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Current time:0:00Total duration:13:49

Foci of an ellipse from equation

Video transcript

let's say we have an ellipse formula x squared over a squared plus y squared over B squared is equal to one and for the sake of our discussion we'll assume that a is greater than B and then well all that does for us is it lets us know this is going to be kind of a short and fat ellipse or that the semi-major axis is going to or the the major axis is going to be along the horizontal and the minor axis along the vertical and let's draw that let me draw this ellipse I want to draw a thicker ellipse let's say that's my ellipse and let me draw my axes okay this is my the horizontal right there and there we have the vertical and we've you know we've we've studied ellipse in pretty good detail so far we know how to figure out it's the the semi-minor the semi-minor radius which in this case we know is B that's the same B right there and that's only the semi-minor radius because B is smaller than a if B was greater it would be the major radius and then of course the major radius is a and that distance is this right here now another super interesting and perhaps the most interesting property of an ellipse is that if you take any point on ellipse on an ellipse and measure the distance from that point to two special points which we for the sake of this discussion and not just for the sake of discussion for pretty much forever we will call the focuses or the foci of this ellipse and these two points they always sit along the major axis so in this case it's the horizontal axis and they're symmetric around the center of the ellipse so let's just call that those these points let me call this one f1 and this is f2 and it's for focus focuses f2 so the super interesting fascinating property of an ellipse and it's often used as the definition of the lips is if you take any point on this ellipse and measure its distance to each of these two points so let's say that I have this distance right here let's call this distance let's call this distance d1 and then I have this distance over here so I'm taking any point on that ellipse or this particular point and I'm measuring the distance to each each of these two foci and this is d2 we have to do it in a different color I'm reusing all the color so this is d2 this whole line right here that's d2 so when you take when you find these two distances you sum them up so this d2 plus d1 this is going to be a constant that it actually turns out is equal to 2a but it turns out that it's true anywhere you go on the ellipse oh let me make that point clear and I'm actually going to prove to you that this constant distance is actually 2a where this a is the same as that a right there so just to make sure you understand what I'm saying so let me take another arbitrary point on this ellipse let's take it right there and if I were to take measure the distance from this point to this focus let's call that let's call that point D 3 D 3 and then measure it from measure the distance from this point to that focus let's call that d 4 d 4 if I were to sum up these two points it's still going to be equal to 2a let me write that down D 3 + d 4 is still going to be equal to 2a that's just neat 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 I'll use the word locus which is kind of the the graphical representation of the set of all points that where the sum of the distances from from to each of these focuses is equal to a constant and we'll play with that a little bit we'll figure out how do you figure out the focuses of an ellipse but the first thing to do is just to 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 kind of these these I guess you could call them the extreme points along the x-axis here and here right 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 is equal to 2ei so let's figure out how to do that so one thing to realize 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 right there right because these two points are symmetric around the origin so this right here is the same distance as that right 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 Green let me write that down so let me write down these let me call this distance I don't know let me call it let me call it G just to have say let's call that G and let's call this H if this is G 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 which is what 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 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 to a fair enough hopefully hopefully that that is good enough for you now the next thing to now that we've realized that is how do we figure out where these faux size-10 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 all of a sudden does that no matter where we go it was easy to do these points but even if we take these put this point right here and we say okay what's this distance and then summit to that distance that should also be equal to 2ei and we could use that information to actually figure out where the folks I lie so let's say I have let me draw another let me draw another one okay now am I using the right okay so this is nope don't want to draw a circle so that's my ellipse and then I want to draw the axes for clarity now let me write down the equation again just so we don't lose that x squared over a squared plus y squared over B squared is equal to one 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 right 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 it each of these points distance from each of the focuses and add them up you get a constant number we figured out that constant number is 2a so we figured out if if you take this distance right here and add it to this distance right here it'll be equal to 2a right so we could say that you know if we call this D d1 this is d2 we know that d1 plus d2 is equal to 2a an interesting thing here is that this is all symmetric right this length is going to be the same d1 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 of the ellipse this ellipse is symmetric around around the 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 fair enough I think we're making progress the other thing to think about and we've 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 what is it we have the focal length and we can do it on this triangle or this drawing I'll do it on this fried one here this focal length is f let's call that F F squared plus B squared is going to be equal to the hypotenuse squared which in this case is d2 or a 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 our from our 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 pretty intuitive way to think about something so you just literally take the difference of these two numbers whichever is larger you you or whichever is smaller you subtract from the other when 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 to some real problems where where they might ask you a find find the focal length or find the coordinates of the focuses so let's say I had the equation X minus 1 squared over 9 plus y plus 2 squared over 4 is equal to 1 so let's let's just graph this first of all this could be interesting so I'll draw the axes that's the x axis this is the y axis and we immediately see what's the center of this the center is going to be at the point 1 negative 2 and if that's confusing you might want to review some of the previous videos centers at 1 X is equal to one y is equal to minus two that's the center and then this the major axis is the x axis because this is larger and so B squared is or a squared is equal to 9 or the semi-minor radius is going to be equal to 3 so if we go one two three one two three go there then you have one two three no one two three one two three I think this let's see one two three you go there roughly and then in the Y direction the semi-minor radius is going to be to write the square root of that so b is equal to two so you go up two then you go down two and this ellipse is going to look something like pick a good color it's going to look something like this fair enough and what we'd want to do is we want to find out the coordinates of the focal points so the focal points are going to sit along the semi-major axis right we need to figure out these focal distances and then we can essentially just add and subtract them from the center and then we'll have the coordinates what we just showed you all right hopefully I showed you that the the focal length or this distance F the focal length is just equal to the square root of the disk of the difference between these two numbers right so it's the square root of nine minus four so the focal length is equal to the square root of five so if this point right here is the point we already showed that this is the point the center of the ellipse is the point 1 comma minus 2 this the coordinate of this focus right there is going to be 1 plus the square root of 5 comma minus 2 and the coordinate of this focus right there is going to be 1 minus the square root of 5 comma minus 2 and all I did is I took the focal length and I subtract since we're along the major axis or the x axis I just add and subtract this from the x coordinate to get these two coordinates right there so anyway this is kind of the really neat thing about conic sections and it has these interesting properties in relation to these foci or in relation to these focus points and then in future videos I'll show you the foci of a hyperbola or the the foci of a well it only has one focus of a of a parabola but this is really starting to get into what makes conic sections need everything we've done up to this point has been much more about the mechanics of graphing and plotting and figure out the Centers of conic sections but now we're getting into the little bit of the the mathematical interesting parts of conic sections anyway see in the next video