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Current time:0:00Total duration:15:24

Foci of a hyperbola from equation

Video transcript

in the last video we learned that any lips can be defined as the locus of all points where the sum of the distances to two special points called foci and let me draw this all out so that's my x-axis the sum of the distance to these two special points called focuses or foci is a constant so if this is my ellipse let me draw it out it looks about I want to be right around that looks about right it's centered at the origin doesn't have to be but for for our purposes let's make it center to the origin if this is one focus point right here and this is the other focus point this ellipse could be defined as the set of all points or the locus of all points or if I take the distance or if any one of these points that that exist on the ellipse and take the distance to each of the locus --is sorry each of the focus is if I take that nope I don't want to do that if I take that distance and add it to this distance so let's call those I don't it's call this d1 d2 thick let's call that d1 this is d2 that that's going to be equal to a constant number along the whole ellipse so if I take a random let's say another point along the lips so I take this point right here and if I were to sum this distance if I were to sum this distance to that distance so let's call this let's call this right here d3 this is d4 the sums of these distances to the focus along this ellipse are going to be a constant so in this case D 2 plus d1 this plus that is going to be equal to D 3 plus d 4 D 3 plus d 4 and this would be true wherever you go along the whole ellipse and we learned in the last video that this quantity is actually going to be equal to 2a where a is the distance the semi-major radius if this is the formula for the ellipse this is where the a comes from x squared over a squared plus y squared over V squared is equal to 1 and we learned that the focus the focal distance or the distance from the center of ellipse of the ellipse which is this distance right here that focal distance is just this the square root of the difference of these two numbers so it's just so we could go if this is the focal distance from here to here it's just equal to the square root of if a is larger then it'd be a squared minus B squared which is the case in this ellipse if we have a vertical ellipse and I really didn't cover it in the last video but let me just show you what it would look like let's say that the ellipse looks something like this let me see all right I'll use blue let's say the ellipse look like that in this case our our semi major radius is now in the Y direction so this is so this is B this is a and in this case B is greater than a right because the the the ellipse is tall and skinny and in this case the focuses are going to alt well they're always going to lie along the major axis in this case the major axis is the vertical axis so the focus is are going to sit here and here and in this case the focal lengths are going to be vertical down from the origin and vertical up from the origin and we get instead of it being a squared minus B squared now since B is larger than a the focal length which is this the focal length is going to be equal to V squared minus a squared fair enough now I did all of that to kind of compare it to what we're going to cover in this video which is the focus points or the foci of a hyperbola and hyperbola it's it's very close to an ellipse I mean you could you could probably guess that because if this is an equation of an ellipse this is the equation of a hyperbola x squared over a squared minus I squared over B squared is equal to 1 or we could switch these around where the minus is in front of the X instead of the Y and we could cover that in a second but this hyperbola looks something like this let me see if I can draw it see if I draw the axes and then I want to draw the asymptotes you could prove it you could look at the some of the previous videos but the asymptotes for this hyperbola are going to be Y is equal to plus or minus B over a X just going to look I'll just draw a Miss kind of tilted lines so it would look something like that something like nope I want to make it something like that those would be asked does this one centered at the origin because it's not it hasn't been shifted and then that's those two lines right there and then this is what I call kind of a horizontal hyperbola the way you could think about that is well if you if you solve for y you'll see that you're always going to be a little bit lower than the asymptote the other option is you say okay can x or y equals 0 well if Y is equal to 0 that puts us along the x axis right and you get x squared over a squared is equal to 1 or X so if Y is equal to 0 you get x squared over a squared is equal to 1 which means that x squared is equal to a squared which means that x is equal to plus or minus a so the points a comma 0 so this is the point a comma 0 and the point minus a minus a comma 0 are both on this hyperbola and since they have to kind of be contained by these asymptotes never go through it you know that this is going to be a a hyperbola that opens to the left and the right so it'll look something like this nope let me use this color so it'll look something this is the hard part it'll get closer and closer on that side and then kind of you this is one of the vertexes or vertices of the hyperbola and we'll go like that or this distance and notice the the similarity here with the ellipse this distance right here let me do it in a more vibrant color this distance right here between these two I guess you can kind of the elbows of the two ellipses that distance right there this is a and this is also a so you have a to a distance which is very similar to this situation where this distance is a and this distance is a so your distance between the two left and right points in a horizontal ellipse is the same as the distance between the two left and right points on a hyperbola is just the hyperbola opens outward while the ellipse opens inward fair enough but the whole point of this video is to discuss foci and you might have guessed and I did touch on it in the last video that hyperbola also have foci but they open they're going to be to the right and the left of these two points so this is a let me do them let me do them in a bright color because I want you to see them let's say that those two those are the two foci and a hyperbola this is and notice the difference an ellipse one of the definitions of an ellipse was the locus of all points or the set of all points where the distance from each of those points the sum of the distance from each of those points to the two foci is a constant now the definition of a hyperbola one of the definitions of a hyperbola can be the locus of all points where you take the the difference not the sum you take the difference of the distances between the two foci so if let me write that down so if this is this is d1 and this is d2 so you have a situation and we want to take we could take the absolute value of the difference because they might be they might at some points d1 will be longer than d2 if you're on this curve if you're on this curve d1 will be shorter than d2 so d1 minus d2 the absolute value is going to be equal to constant in the ellipse situation d1 plus d2 was a constant so they're very closely related you're in the ellipse you're taking the sum of the distances to the focus points and saying it's a constant and a hyperbola you're taking the difference of the distances to the focus points and saying that's a constant so this number right here is going to be the exact same thing as if I took a point right here and I'm picking these points arbitrarily as long as they're on the hyperbola and if I call these two points let's call these d3 and d4 the difference between D 1 and D 2 is the same thing as the the distant the difference between D 3 minus D 4 this is going to be a constant the entire way around the ellipse and so the next question is well what what is this constant going to be equal to and this is where it's it's useful to find a point where it's you can kind of get the intuition and we did it with the ellipses where we said oh if we take these points we used logic in the last video to say oh the distance between the sum of the distance between this and this that sum is going to be equal to we saw is going to be equal to 2ei or the distance of the semi-major axis right because this distance was the same as this distance so this Plus this is the same thing as this Plus that which is 2a right so the entire time the constants the sum of the distances to the two foci was equal to 2a now in the hyperbola what is the difference of the distances to the two foci so let's take this point right here on the hyperbola and we're saying so what is what is let me do a good color what is the magenta distance that's the distance to that foe sy- minus this or let me pick another color minus this and a light blue distance this magenta distance minus this light blue distance so we can make a very similar argument that we made in the ellipse situation this light blue distance is the same this the distance from this vertex or from this I guess you know this leftmost point of this rightward opening hyperbola to this focus is the same as this distance because a hyperbola is symmetric around the origin or the focal length is the same on either side of the center of the hyperbola depending on when you view it but I think that's a that that's not too much of a stretch of a statement for you to for you to accept so if this distance is the same as this distance then the magenta distance minus this blue distance it's going to be equal to this green distance right and this green distance is what that's 2a we saw that at the beginning of this video so this once again is also equal to 2a anyway I'll leave you there right now actually well look and let's actually just do one problem just because I like to make one concrete because I told you at the beginning that if you wanted to find let's say D so what if you have an ellipse so if you this is an ellipse x squared over a squared plus y squared over B squared is equal to 1 we learned that the over B squared this is an ellipse we learned that the focal length is equal to the square root of a squared minus B squared now for a hyperbola you kind of see that there's a very close relation between the ellipse and hyperbola this is kind of a fun thing to ponder about you know in a hyperbolas equation looks like this x squared over a squared minus y squared over B squared or could be Y squared over B squared minus x squared over a squared is equal to 1 it turns out and I'll prove this to you in the next video it's a little it's a bit of a hairy math problem that the focal length of a hyperbola is equal to the square root of the sum of these two numbers is equal to the sum of a squared plus B squared so if I were to give you so notice the difference it's just a difference in sign you're taking the difference of those kind of two denominators and now you're taking the sum of the two denominators so if I were to give you the following hyperbola x squared x squared over 9 plus y squared over 16 is equal to 1 well the first thing you do it don't ever well we could just figure out the focal length just by plugging into the formula the focal length is equal to the square root of a squared plus B squared this is a squared right a is 3 B is 4 so 9 plus 16 is 25 which is equal to 5 and so if we were to graph this let me see so that's my y-axis that's my x-axis its and the focal length is the distance to the in this case to the left and the right of the origin if it was kind of an up and down opening hyperbola would be above and below the origin so this is a this is a oh sorry this should this shouldn't be a plus we're doing with a hyperbola that should be a minus don't want to confuse you what I had written before with a plus I would've been any lips a minus is a hyperbola so the two asymptotes this is centered at the origin it hasn't been shifted are going to be 16 over 9 so it's going to be fairly steep asymptotes it's going to look something like that and that those are the two asymptotes the two vertex points are at two times a a is three right a squared is equal to nine B squared is equal to sixteen so this is the center so the two vertex points this is three minus three and then the focal points are going to be at pot from the center five to the right so it's going to be right here so that's five comma zero so it's five comma 0 and minus 5 comma 0 this is minus 3 and this is 3 so if we were to graph it it would look something like this there you go and if you were to take an arbitrary point on that hyperbola and take this distance and subtract that from that distance that will be a constant number that would be exactly equal to 2a or exactly equal to 6 in this particular example anyway hopefully I didn't confuse you too much with that little sign error that near the end of the video but in the next video I'll prove to you this formula which is a little bit of a hairy algebra but it's fun to do regardless