vertices are the farthest away from the center, a is the largest of the three lengths, and the Pythagorean relationship is: Hyperbola A hyperbola is "the set of all points in a plane such that the difference of the distances from two fixed points foci is constant".

The difference of the distances to any point on the hyperbola x,y from the two foci c,0 and -c,0 is a constant. The absolute value is around the difference so that it is always positive. You can use that definition to derive the equation of a hyperbolabut I'll give you the short form below.

The only difference in the definition of a hyperbola and that of an ellipse is that the hyperbola is the difference of the distances from the foci that is constant and the ellipse is the sum of the distances from the foci that is constant.

The graphs, however, are very different. The Transverse axis contains the foci and the vertices. This is also the constant that the difference of the distances must be. Since the foci are the farthest away from the center, c is the largest of the three lengths, and the Pythagorean relationship is: