in this video we're going to focus on hyperbolas now you may want to get a sheet of paper and a pen to write down some notes so we have a horizontal hyperbola on the left and a vertical hyperbola on the right the formula that corresponds to the horizontal hyperbola is x squared over a squared minus y squared over b squared which is equal to one this is the case when the hyperbola is centered at the origin a is the distance between the center of the hyperbola and the vertex so to get the two vertices the coordinates are plus or minus a comma zero for this type of hyperbola and the same is true for the other one a is the distance between the vertex and the center of the hyperbola now the equation that corresponds to the vertical hyperbola is as follows it's y squared over a squared minus x squared over b squared which equals one so when a squared is under x squared it's going to open left and right or when there's a positive sign in front of x squared when there's a positive sign in front of y squared it opens up and down and a squared is always under the positive term b squared is under the negative term now the coordinates for the vertices for the vertical hyperbola are zero comma plus or minus a the coordinates for the foci are plus or minus c comma zero on the left and for the hyperbola on the right it's zero comma plus or minus c the hyperbola i mean the foci is always located towards where the hyperbola opens up and it's c units away from the center now the equation that correlates a b and c for the hyperbola is c squared is equal to a squared plus b squared in contrast for ellipses the equation is different c squared is equal to a squared minus b squared for an ellipse a is always larger than b but for hyperbolas a may be larger than b and sometimes b may be larger than a so the relative sizes of those two are not important they are important for ellipses but not for hyperbolas so make sure you're aware of that key difference and we'll go over plenty of examples so you'll see this in action now the next thing you need to be familiar with is the transverse axis the transverse axis is a line segment that connects the two vertices together so the length of the transverse axis is 2a here we have a vertical transverse axis and here on the left is the horizontal transverse axis the distance between the foci or the two focal points is always going to be 2c the length of the transverse axis will be 2a now let's talk about the equation of the asymptotes for the horizontal hyperbola it's y is equal to plus or minus b over a times x for the vertical hyperbola it's y is equal to plus or minus a over b times x and this is true when the hyperbola is centered at the origin the equation is different if the center is not at the origin and we'll talk about that later in the video number one graph the hyperbola identify the coordinates of the center vertices and foci and then write the equations of the asymptotes so what's the value of a squared and b squared in this problem a squared is going to be under the variable that has a positive coefficient the one if the negative sign is going to be associated with b squared so in this problem a squared is the number under x squared a squared is four a is going to be the square root of four which is two b squared is equal to nine and b is going to be the square root of nine which is three so with this information we can go ahead and draw a graph by the way the center is going to be the origin for this particular problem so a is 2 and it's under x squared so we're going to do is we're going to travel two units to the right from the center and two units to the left let's put some marks on this graph first so two units to the right and then two units to the left now b is three and it's under y so we're going to go up three units along the y axis from the center and then down three units then what we're going to do is we're going to create a rectangle around these four points the purpose of the rectangle is to help us to draw the asymptotes the asymptotes will be diagonals that will go right through the center of the box so those are the two that the two asymptotes and what we have here represents the vertices of the hyperbola so whenever you have a positive x squared term the hyperbola is going to open towards the left and towards the right so it's going to look something like this and then it's going to follow the asymptote it's curved by the vertices but once it follows the asymptote it becomes almost linear so that's how we can graph the hyperbola to the best of our ability so it's initially curved and then it becomes linear as it approaches the asymptotes now let's write the coordinates of the vertices so the vertices is going to be plus or minus a comma 0 for this type of graph and a is 2 so it's plus or minus two comma zero so this one here is at two comma zero and this one here is negative two zero so those are the coordinates of the vertices now let's focus on the coordinates of the foci this is going to be plus or minus c comma 0. so we need to calculate the value of c and we could use this formula to achieve it c squared is equal to a squared plus b squared a squared is 4 b squared is 9 so c squared is 13 which means c is equal to the square root of 13. so the coordinates of the foci is going to be plus or minus root 13 comma zero the square root of 13 as a decimal value is approximately 3.6 so we have a focus there and the focus here so those are the two focal points of the hyperbola as you can see the hyperbola always opens towards the focal points now let's write the equations of the asymptotes now when you have an asymptote centered at the origin and if it opens to the right and to the left the equation is going to be y is equal to plus or minus b over a times x this value in front of x is the slope it's rise over run going from the origin to this point we have a rise of 3 that's our b value and a run of a so the slope is rise over run b over a so that can help you determine if it's b over a or a over b if you ever forget so b is three a is two so y is equal to plus or minus three over two x now there's two equations for the two asymptotes that we have here the one on the right this is going to be y is equal to positive three over two x as for the one on the left this is y is equal to negative 3 over 2x and it makes sense because this line here the slope is positive because it's increasing as you move to the right as for the other line the slope is negative because it's decreasing as you move to the right so those are the two equations that correspond to the two asymptotes that we have in this graph and the asymptotes are linear which means that the hyperbola becomes linear as x increases near the vertices it's kind of curved but as x gets larger and it approaches the asymptote it becomes it behaves like a linear function now let's work on a similar problem so we're going to do the same thing we're going to graph the hyperbola identify the coordinates of the center of vertices and the foci and then we're going to write the equations of the asymptotes right now this hyperbola is not in standard form we need this number to be a one so what we're going to do is we're going to divide everything by 144 144 divided by 9 is 16 so we're going to get y squared over 16. 144 divided by 16 is 9. so we're going to get x squared over 9. and 144 divided by itself is one so i'm just going to rewrite this here so now that our equation is in standard form we can go ahead and find the details that we need so what's a squared and what's b squared in this equation a squared is always going to be the one that's has the positive coefficient so a squared is going to be 16. so that means that a is the square root of 16 which is 4. b squared is 9. b is going to be the square root of 9 which is three so now let's calculate c c squared is equal to a squared plus b squared a squared is 16 b squared is nine 16 plus nine is 25 and the square root of 25 is 5. now let's go ahead and draw a rough sketch of this graph now since y squared is listed first this is going to be a vertical hyperbola so we're going to travel a units along the y axis since a squared is under y squared and then a units down from the center so just like before the center is going to be the origin it's zero comma zero b squared is under x squared and since b is 3 we're going to travel three units to the left and to the right along the x axis now let's create our rectangle and then let's draw the asymptotes which is going to go from one side of the rectangle through the center to the other side now the coordinates of the vertices it's going to be zero plus or minus a and a is four so it's zero plus or minus four here's the first vertex at 0 comma 4 and here's the second vertex at 0 negative 4. and so the graph is going to look like this so that's how we can graph this particular hyperbola so now that we have the coordinates of the vertices and the center we need to find the coordinates of the foci so it's going to be zero comma plus or minus c and c is five so at zero five and at zero negative five we have two focal points so that's the coordinates for the foci now the last thing we need to do is write the equations of the asymptotes so it's going to be y is equal to plus or minus the slope times x the slope is going to be rise over run so the rise is a units the run is b units so rise over run is going to be a over b a is 4 b is 3 and so it's going to be y is equal to plus or minus 4 over 3 x so that's how we can graph the hyperbola identify the coordinates of the center of vertices and foci and that's how we can write the equations of the asymptotes for this particular equation now let's write down some more notes so let's focus on the horizontal hyperbola first when this hyperbola is centered at the origin the equation that correlates to it is x squared over a squared minus y squared over b squared but what happens if the center is not at the origin let's say it's at some point h comma k the equation changes to this it becomes x minus h squared over a squared minus y minus k squared over b squared is equal to one so you may want to write some notes down even if you don't know how to apply it you may just want to jot these down and then use it later for reference now the next thing that we mentioned is that the vertices for horizontal hyperbola centered at the origin is plus or minus a comma zero when it's not centered at the origin it's going to change all we need to do is add the center to this to those coordinates so we're going to add h to plus or minus k i mean plus or minus a rather so it's h plus or minus a and then we're going to add 0 with k so it just becomes k so that's how you can write the new vertices simply add the center to the vertices and then you get the the new vertices when the center is not at the origin the coordinates of the foci is plus or minus c comma zero so when it's shifted and the center is nothing not at the origin simply add h and k to this so it becomes h plus or minus c comma k now the equation of the asymptotes is plus or minus b over a times x when the graph is shifted this becomes y minus k is equal to plus or minus b over a times x minus h so that's how the equations will shift when the center moves from the origin to some point h comma k now what if we have the vertical asymptote how will the equations change so first let's write the equations when it's centered at the origin so for this one it's going to be y squared over a squared minus x squared over b squared and when the center shifts from the origin to h comma k the equation will be adjusted so it's going to be y minus k squared over a squared minus x minus h squared over b squared which equals 1. the vertices will be zero comma plus or minus a the new vertices all we're going to do is add the center to this so it becomes h comma k plus or minus a the foci will be zero plus or minus c the new coordinates of the foci will be h comma k plus or minus c the equation for the asymptotes will be plus or minus a over b x but the new equation becomes y minus k is equal to plus or minus a over b x minus h so feel free to write those equations this equation remains the same c squared is equal to a squared over b squared in all cases number three graph the hyperbola so we're going to follow the same instructions that we did for the first two problems but this time we could see that it's not centered at the origin so let's begin by finding the center here we have x minus 3. we're going to change negative 3 to positive 3 and here we have y plus 2 we're going to change positive 2 to negative 3. so the coordinates of the center are 3 negative 2. a squared is typically the first number that we see here a squared is four so a is going to be two b squared is nine so b is going to be three next we need to calculate c a squared is four b squared is nine four plus nine is thirteen so c is going to be the square root of thirteen now let's go ahead and graph the hyperbola so let's begin by plotting the center which is that 3 negative 2. now a is 2 a is associated with the x variable so we're going to travel 2 units to the right parallel to the x axis and 2 units to the left b is three b is b squared was under y squared so it's associated with y so we're going to go three units up parallel to the y axis and three units down from the center parallel to the y axis now let's draw a rectangle and then let's draw the asymptotes now this particular hyperbola will it open to the left and right or will it open up and down do we have a horizontal hyperbola or would you say we have a vertical hyperbola now we have a positive sign in front of x squared so this is going to open left and right if y squared was positive it would open up and down so the graph is going to look something like this so these two points represent the vertices of the hyperbola and if you recall the vertices for a horizontal hyperbola that is not centered at the origin is going to be h plus or minus a comma k h is 3 a is 2 and k is negative two so you can always look at the center to find h and k so first we have three plus two which is five comma negative two so that's the vertex on the right side for this one it's going to be three minus two comma negative two three minus two is one and so we get the point one negative two for this vertex on the left so those are the coordinates of the vertices now let's find the coordinates of the foci so it's going to be h plus or minus c comma k h is 3 c is the square root of 13 k is negative 2. so this represents the coordinates of the foci now to get the x value of the foci we have 3 plus the square root of 13. the square root of 13 is 3.6 so 3 plus 3.6 is 6.6 so somewhere in this region we have the focus and then 3 minus the square root of 13 which is negative 0.6 so somewhere in this region is the other focal point so that's the location of the foci on the graph the last thing we need to do is write the equations of the asymptotes so it's going to be y minus k is equal to plus or minus now the rise is v units the run is a so the slope is going to be b over a and then it's x minus h so k is negative two y minus negative two becomes y plus two so what we have here is similar to the original equation this is going to be equal to plus or minus b over a b is 3 a is 2 and then x minus h so we have x minus 3. so this represent the equations of the two asymptotes now let's work on another problem so we're going to do what we did in the last problem we're going to graph the hyperbola identify the coordinates of the center vertices and foci we're going to write the equations of the asymptotes but this time we're also going to determine the domain and range of the graph so what's a squared and what's b squared a squared is usually the number in front so a squared is going to be 9 which means a is the square root of 9 which is 3. b squared is going to be the other one 16 b the square root of 16 is 4. now let's calculate c c squared is a squared plus b squared a squared is 9 b squared is 16. 9 plus 16 is 25 and the square root of 25 is 5. now let's determine the center of this particular hyperbola here we see negative two so the x coordinate of the center will be positive two here we see negative one the y coordinate of the center will be positive one now let's go ahead and plot it so let's begin by plotting the center which is at two comma one and then we see that a is three a is under y squared so we're gonna go up three and then we're gonna go down three now b is four b is under the x variable to think of left and right when dealing with x so we're going to go right 4 from the center and then left 4 units now let's draw a rectangle next let's write out or draw the asymptotes now let's do the same for the other one i think i messed it up let's do this one more time i know my graph is not drawn to scale but we'll make the best of that so now the graphene part so we know it's going to be curved at the the vertex and then we need to follow the asymptote so let's try to draw an accurate graph so that's as best as we can do for it now as long as you draw a rough sketch most teachers will give you you know the correct answer but ideally you want to follow the asymptote as you move away from the vertices so that's the hyperbola that we can draw for this particular problem now we have the coordinates of the center so now let's focus on the vertices so for this particular type of graph where we have a vertical hyperbola the vertices is going to be the coordinates of the vertices will be h comma k plus or minus a so h is 2 k is 1 so it's 2 comma 1 plus or minus a which is 3. so this point here this vertex and that's going to have an x value of 2 and then a y value of 1 plus 3 which is 4. now the vertex below that will have the same x value but the y value will be one minus three so the point will be negative two two negative two so those are the coordinates of the vertices two comma four and two negative two now let's focus on finding the coordinates of the foci so the coordinates will be h comma k plus or minus c h is two k is one and c is five so the first one is going to be at two comma six which will be somewhere here the next one will be 2 and then it's going to be 1 minus 5 which is negative 4. so i'm just going to write that here so this one is 2 comma 6 and this will be 2 negative 4. so those are the coordinates of the foci now let's focus on the equations of the asymptotes the formula that we need is y minus k is going to equal plus or minus the slope and so we have a rise of a a run of b so it's going to be a over b and then x minus h so y minus k is going to be what we see here that's y minus 1 is equal to plus or minus a is 3 b is 4. and then x minus h is what we see here so x minus 2. so those are the equations of the asymptotes now the last thing we need to do is find the domain and range of the graph for a vertical hyperbola the domain is always going to be all real numbers when you need to find the domain look at the x values analyze it from left to right the lowest x value will be negative infinity and the highest x value is positive infinity so x can be anything along this curve this curve is continuous there are no jumps as you go from left to right so the domain is all real numbers now the range is a different story we need to look at the y values so view it from going from the bottom to the top this can go all the way down to negative infinity so the lowest y value is negative infinity as we go up this stops at a y value of negative two now between a y value of negative two and four there is no curve there is no hyperbola so we're going to pick it up back at four and then this goes up to infinity so the range from bottom to top is negative infinity to negative 2 negative 2 is included because that's a vertex so we're going to use a bracket union we're going to start back up at 4 and continue to infinity so that's how we can write the range of a hyperbola and this is the domain for a vertical hyperbola you