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let's do another conic section identification problem so I have four Y squared minus 50 X is equal to 25 x squared plus 16 y plus 109 so the first thing I like to do is to group all of the x and y terms onto one side of the equation and leave all the constants on the other side so let's do that so on the left hand side I'll put the 4y squared 4y squared and actually I'm also going to group all the X and y terms in this step so u 4y squared let's move this 16 Y on to the left-hand side so if I subtract 16 Y from both sides of this equation I get minus 16 Y minus 16 Y on the left-hand side and of course it will disappear on the right-hand side and then I want to subtract the 25 x squared from both sides of this equation so I get minus 25 x squared minus 50 X that's that right there and then I'll leave this 109 on the right-hand side is equal to 109 and now that we have the XS and the Y's on the same side of the equation we know what type of we know the general direction we're going to go in because they're on the same side they have different coefficients and one is positive and one is negative so that lets us know that we're dealing with a hyperbola so let's complete the square and get it into the standard form so the easiest way to complete the square is if you have a 1 coefficient on the y squared of the x squared term so let's factor out a 4 in this case so you get 4 times y squared minus 4y I'm going to add something later when I complete the square minus 25 times x squared plus let's see minus 50 divided by minus 25 is 2 plus 2x I'm going to add something later is equal to 109 and these things we're going to add those are what complete the square make these things a perfect square so if I take this you have a minus four here I take half of that number and this is just completing the square I encourage you to watch the video on completing the square where I explain why this works but I take I have a minus four I take half of that it's minus 2 and then minus 2 squared is plus 4 now I can't do one thing to one side of the equation without doing to the other and I didn't add a 4 to the left-hand side of the equation I actually added a 4 times 4 right because you have this 4 multiplying it out in front so I did a 16 to decide to the left side of the equation so I have to also add it to the right-hand side of the equation right this is equivalent to also having a plus 16 here that might make it a little bit clearer right when you factor it out then it becomes a 4 we would have added a 16 up here as well likewise if we take half of this number here half of 2 is 1 1 squared is 1 we didn't add a 1 to the left-hand side of the equation we added a 1 times minus 25 so you want to put a minus 25 here and likewise this would have been the same thing as adding a minus 25 up here and you do a minus 25 over here and now what does this become this Y the Y terms become 4 times y minus 2 squared Y minus 2 squared what I want to review factoring a polynomial if that if you found that a little confusing that step minus 25 times X plus 1 squared that's that right there X plus 1 squared is equal to let's see 109 plus 16 is 25 minus 25 is equals 100 we're almost there so we want a 1 here so let's divide both sides of this equation by 100 so if you will get Y minus 2 squared 4 divided by 100 is the same thing as 1 over 25 so this becomes over 25 minus C 25 over 100 the same thing as 1 over 4 so this becomes x plus 1 squared over 4 is equal to one and there you have it we have it in standard form and yes indeed we do have we do have a hyperbola now let's graph this hyperbola so the first thing we know is where where the center of this hyperbola is the center of this hyperbola is at the point X is equal to minus 1 so it's at X is equal to minus 1 Y is equal to 2 and let's figure out the asymptotes of this hyperbola so if this was this is Way IO is doing cuz I always forget the actual formula if this was kind of in it centered at 0 and it looks something like this Y squared over 25 minus x squared over 4 is equal to 1 I do this to figure out what the asymptotes would have been if we were centered at 0 because it's a lot easier to deal with these equations than to deal with these so we could solve we multiply both sides by 100 we're kind of unwinding what we just did so if you actually let's multiply both sides by let's multiply both sides by 25 so then you get Y squared minus 25 over 4x squared is equal to 25 and then I'll just go right here and then if I add 25 over 4x squared to both sides I get Y squared is equal to 25 over 4x squared plus 25 and so Y is equal to the plus or minus square root of 25 over 4x squared plus 25 and like always the asymptotes the hyperbola will never equal the asymptotes or intersect the asymptotes but it's what the the graph approaches as X approaches positive and negative infinity so as X as X approaches positive and negative infinity and you learn the concept of limits later on but I think you get it at this point because that's what the idea of what asymptotes even is is that you know as X gets really large you're approaching this line so as X approaches positive or negative infinity as we've done in the previous videos this term starts to matter a lot less because this term is huge so then Y is approximately equal to plus or minus the square root of just this and the square root of just this term is five halves X so though that would those would be our asymptotes if we were centered at zero but of course we're centered at negative 1/2 so let's graph that and then we could figure out if it's an upward opening or a downward opening upward opening or downward opening graph okay we're at Center negative 1/2 let me see so I want to be in D it's my y-axis this is my x-axis and we're centered at -1 and then 1/2 that's the center and then you know this would have been the two lines if we were if we were of the asymptotes if we were centered at zero but now this tells us the slope of the two asymptotes so the asymptotes are going to intersect at the center of our hyperbola so to speak so these are the slopes of the two asymptotes so one is positive five over two so positive five over two means if we go over to so one two and X we go up five so 1 2 3 4 5 so we'll end up right over there so I can draw that line you just need two points for a line so that line would look like that and then the other asymptote is minus 5 over 2 so for every 2 we go over to the right we go down 5 so 1 2 1 2 3 4 5 so we end up right about there and so that line would look like it would look like that good enough so those are the two asymptotes and they go on forever in those directions and now we could think of it two ways we could either say okay does if we look at well let's look at the original one I'll actually look at this one Ken and if we look at if it was centered at 0 could X equal 0 well sure X could equal 0 if x is 0 then Y squared over 25 equals 1 Y squared equals 25 Y would be plus or minus 5 five right so in this case this term could be equals zero so we could say that X could equal negative one right if X is equal to negative one then Y minus two squared over 25 will equal one let's do that let's set if X is equal to negative one X I was using the line tool is equal to negative one then what is this expression become I don't want to lose it so I'll write it right there so then you get Y minus two squared over 25 this becomes zero minus zero is equal to one so you get Y minus two squared over 25 is equal to one y minus two squared is equal to 25 just multiply both sides by 25 y minus 2 is equal to plus or minus I'm just taking the square root of both sides 5 so y minus 2 is equal to a positive 5 or Y minus 2 is equal to minus 5 add 2 to both sides of this you get Y is equal to 7 add 2 to both sides of this you get Y is equal to minus 3 so we know that the points minus 1 7 and minus 1 minus 3 are both on this graph so minus 1 that's here 1 2 3 4 5 6 7 minus 1 7 and minus 1 1 2 3 are both on this graph so let that lets us know since we're inside here that this tells us that this is kind of a vertical asymptote and another way to guess it is if you see that oh the Y term is the Y squared term is positive but either way or the other way to think about it is when you take the positive square root right when you take the positive square root you're always going to be a little bit above the asymptote right that's the other way to think about this that we're always going to be a little bit and this is this is the positive square the positive square root is the top line so we're always going to be a little bit above the asymptote right these this is the asymptote we're always a little bit above it and as obviously this number gets larger this starts to matter a lot less so the graph is going to look something like this it's going to come down and then go off and never quite touch the asymptotes but approach it so it's going to get really close to the asymptotes and then go off and go off in that direction anyway hope you found that helpful this was a slightly hairier problem so it should be instructive