 Well let's see this is episode 30 and in this episode we talk about the third of the three conic sections. We'll talk about hyperbolas. Let's go to the first graphic and just recall what how the three conic sections are formed. Let's see you notice that in the double cone on the left if we cut parallel to the edge if we cut through the lower cone we get a parabola but if we tilt it just a little bit so that we come out the other side of the double cone we get an ellipse and now today we'll be talking about the hyperbola. Now this is where you cut the double cone steeper steeper than this inclination. Here we have shown a vertical cut and you see we're going to cut both the upper and the lower cone. If we come to the green board let me just rotate that last figure for 90 degrees and let's say that this is the double cone that we were just looking at and rather than cutting it over here on the side what if I cut it right here in front and so when I cut through in the front we're going to get a curve that comes down right about there and then the other one the other branch will be down here and you can see that in the background we have the edges of the double cone but then in front we have the two branches of the hyperbola. Okay so with this introduction let's go to our list of objectives for today. Let's go to the next graphic. Today we'll be looking at first of all the fundamental equations of the hyperbolas then we'll look at transformations of hyperbolas where we shift them off of the origin and finally we'll look at the unique reflective property for hyperbolas just like parabolas and ellipses had a reflective property so do the hyperbolas and we'll look at an application. Okay well you remember that when we were looking at an ellipse we said if we go to the green screen we said that we were going to pick two points on the x-axis we call these the foci focus number one and focus number two and we were going to locate these equidistant from the origin so we were going to put focus number one at the point c0 and focus number two at the point negative c0 and then we were going to choose points that now this is for the ellipse we were going to choose points out here on the plane xy so that the sum of the distances to the two fixed points was constant and we call that the length of our string you remember we drew an ellipse using a string and the length of the string we said was 2a well the difference now for the hyperbola is that rather than the sum of the distances being 2a it's the difference of the distances being 2a so what i'm going to do is choose a point say over here a point xy and i'll find the distance let's take the larger distance over here to f2 and i'll call that distance d2 and the shorter distance in this case to f1 and i'll call that distance d1 and this time i want the difference of the distances to be 2a although i could i could put this point xy on either side and i'll always take the larger distance minus the smaller distance to be 2a so if i had chosen a point say over here and call this one xy then i would take the larger distance this time d1 minus the smaller distance this time d2 and the difference of the distance this is 2a for a hyperbola okay so in other words if i write this down in a more algebraic form i would say that the absolute value of d2 minus d1 is equal to 2a i'm putting an absolute value because sometimes the distance d1 is larger than d2 but the absolute value should be 2a and a hyperbola is the set of all points such that the difference of the distances to the two fixed points or the foci is equal to 2a uh let's look at that at that next graphic and we'll see this with this written out a hyperbola is the set of points in the plane with the property that the difference of the distances to two fixed points or the foci is constant and we're calling that constant 2a and the center of the hyperbola is the midpoint between the foci uh now coming back to the green board if i were to substitute for distance d2 i would put the square root of x plus c squared plus y squared minus and then for d1 i'll put the square root of x minus c squared plus y squared that absolute value is equal to 2a now if i square both sides i could eliminate the absolute values because whether this difference is positive or negative the square will be the same so when i square on the left and when i square on the right on the right that's going to be a 4a squared now without going through all the algebraic manipulation that we did last time for the ellipse uh let me just let me just summarize it by saying if you follow a similar procedure for the hyperbola um here's what here's what you come to as your final answer in this case you'll get x squared over a squared minus y squared over c squared minus a squared is equal to one so that this is this is after going through uh quite a bit of algebra you can make this equation become this equation now some of the differences between this result and the in the result that i got with an ellip with the ellipses is number one there's a minus in the middle instead of a plus uh also this difference in the denominator is a little different you do you know what's different about that do you recognize the difference there c squared yeah c squared comes first instead of second so this is c squared minus a squared rather than the other way around so when i make my substitution for b squared you remember for the ellipse b squared was substituted for a squared minus c squared now i'm going to substitute it for c squared minus a squared and if i substitute that into this denominator i'll just put a i'll just put a b squared right there and this is the fundamental equation of a hyperbola if the foci are on the x-axis okay so we've we've left out quite a bit of the verification in this case to save a little time but let's go to the next graph you can we'll see a summary of this information uh a hyperbola with foci at plus or minus c zero so those are both on the x-axis and the difference of distances being 2a has the equation x squared over a squared minus y squared over b squared equals one just like we wrote down here where b squared has been substituted for c squared minus a squared now you notice that the difference uh that the that the center rather is at the origin and you notice there are two dotted lines in the background those dotted lines represent the sides of the double cone you see in this illustration our double cone is being laid down sideways and we've cut the double cone in the front and those dotted lines represent the sides of the double cone those are called acid well those are asymptotes like we've seen asymptotes earlier in this course and their equations are y equals plus or minus b over a times x what we're going to verify that in just a moment and finally the eccentricity e is still the ratio c over a but in this in this situation c is bigger than a and therefore the eccentricity will be bigger than one okay uh let's come back to the uh green board and let's talk about those asymptotes for a moment now the fundamental equation of the hyperbola that we're referring to here has the equation x squared over a squared minus y squared over b squared equals one and if i draw the graph of it over here we have foci at c and at negative c on the x axis now the actual curve of the hyperbola comes in sideways comes inside of this focus and turns back out same thing on the other side the hyperbola comes in this this branch they call it this branch comes in comes inside the focus and it turns back out so it turns out that the foci are actually on the outsides of the hyperbola whereas on the ellipse the foci were on the inside of the hyperbola now where does this where does this hyperbola cross the x axis let's see it would cross the x axis when what's zero when y is zero when y is zero okay now if you imagine that if i cover up the y term y is zero then what we have here is x squared over a squared is equal to one our x squared is equal to a squared our x is equal to plus or minus a so it's going to be crossing the x axis at plus or minus a that's easy to remember because that's the that's the square that's underneath the x term now is a bigger or smaller than c well you remember in our substitution we have that b squared is equal to c squared minus a squared that's the substitution we made and what that tells me is that the c is bigger than the a and therefore the intercepts at a and negative a here's a and here's negative a these intercepts are inside of the foci just like we had indicated now what are the y intercepts for this hyperbola let's see how do i find y intercepts x equal to zero we're going to set x equal to zero and when i said x equal to zero in other words if i just cover that up i'm going to get minus y squared over v squared is equal to one and uh steven what do you notice about that equation um and it's a negative number is equal to a positive number which can't happen because yeah we've got a negative on the left we have a positive on the right you see these squares can't be negative can they because they're squares and so y squared over b squared can't be negative and then we put a negative on that which says it can't be positive and it's equal to a positive one so this is impossible and therefore there are no y intercepts now if you go back and think about the graph that you just saw a moment ago remember the two branches are going like this so they don't they don't cross the y axis and now we've verified that with the um uh by by trying to find the y intercepts okay now what about those asymptotes well let's solve this equation for y uh minus y squared over b squared is equal to one minus x squared over a squared and uh let's multiply both sides by negative one to get rid of the negative y squared over b squared is equal to now if i multiply this by negative one i'll just reverse the order because this this ratio this rational expression becomes positive and i get x squared over a squared minus one so that says that y squared is equal to b squared times this that'll be b squared x squared over a squared minus b squared and if i write this in a slightly different form let's get a common denominator the common denominator is a squared so this will be b squared x squared minus a squared b squared okay now to solve for y y would be i'd have to take the square root of this and that'll be the positive and the negative square roots because there's an upper there's an upper half and a lower half uh square root of uh let's see well let's just rewrite that inside b squared x squared minus a squared b squared all over a squared now i can simplify this expression a bit if i factor out some square factors uh for example i could factor out the a squared in the denominator and just put an a out in front and what can i factor out of the numerator a b squared a b squared so i'll put a b out in front that's a plus or minus b over a and then i'm left with a simpler square root um susan what would be under the radical when i factor those out mm-hmm yeah minus let's see now we took out the b squares one uh not one there's an a squared left right here we factored out the common factor of b squared so there'll be an a squared left and i have to leave the square on there because it's still into the radical okay now what i'm thinking is as as i follow the graph further and further out here to the right hand side the x's are getting bigger and so are the y's as a matter of fact now as the x's get bigger x squared minus a squared this a squared tends to look relatively small as the x's get really large and so this this square root becomes approximately just the square root of x squared because when x squared gets big the a squared stays the same this difference is roughly the same thing as x squared and so at that moment this is approximately the same thing as plus or minus b over a times now if this is going to become approximately x squared the square root of x squared is x and so to kind of summarize this i'm saying that as x gets large that is if i go further further out to the right or if i go further out to the left then this graph is approximately plus or minus b over a times x so if i draw in the line y equals plus or minus b over a times x the hyperbola will approach that line and so that becomes an asymptote now let me just erase some of this work and and we can we can illustrate this further i'm going to move my graph down to the middle of the screen so that i can i can do it better justice by drawing it larger um let's see now we said that the foci were at plus or minus c okay so here's plus c and minus c we said that the intercepts were at let's see that would be a and this would be negative a and we said that the asymptotes by the discussion there would be plus or minus b over a times x now that says the slope is b over a so i should go over a up b well now i've already gone over a right there so let's go up b let's say b is right about here and so if i go over a and up b i'll get a point right there and my asymptote should go let's see through the origin because the the y intercept is is zero so from the origin through the point um ab and just keep right on going because the slope is b over a now here's what you'll normally see in textbooks of what most people do they actually make a little box right here at b comes down through negative a goes back through negative b this is negative b here and then comes up through positive a okay so this is a dotted box and its width along the x axis is two a and its height along the y axis is two b and then draw a diagonal line through the corners of that box and that should go right through the origin and this is the line y equals b over a times x that's one of the asymptotes and then the other line that i draw is also through the box but it's the other diagonal i'll draw it this way and what's the equation of that line negative b over a times exactly y equals negative b over a times x and i'm drawing those lines in because the hyperbola is going to be approaching those dotted lines as you go out to the to the right and to the left now we said the x intercept was at a so i'm gonna i'm gonna actually touch the box right here and my graph turns and goes out and approaches that asymptote and on the other side maybe i should take out my uh my little arrow there now and my graph approaches the asymptote over here on the other side a similar thing happens it crosses the x axis at negative a and this branch goes up and approaches this asymptote y equals negative b over a and it approaches the asymptote over here y equals positive b over a now you notice this is not a parabola that i'm drawing although you might mistake it for a parabola but parabolas don't have these diagonal asymptotes that they approach parabolas actually turn if i were going to draw a parabola the parabola if it were a parabola it would actually turn and go out like that and there's no diagonal line that that would approach so uh it has this graph has some similarities to a parabola but it's it's surely not a a parabola okay um let's see now you know the base of this little triangle right here that was a and the height of that little triangle was b now i tell you what can anyone tell me how long is the hypotenuse of this little triangle let me just darken the triangle in right there a squared plus b squared yeah that should be a squared plus b squared now you know we've seen that before because there was a substitution that we made just a few minutes ago and that was that b squared equals c squared minus a squared now if you look at this from another point of view c squared is equal to a squared plus b squared and so c is the square root of a squared plus b squared so i think we have something rather rather relevant here if the base is a and the height is b the diagonal is the square root of a squared plus b squared that's c this is c right here it's the diagonal of the triangle sitting in that box now if you had a compass and if you could open your compass from the origin out to there and just swing that down if you were to just rotate that down you would intersect at the focus down below so the length of that diagonal is the same as the distance out to the focus which is clearly then bigger than bigger than a yeah steven i was just looking at your graph of the hyperbola that you drew and it almost seems like it's um the equation y equals one over x rotated 45 degrees you know as a matter of fact uh when we looked at rational functions one of our fundamental graphs was f of x equals one over x let me just write that over here f of x equals one over x yeah this this is uh interesting that you say this because uh this graph if you remember looks sort of like this this is a hyperbola just like this but it's been rotated 45 degrees and those uh two axes are the two diagonal asymptotes now rotated into the axes this is a rotated hyperbola and you'll talk about this in trigonometry but you have to know a little bit about sines and cosines and trigonometry to explain this aspect but one over x is a rotated hyperbola uh one over x squared is not it has two branches but you remember it's in the first and the second quadrants okay one more thing to point out in this illustration um i have the two branches of the hyperbola i have the two dotted lines and if you think back to that double cone illustration this is the double cone set sideways and we've taken a cross section of the double cone out here in front of it cut through both sides and the dotted lines actually represent the sides of the double cone behind it and so as you cut through the double cone as you go further out the actual curve approaches the edge of the double cone although there actually one is one is out in front of the other if you think of this is sort of a three-dimensional illustration the hyperbola has been shifted out in front a bit but uh looking down on it it appears that the branch of the hyperbola approaches the side of the double cone even though one's actually above the other okay now let's go to the next graphic and look at what happens if we put the foci on the y-axis now in this case it says a hyperbola with foci at the points zero plus or minus c and once again the difference of the distance this is two a its equation will be y squared over a squared minus x squared over b squared is one uh what is the difference in that equation in the equation that we just saw before it you know at first glance you may say isn't that the same equation but there's actually a subtle difference in it the x's and y's switched the x's and y's have been interchanged yeah we have x squared we have y squared over a squared minus x squared over b squared so the way you decide which axis the foci are on is to look at the positive term y squared in this case over a squared and whatever is the denominator under that positive term that's the square of the intercepts so this graph is going to cross the the y-axis at plus or minus a uh once again the um the substitution is b squared equal c squared over a squared that equation is derived in the very same way as the first equation which we actually didn't derive um is derived in the same way and you make the same substitution you'll notice there is a difference also in the asymptotes the asymptotes are now y equals plus or minus a over b i think in the previous one it was y equals plus or minus b over a let's come to the green screen and just discuss this equation for a moment we have y squared over a squared minus x squared over b squared is equal to one and so what we're going to do here is um draw our coordinate system and our foci are at c and uh negative c and to find the y intercepts to find the y intercepts here i let x be zero so if i cover up the zero because x is zero then that says y squared over a squared is one or in other words y is equal to plus or minus a so i'll put my my a here and my negative a right there and again i know that a is smaller than c because b squared equal c squared minus a squared so this number has to be bigger than that number so that i don't get a negative in that case um now when i go to draw my asymptotes on my on the other axis i'm going to go out b and negative b and i'm going to make my little box now and through those um through those vertices through those corners i'm going to draw my asymptotes there's an asymptote and here's an asymptote uh now you notice this time the rise is a and the run is b so in other words if i take my a and b out this is actually a and this is equal to b and therefore this is the equation y equals a over b times x that's not the same ratio we saw when the foci were on the x axis that was b over a before so there are a few things you have to kind of keep straight about hyperbolas that um that we didn't seem to have to take so much time about when we talked about ellipses the other asymptote is y equals negative a over b times x uh once again what is the length what is the length of the diagonal in this right triangle it's going to be c again yeah you see uh this this length is going to be the square root of a squared plus b squared which is c and therefore if you take that line segment and just pivot it and rotate it over under the y axis that's where your focus will be um oh we we've done everything except draw the hyperbola haven't we okay here's the hyperbola it's going to come up this way and turn and go back up that way and on the other side it's going to just touch the box and it's going to come out and approach approach but not actually cross the asymptotes there you know one of the mistakes that i sometimes see students make is when they go to draw the hyperbola they have it they have the vertex going through the focus but you see it actually comes up and it's tangent to that box and then and then goes back out okay i think it's time that we actually work some concrete examples here so let's go to the next graphic and look at look at three problems that ask us to sketch graphs of these hyperbolas okay first of all we have uh x squared over nine minus y squared over sixteen is one now let me ask the class here what is it that tells you in part a that that's a hyperbola if it hadn't mentioned it in the instructions i mean you might say isn't that an isn't that one of those ellipses you talked about last time well there's a minus in between there's a minus instead of a plus yeah so in that first equation uh you notice it's in the form x squared over a squared minus y squared over b squared is one so already this tells me which axis the foci are on you see the positive term here is the x term so the foci are on the x axis okay so i know that the branches will be opening to the left and to the right furthermore looking at this this tells me that a squared is nine so a is equal to three and it tells me b squared is uh sixteen so b is equal to four and you say wait a minute you've got a smaller than b well you know for ellipses a was always bigger than b but for hyperbolas a and b can have any relationship the the number that's the biggest is always c is always the focus number when you're talking about hyperbolas while we're at it let's go ahead and calculate c uh b squared is equal to c squared minus a squared and so sixteen is equal to c squared minus nine and so c squared is equal to twenty five and therefore c is equal to five i'm taking the positive square root here because a b and c are always taken to be positive so now we know a we know b and we know c let me just add that to my list up here c is equal to five and i think we're ready to draw the graph and the graph will look like this i'm going to go out uh five units to locate the focus okay so there's focus number one that's it that's at c and i'll go out five units in the other direction and that'll be that'll be focus number number two right here and right here uh but while i'm at it i'll go out three units to locate my uh x intercepts at three and at negative three and i want to draw my box to locate my asymptote so i'm going to go up and down four one two three four and one two three four and i'll make my box right here and you remember that the dimensions of the box are two a and two b and i'll draw the asymptotes through here oops i missed that one let's just make it bend a little bit sort of artistic license you might say and now my hyperbola branches open left and right and i start right at the box at that intercept and and the first branch comes up like this and it comes down like that and the other branch looks like this okay and that's the graph of x squared over nine minus uh y squared over 16 equals one you know something i i failed to mention earlier is that the axis that you placed the foci on uh when we talked about ellipses that was called the major axis uh when you talk about hyperbolas this is referred to as the transverse axis don't ask me why uh it has a different name i could see why you would want to let me move that over a little bit so you can read that i'll write it on this side this is the transverse axis that's the axis that the foci are on and the other axis the axis that the hyperbola doesn't cross is referred to as the conjugate axis the conjugate axis whereas if this had been an ellipse that would be called the minor axis the major in the minor axis for hyperbolas there's a different vocabulary so it makes it a little bit confusing uh let's go to let's go to b problem b and uh this one is written in a different form we have uh four y squared minus 36 x squared is 36 this is not in standard form this is what we call standard form up here where i have a one on the right what should i do to this equation and b to put it in standard form divide everything by 36 divide by 36 okay so that's going to be um let's see four y squared over 36 minus 36 x squared over 36 is equal to one 36 over 36 or one so this becomes y squared over nine minus x squared over one or if you prefer just say x squared is equal to one uh now which which axis is the transverse axis or the the axis on which the foci lie the y axis on the y axis because that's the term that's positive and that also identifies a a is equal to three because a squared is nine uh b squared is one and so b is equal to one and c is equal to well i'll have to use my identity b squared equals c squared minus a squared and so one is equal to c squared minus nine it looks like c squared is equal to ten so what is c going to equal square root of ten square root of ten okay i'll just fill that one in right here square root of ten okay so when i go to draw my graph we know a is a is three b is one c is the square root of ten let me just record that over here on the side to make a little room a is three b is one c is the square root of ten now let's draw our graph right below this and uh i'm going to be locating my foci on the positive uh y on the on the y axis so i'll go up three so and i'll go down three and then from the origin i'll go over one and i'll go back one and that's where i'll make my box right there i'm leaving out a few tick marks on the axes to kind of speed things along and then i'll draw my asymptotes through there now these asymptotes are fairly steep and therefore the hyperbola looks like this and you see the asymptotes have come closer together they're both steeper and uh so i tend to get a rather thin looking hyperbola but i still approach those asymptotes just like just like before um can anyone tell me the length of the diagonal along here the length of that diagonal root ten it's the square root of ten because it's the value of c and you see if i rotate that if i rotate that over onto the y axis i would get a point right about there that's where the focus is if the focus and the the intercepts uh these are called vertices if the focus and the vertices are close together you tend to get a rather thin hyperbola if the focus is further away from the intercepts you tend to get a wider a wider hyperbola in that case okay there's one more example on this screen let's just uh look at it quickly um we have x squared over 16 let's go back to that screen that we just had up there was one more example we wanted to work here we go uh x square 16 x squared minus 25 y squared is negative 100 now because the x squared is positive you would think that the foci are on the x axis but i don't think that's the case the foci are not on the x axis why not because we're going to have to buy a negative to yeah you see i've i've i've tried to hide the axis that is the transverse axis by putting a negative on the 100 and a lot of times in textbooks they'll put a negative over here to take your attention uh away from the fact that this term is positive but that's actually not the axis that is the major or the transverse axis if i divide by negative 100 this guy becomes positive that's going to be y squared over four and uh this term becomes negative that's going to be 16 over 100 or four over 25 four x squared over 25 and then if i divide by negative 100 this will be a one now we we face this problem when we talked about ellipsis two what do you do when there's a four in the numerator that's not in what you'd call standard form what should we do divide the denominator by four and as well as the numerator exactly right so we're going to divide top and bottom by four and this is y squared over four minus x squared over 25 fourths is equal to one now this tells me a lot of information it tells me first of all that the y axis is the transverse axis so this hyperbola opens up and down it also tells me that a squared is four so a is two because a squared is always the denominator under the positive or in the positive term it tells me that b is five halves because b is always the denominator of the other term which is the conjugate axis and finally c is equal to well we have to work that one out b squared is equal to c squared minus a squared so 25 fourths is equal to c squared minus four so c squared is equal to 25 over four plus four how many fourths will that be when you add those together 41 41 fourth yeah it's kind of awkward 41 fourth so c is the square root of that c is the square root of 41 all over all over two okay now let's try let's try going back well I tell what I'm going to put my information over here and I'll open a space for us to draw our graph five halves and c is the square root of 41 over two now let's come over here and draw the graph of this hyperbola and I'm going to go up two units and down two units I'm going to go over two and a half units and two and a half units and I'll make my box right there parallel to the conjugate and parallel to the transverse axes okay here's the x axis here's the y axis so my asymptotes come through here and my hyperbola has a branch that opens out to the to the uh oh let's see no yeah I I'd I'd put it the wrong place right remember the y axis of the transverse axis so I need to put that those vertices right here and the branches open up thank you Stephen for I didn't draw the graph to readily too quickly and the hyperbola looks like this now we have just graphed the hyperbola 16 x squared minus 25 y squared is equal to negative 100 yeah that's what we're graphing over here we're graphing this hyperbola um and it opens up and down like that okay let's go to the next example this problem says find the equation of the hyperbola with intercepts zero plus or minus two squares of three and asymptotes y equals uh y equals one half plus or minus one half x now you'll have a lot of problems similar to this in your homework you're supposed to find the equation of that perbola and you're given some information about it in this case we're given the intercepts and the asymptotes in a different situation you could be given the um you could be given the foci and you could be given say the length of the transverse axis now when they say the length of the transverse axis they mean the distance between the two foci okay well in this case we're given that the intercepts first of all are on the y axis so I know that the equation should be of the form y squared over a squared minus x squared over b squared is equal to one uh furthermore I know the value of a because a is uh it looks like kind of a comma there I don't think that should be a comma that's two times the square root of three okay so we know that a is equal to two times the square root of three and I know that the ratio either a over b or b over a is one half now let's see when we're when we have our our intercepts our vertices on the y axis is this ratio a over b or is that b over a a over b is a over b okay so uh the ratio of a over b is the same as one over two if the if the uh if the foci and the intercepts had been on the x axis this ratio would be b over a and I know that a is two times the square root of three so this allows me to solve for b and if I cross multiply b is equal to uh four times the square root of three so let's add that to our list up here b is equal to four times the square root of three but that allows me to write down these equations y squared over a squared well now what is this number squared what is a squared you have to square the two and square the square root if you square the two you get four and if you square the square root you get three four times three is 12 so this denominator is 12 minus b squared now let's see to get here's b so b squared would be let's see four squared times the square root of three squared that'll be 16 times three is how much 50 let's see 16 times three um 48 yeah 48 yeah thank you 48 is equal to one now this problem said find the equation on a perbola and this is the equation on that perbola now we could do other things we could find the foci we could draw the graph of it and so forth but this problem didn't ask us to do that so let's move on to a different a different example okay what's the next example we have here um in this next problem we have three very similar looking equations and it says sketch each one of these conics now these are not all all hyperbolas which one of those is not a hyperbola the last one yeah and what is it uh jeff it's an ellipse because we have uh because we have a plus these others have a minus now this says to sketch all three of these conics you notice the terms are all the same x squared over four x squared over four x squared over four etc ones over here on the right hand side i'm going to draw these all on the same coordinate system and let's compare what they look like so um let's take this first let's let's take this first hyperbola what's the value of a in the first hyperbola two is equal to two and what's the what's the transverse or the major axis the x axis so i'm going to go out two on the x axis and the conjugate axis is the y axis and i'll go up and down one so i'll go up and down one and this is negative one and i'll make my little box right there to set up drawing my hyperbola and this allows me to set up the asymptotes okay there's one asymptote this is the line y equals one half x and here's another asymptote and this is the line y equals negative one half x and when i draw the hyperbola uh i will just touch the box right there at this vertex and it'll turn and go out and over here it'll turn and go down and on the other side at negative two it looks like this okay so let's just keep in mind that on this particular graph a was equal to two b was equal to one okay now in the second graph this is a hyperbola which axis which axis is the transverse axis the y axis the y axis and what's the value of a in this problem one a is one okay a is one so i go up one on the y axis i go up one on the y axis there we have it and i'll have a vertex right there and uh oh also at negative one better put two of those in and the conjugate axis or the minor axis is the x axis and on that axis i should go out two so in this case b is two i go out two units so you notice i'm going to get the same box but technically the values of a and b have switched because a is now one and before a was two and my hyperbola is going to look like this it's going to be a very wide hyperbola and it's just the one that fills in the other side of the asymptotes now let's look at how this came about i started off with x squared over four minus y squared over one is one and if you switch the two terms you get you get the other hyperbola on the other side of the asymptotes this is referred to as the conjugate hyperbola so i have this hyperbola and i have this conjugate hyperbola and they have the same asymptotes the same box but they have the opposite transverse and conjugate axes and they have the opposite values for a and b now finally what if i put a plus in here this is an ellipse where does this ellipse cross the x axis at x equals two and negative two two and negative two okay so crosses the x axis here and here it crosses the y axis at plus or minus one here and here and if i draw an ellipse the ellipse is inside the box and it's just tangent at the four on the four sides so if you put a plus in the middle it doesn't matter in which order those terms are because when you put a plus in you can add them in either order you get the ellipse inside the box so it's rather curious how closely knit these graphs are i have a hyperbola i have the conjugate of the hyperbola and then i have the ellipse inside the box that i used to draw the asymptotes so i think that's an interesting relationship that we have okay next graphic here we have an example of a hyperbola that's been shifted off of the origin has a has a translation in it but i've multiplied out the terms so we can't tell how it's been translated we've seen problems like this before when we talked about ellipses i'm just going to write this one here on the green screen so that i can work it out um nine x squared minus sixteen y squared minus seventy two x minus thirty two y minus sixteen is zero um can anyone tell me what you think the first step is going to be here to determine the center of this hyperbola group like group like terms and for what the terms so that you can complete the square right right we're going to have to complete the square um what tells me this will be a hyperbola is you notice that the two terms that are squared there's a minus in there if this had if there had been a plus in here i would think this would be an ellipse but because there's a difference of the squares then i'm thinking it's probably going to be of this form x squared over a squared minus y squared over b squared is one because here's a minus of squares a difference of squares and here's a difference of squares and it's in the same order so i want to make that look to i want to put that in this in this form okay so steven says group the terms group the x's and group the y's so we have nine x squared minus seventy two uh x squared minus i'll factor out the minus and then sixteen y squared and plus thirty two y because i'm factoring out the minus and the sixteen the negative sixteen which is sort of a tag along i'll just put on the other side sixteen over there now what do i need to do before i complete the square here uh susan what what do you see is going to have to be done find a constant uh well we have to find the constant but uh there's a problem here david yeah take out nine take out the nine take out the sixteen yeah we have to get this down to a coefficient of one so this would be nine times x squared minus eight x and then minus sixteen times y squared plus two y and nothing's happening over on the right that's still just uh sixteen okay i need to add a constant here to complete the square and uh let's see the rule says once you have a one on the square term you take half this and square it so let's see half a negative eight is negative four so i'll add sixteen and over here i need to add sixteen times nine oh times nine yeah thank you very much of course times nine that's that's that's something i wanted to point out that students frequently overlooking now i've overlooked it say i'll put in nine of sixteen's not just one so over here i need to be adding on a hundred and forty four that's nine times sixteen and uh in this term let's see half of two is one so when i square it i get a one i'm actually adding on a negative sixteen there so over here i'll have to add on a negative sixteen so this gives me nine times x minus four squared minus sixteen times y plus one squared equals and if i total all this up the sixteen's cancel that's a hundred and forty four okay now this is still not in standard form what do i need to do now divide by a hundred forty four divide by a hundred and forty four yeah so that's going to be a one uh now you know let me just make a note over here to kind of speed these things along a hundred and forty four is nine times sixteen so when i divide by a hundred and forty four here the nine cancels but the sixteen is left so this is x minus four squared over sixteen because the nine canceled off and in this case the sixteen cancels and i'm left with a nine so this is my hyperbola in standard form except i can see the shift what shift is being made here to the right four to the right four and what shifts being made here down one uh up one actually because oh i'm sorry down one of course down one there's a plus i've got to go down so uh yeah thank you and so now let's draw that graph okay so what i've done here is to move the problem up to give me a little room to draw the graph let's come back to the green screen and uh let's see now the center of this hyperbola is going to be at the point four negative one four negative one now this is the center you know all of the graphs we've drawn so far have been centered at the origin this was now centered this is now shifted off center uh a is equal to four and b is equal to three so if i go over four more that's going to be my transverse axis intercept on the right and if i go back four that'll put me right here that's my transverse axis intercept on the left now to make my asymptotic box i'm going to go up three and i'm going to go down three right here and now here is the box that i'll use to draw the asymptotes okay we've got the center you know maybe i shouldn't call it c because that we use c for focus so i'll just eliminate that and my asymptote comes through here and the other asymptote comes through like this now when i draw my hyperbola will i be drawing the upper and lower branches or left and right branches left and right left and right exactly and here's my vertex so at that point my graph turns and goes out and approaches those asymptotes that asymptote and it comes down like this and on the other side it turns and approaches asymptotes on either side um are there any questions about that let me ask you this how long is the um how long is this diagonal how long is the diagonal right there this was five this was three this is five so what that tells me is the focus instead of going over four the focus i should go over five you get a focus there and if i go back five i'll get a focus right here let me ask you another question if i had drawn in the hyperbola that had a branch above and below what would have been its equation y plus one squared over nine minus x minus four squared over sixteen right equals one equals one of course all you do is just reverse those terms and by the way if i drew the ellipse that's inside this box what would be its equation x minus four squared over sixteen plus y plus one squared over nine equals one exactly so you just put a plus a plus in there and now you have the ellipse inside the box okay very good now we have one application uh that i wanted to mention here before we run out of time i think this is kind of an interesting one it sort of sets up a sort of a sort of a mystery can we go to the next graphic okay bill says ted what was that ted says a gunshot now ted and bill are standing a thousand feet apart assume bill hears the shot one second after ted hears the shot show that the gunshot occurred somewhere along a branch of this hyperbola and there's a hint that the speed of sound is about 800 feet per second okay let me just show you how we could solve that problem uh here on a coordinate system let's say we put ted and bill on the x axis i'm going to put ted over here and i'll put bill over here and they're 1000 feet apart so that makes this uh the point 500 zero and this is the point negative 500 zero okay now ted hears the shot before bill does so it sounds like the shot should be somewhere over on this side of the graph on this side of the coordinate system now let's say the gunshot took place right here the sound travels to ted and the sound travels to bill but it takes longer to get to bill in fact it takes one second longer now if sound travels at 800 feet per second uh then how much further away does or how much further does the sound travel to get to bill than to ted if it took one extra second well it would have to take 800 it would travel 800 extra feet because in one second it'll travel 800 feet so what that tells me is 2a is equal to 800 because remember 2a was the uh difference in the distances and if i think of these as being foci then i have that c is equal to 500 and a is equal to 400 a here being 400 and if i know a and c i can calculate b squared b squared is c squared minus a squared so c squared is 25 250 000 you know when you square a number that ends in zeros you just double the number of zeros so it's 25 and four zeros and the same thing for a squared that'll be 16 with four zeros and therefore that difference is a nine with four zeros that's going to be 90 000 so therefore if uh if i have ted and bill as the two fixed points and the fixed difference of distance is 800 then this must be the equation x squared over a squared which is 250 000 minus y squared over b squared which is 90 000 is equal to one that's a hyperbola and this hyperbola comes in like so i'll leave out the asymptotes here in the interest of time and it's so the gunshot took place on one branch of a hyperbola and this is the hyperbola that we were that we were given in that problem okay uh now one other thing about the reflective property of hyperbola we just have a few seconds left but let me just mention that if the hyperbola looks like looks like this and if the foci are here then if uh if you have light coming in toward one focus it's reflected to the other focus point if this is a reflective surface now if this is a reflective surface this will be reflected back to the other focus and so it goes the light just keeps being reflected back and forth toward one focus and then toward another well you know i i think we're just about out of time so uh we have just finished the three conic sections next time in episode 31 we'll talk about sequences and series and i'll see you then