Graphing Conic Sections: Circles, Ellipses, Parabolas, and Hyperbolas

in this video we're going to focus on how to graph conic sections like ellipses parabolas hyperbolas and also circles as well and in addition to that we're going to talk about how to look at equation and to tell if it's going to be a circle ellipse Parabola and hyperbola and how to put it in standard form so let's focus on the graphing part let's say if we have this equation squared plus y squared equals 9. and this is a circle but how can we graph this circle so the standard equation for a circle is it's x minus H squared plus y minus K squared is equal to r squared the center of the circle is H comma k but there's no H and K in this problem so the center is based on the origin it's zero zero now r squared is basically 9. so if r squared is equal to 9 if you take the square root of both sides um you'll see that the radius is 3. so starting from the origin which is 0 0 you need to go three to the right three to the left up three and then down three and then draw a circle connecting those points so that's how you can graph that particular Circle or that equation let's try another example so let's say if we have this equation x minus 3 squared plus y minus 4 squared actually let's say y plus 4 squared equals 16. so we need to find H and K so the center is basically if you set x minus 3 equal to zero and you solve for x you'll get three so H is 3. so basically you change the negative 3 to a positive three and the same thing with the four you change the positive 4 into a negative 4. so the center is three negative four and since r squared is equal to 16 the radius of the circle is 4. and then we just have to graph it so we need to go three units to the right and up four I mean down four units so here's the center it's at 3 negative 4. now because the radius is 4 and we need to go four units to the left which will give us this point four units to the right and then up four units and then down four units and so that's the circle let's try one more example so let's say if we have this equation x plus two squared plus y minus 3 squared is equal to four so the center is negative two comma three you change the positive two into a negative two and the negative three into a positive three and the radius is the square root of four so it's two and then plot the center so we need to go 2 to the left of three so it's over here and in radius is two so we need to go down two and then up to 2 to the right and then two to the left and then connect those four points and so that's how you can graph circles so now let's move on to an ellipse an ellipse looks similar to a circle but it's uneven it's not equal in all sides let's start with this equation x squared over 25 plus y squared over 49 and let's say that's equal to 1. now the general equation for an ellipse it's x minus H squared a squared and then y minus K squared over B squared is equal to 1. now the a squared is not always under the X a squared could be under y minus k we need to know to distinguish between A and B A Squared is the larger of the two so in this case um for this problem that we have this is actually a squared because 49 is larger than 25. 25 is actually B squared now we need to realize that the center is zero comma zero because if you see just x squared and Y squared if it's not like x minus 2 or Y plus 3 then the center is 0 0. so it's based on the origin now since a squared is equal to 49 what this means is that a is equal to 7. and 7 is under y squared so from the origin we need to go seven units up and seven units down the way we plot is very similar to a circle but um there's no radius A and B are different if a and b were the same then it would be a circle now B is uh B squared is 25. so if you take the square root of 25 that's 5 and B is under X so we need to go 5 units to the right and five units to the left and then we just connect those points so that's how you can graph an ellipse now you need to know that this axis right here is called the major axis but it's vertical so it's the major vertical axis this line here because it's shorter it is known as the minor axis but it's horizontal so it's the minor horizontal axis and this one is the major vertical axis so let's say if you get a question and they ask you for what is the major vertical axis length how long is it the major vertical axis it's always equal to twice the value of a and so in this case I'm going to write M A for major axis since it's equal to 2A it's 14. the minor axis they're both n the left of the minor axis is twice the value of B so 2 times V is 10. by the way the endpoints of the major axis which is um zero seven and zero negative seven those endpoints are known as vertices which is the plural form of vertex so some books will call it the major vertices and other textbooks we'll call the endpoints of the minor axis like the minor vertices if they don't use the word minor then the vertices corresponds to the endpoint of the major axis now the only other thing we need to find is the foci the foci is it's along the major axis but whenever you have a graph that opens up the foci is somewhere in the middle so this should be a full guy somewhere between along this major axis between 0 and 7. to find it you need to use this equation c squared is equal to a squared not plus but minus B squared for a hyperbola it's a squared plus b squared kind of like the Pythagorean theorem but for um for an ellipse it's a squared minus B squared because um C is less than a and I'll explain why shortly but let's solve for c squared we already know the value of a squared and this is B squared so it's 20 not 25 but 49 minus 25. and that's 24 but if you take the square root of both sides the square root of 24 we can reduce that to root 4 and root 6. so it's 2 root 6. so the foci the coordinates of the foci x is 0 Y is C so it's plus or minus 2 root 6. as a decimal two root 6 is about 4.89 so just under 5. so the forecast should be somewhere over here and somewhere over here so it's inside imagine this ellipse has like a half Parabola it's inside of it so that's why the foci that's why C is less than a and so therefore it's c squared equals a squared minus B squared because a is bigger than C as you can see this is the value of a which is longer and this is the value of C is the distance between the center of the ellipse and the foci so if the origin is not 0 0 then the coordinates of the foci will be affected by the way sometimes you may get a question to ask you for the X and Y intercepts for this particular example the y-intercepts are basically the endpoints of the major axis which is 0 7 and 0 negative seven so those are the Y intercepts the x-intercepts are the endpoints of the minor axis so in this case it's a plus or minus five comma zero that's a nice way to write it if you don't want to write it twice but let's try another example so let's say if we have this problem x minus 2 squared divided by 16. Plus y plus 3 squared divided by 9. and let's say that's equal to one now the first thing you need to do is find the center so since we have x minus 2 we need to reverse it so it's gonna be plus 2 and for y plus 3 change it to the minus 3. so we need to travel two units to the right and down three so that's where the center is but keep in mind the center is not actually on the graph so now we need to know what's a squared and what's B squared a squared is the larger of the two this is a squared and this is B squared so if a squared is 16 that means a is the square root of that which is four and the square root of 9 is 3. so notice that a is under um the it's under X so you need to travel four units along the x-axis so four units to the right and four units to the left and B is under the the Y variable so in V is three go up three units and then down three units so it's not very difficult to graph an ellipse it's finding the other details about it that you have to be careful so let's identify what's the major axis and what's the minor axis the major axis is it vertical or horizontal the major axis is wherever a is a is along the x-axis so the longer side is the major axis the major axis is that is basically the x coordinate of this actually the y-coordinate because it's horizontal so the major axis is that y equals negative three that's the equation for it the minor axis is wherever B is this is the minor axis because B was over here so the minor axis if you want to write that as an equation it's x equals 2. the length of the major axis is equal to 2A twice the value of a and since a is four it's about eight units so this is positive six this is negative 2. they differ by eight units and the length of the minor axis that's twice the value of B and so B is three so it's six this this this is negative six here and Y is zero at this point so you can see this is a distance of six and here we have a distance of eight now if you want to find the endpoints of the major axis since you already graphed it you can see that it's going to be negative 2 3. and 6 3. but if you wish to find it from the center the endpoints of the major axis is basically the vertices and it's for this particular example is h plus or minus a comma k because a is along the uh the major axis which is uh it's horizontal so we add or and subtract a to the x coordinate of the center so basically we do 2 plus or minus 4 comma negative 3. so keep in mind H is 2 K is negative 3. so basically what this means is that we go four to the right along the x axis and 4 to the left from the center and so we get these two points two plus four is six and two minus 4 is uh negative two which correlates to this point this is 6 negative 3 and negative 2 negative 3. so that's the vertices or the endpoints of the major axis now if we wish to find the minor vertices or the endpoints of the minor axis which is right here we can use this equation so I'm going to put MV for like minor vertices so it's going to be H comma K plus or minus B because we added and subtracted B to the y coordinate of the center so it's 2 comma the y coordinate which is negative 3 plus or minus the value of B which is three so the two answers that we get is a two comma negative three plus three is zero so that's basically this point right here two zero and negative three minus 3 is negative six so we get this point two negative six so these are our minor vertices now if you need to find the intercepts as we can see we have one x intercept only in this graph which is 2 0. and there is a y-intercept which is here and here now we don't know exactly what those points are just by looking at this graph because this is a rough sketch but for any equation if you have to find the x-intercept plug in 0 for y and then algebraically solve for x if you need to find a y intercept plug in 0 for x and then solve for y now the last thing we need to do is find the foci the focus should be somewhere along the major axis now those points aren't precise but it's a rough estimate the first thing you need to do is find the value of C so c squared is a squared minus V squared which is um a squared is 16 B squared is nine so that's seven so if you take the square root of both sides C is radical seven so to write the coordinates of the foci since it's along the since the major axis is horizontal you need to add and subtract C to the x coordinate of the center so the equation for this particular graph it's going to be h plus or minus C comma k so let's see if I could put that here so the foci is going to be H which is 2 plus or minus C which is radical seven comma negative 3. if you want you can convert that to a decimal value but the full class should be somewhere around here and some way there as you can see it's inside of the ellipse so for the sake of practice let's try one more example let's say if we have a X Plus 1 squared over nine Plus y minus 2 squared over uh 25 and let's say that's equal to 1. so first let's find a center it's a negative one comma two which is right here and a squared actually that's a b squared B squared is a smaller of the two B squared is nine a squared is 25 since 25 is larger than 9. so that means a is 5 and B is 3. so a is under the Y so we need to go up five units one two three four five and then down five three four five B is associated with X so we need to go three units to the right and then three units to the left and then connect the the auto points so that's our ellipse and if you want to find a foci solve for C so since c squared is a squared minus V squared we know a squared is 25 B squared is 9. 25 minus 9 is 16. the square root of 16 is 4 so C is 4. so we need to go up four units which is over here and down four units from the center so that's the coordinates of the foci but to write it this time it's not h plus or minus C it's h comma K plus minus C because the major axis it's vertical the foci the center and the major vertices they all pass through the major axis so therefore the foci is H which is negative one comma K which is 2 plus or minus the value of C which is four so two plus four is six that gives us this point it's at negative 1 6. and then the other one two minus 4 is negative 2 which is over here it's at negative one comma well actually this one negative 1 comma negative two so those are the coordinates of the foci so now let's find the endpoints of the vertices or the endpoints of the major axis so those two points you can probably see it what it is graphically but I'm going to do it this way it's h comma K plus or minus a since the major axis is vertical so it's negative one two plus or minus the value of a which was five so two plus five is seven and two minus 5 is negative 3. so these are the major vertices which are the endpoints of the major axis now keep in mind the length of the major axis is two times the value of a so in this problem it's ten that's the length of the major axis the life of the minor axis is two times the value of B which is six so let's try an example where we have to find the X and Y intercepts just for the sake of practice so let's say if you want to find the x-intercept plug in 0 for 1 I mean zero for y and then solve for x so what I would do to get rid of all the fractions is I would multiply by 9 and 25. so if you take this term and you multiply by 9 and 25 the nines cancel and so you get 25 X Plus 1 squared negative 2 squared is 4. and this term times 9 and 25 the 25s cancel so you just get 9. and then here 1 times 9 times 25 the value of nine quarters that's two dollars and 25 cents so we know it's 225. so 4 times 9 is 36. and 225 minus 36. that's 289 I believe and then divide both sides by 25. so X Plus 1 squared is 289 over 25. and then take the square root of both sides the square root of 289 I believe that's 17. and so you get X Plus 1 is equal to plus or minus 17 over the square root of 25 which is 5. so X is negative 1 plus the minus 17 over 5. now if you go back to the example that we had notice that we have two x-intercepts because the ellipse it touches the x-axis twice if you want to find a y intercept pretty much plug in 0 for X solve for y and just basically follow the example that we just did but now let's move on to hyperbolus so let's say if we have x squared over nine minus y squared over four equal one the general equation of the hyperbola it's x minus H squared over a squared minus y minus K squared over B squared equals one now a is not the larger of this 2 this time a squared is simply whichever comes first you can also have the other form of this equation which is y minus K squared over a squared minus x minus H squared over B squared equals one so whichever one is positive is associated with a whichever one is in front of the negative sign or whichever one is negative um has the B squared now we need to realize that for a hyperbola if x squared comes first it opens uh left and right if y squared comes first it opens up and down but we'll go over a few examples so let's solve this one the center is 0 0 because there's no H and K for this example so that's where the uh that's the midpoint of the two vertices since a is 3 we need to go three to the left and 3 to the right and B squared isn't 4 so that means B is two so we need to go up two and down two and then make a box in this case like a rectangle and then draw two lines these are the asymptotes make sure that these two lines crosses the uh make sure it forms like a diagonal through the rectangle it has to pass through the center as well now these two points are where the graph is actually going to touch it's going to meet at those two points it doesn't touch these two points here those are imaginary points so this graph is going to look like this it follows the asymptote it touches the vertex and then it follows the other asymptote so that's how you graph a hyperbola these are called the uh the vertices those are the vertex of the hyperbola so the vertices are plus or minus 3 comma zero the foci is somewhere here it it's always where the graph opens towards so if the graph opens this way the foci is like towards it if it opens this way the foci is over here now to find it it's a little different instead of it's c squared equals a squared not minus but plus b squared for an ellipse it's a squared minus B squared this time as you can see the value of C is larger than the value of a so that's why c squared is a squared plus b squared so it's going to be 9 plus 4. that's 13 so C is the square root of 13. so the foci therefore is plus or minus radical 13 comma zero now the only other thing that we need to do for this type of graph is we need to find the asymptotes the equation for the asymptotes it's y minus K is equal to now it can be B over a or A over B plus or minus and then x minus h here's how you tell so what we have is the asymptote is basically a line and whenever you have a linear equation in slope intercept form it's MX plus b notice that the letter in front of X is M which represents a slope so B over a is the slope of the line which is rise over run so from the center to this point you need to go up B units that's your rise and then you run a units so it's a rise over run B over a for this particular example when it opens this way it's actually going to be A over B but you can tell by just doing Vise over run so for this actual equation we know that H and K are zero so to write the asymptotes it's going to be just y there's no k is equal to plus or minus the value of B which is um B is a two B squared is four so B is 2. and a if a squared is nine a is three so it's Y is equal to plus or minus two thirds X that's the those are the asymptotes for this particular hyperbola so let's try another example so let's say if we have X plus one squared over 25 minus y minus 2 squared over 16. so a squared is the first number it's not the bigger or the smaller one whichever comes first whichever is positive and the second one is usually negative so that's B squared so we know that a is 5 and B is 4. and the center is negative one positive two just don't forget to change the signs to find a full guy it's c squared is equal to a squared plus b squared so a squared is 25 B squared is 16. so that's 41 so C is the square root of 41. so let's plot the center first it's at negative 1 positive two so it's over here now a is under the X variable so we need to go five units to the left and five units to the right so this coordinate is at negative six this is four and B is four so we need to go up four units so that's at positive six and down four units to negative two and then draw the rectangle and then draw the asymptotes now this is the value of a same as here so a is horizontal this time just like last time by the way this is called the transverse axis and the graph is going to touch at those two points so it should look something like this so the vertices which are these two points but let me write the general equation it's a h plus or minus a comma k so H is uh the x coordinate of the center and a is five and K is 2. so negative one plus five that's four and negative 1 minus five is negative six so this point as we can see it's um it's four comma two here this is two and this is four so that's four comma two and this one it's at negative six comma two which is what we have so those are the vertices of the graph now to find the foci it's also along the transverse axis is horizontal so it's going to be H but plus or minus C comma K so we've got to find the value of C which we already know it to be root 41 so it's negative 1 plus or minus root 41 comma 2. now we can't add negative 1 plus root 41 so we're just going to leave it like this so we know the foci is somewhere over here let's and it's somewhere on the outside it's past the vertices now the only other thing we need to find is um the asymptotes which is y minus K is equal to plus or minus now rise starting from here we need to rise B units because B is a vertical a is horizontal so the run is still a so it's a b over a rise over run and then x minus h so that's y minus K is negative two action okay is a case Plus 2. so it's a y minus 2 is equal to the plus or minus B over a this is B this that's a so four over five x minus h so H is um negative one so negative negative one you basically get what's here so X plus one so notice that what you see here is what's here and what you see there is the same as that but this is the equation of the asymptotes now the reason why we have plus and minus is because there's two of them if it's plus four over five the slope is positive so it's going up minus four over five the slope is negative so it's going down the negative one is four the one that decreases as you go from left to right the positive one is for the one that goes up has to go from left to right now for the sake of practice we're going to try one more example let's try this one X minus 2 squared actually not let's start with Y this time so y minus 2 squared over 4 minus X plus 3 squared over nine and that's equal to 1. so a squared is whatever comes first and this is B squared so a therefore is the square root of 4 which is two B is the square root of nine which is three and to find c c squared is a squared plus b squared so 4 plus 9 which is 13. so therefore C is root 13. the center is uh let's look at the x value first so it's negative 3 and the Y Value Plus 2. so it's at negative three comma two now since a is 2 but it's under the Y we need to go up two units and then down to units so wherever um the points that a give you is the vertices so that's where the graph is going to be so the transverse axis this time is actually vertical now to find we know B has to be so we need to go three units to the right three units to the left and let's draw our rectangular box next let's try the asymptotes so keep in mind the major the vertices are here so this graph is going to open like this it opens up and down because Y is in front this time so the foci is somewhere over here it's up and down rather than left and right and to find the coordinates of the foci it's we need to add and subtract C to the y-coordinate of the center because we went up and down and Y is associated with up and down X is associated with left and right so the foci is the center negative three comma two but plus or minus C or plus or minus root 13. so that's the coordinates of the foci and for the vertices which are basically these two points it's h comma K plus or minus a so it's negative three and K is 2 so 2 plus 2 is 4. and 2 minus 2 is 0. so those are the major vertices now the last thing we need to find is the asymptote which is y minus K is equal to plus or minus okay so we went up a units the Rises a the run is B and since the slope is rise over run It's A over B and then x minus H so y minus K is basically exactly what you see here so it's y minus 2. and a is 2 B is 3 so 2 3. x minus H is basically X plus three and that's how you find the equation of the asymptotes and I believe we're not missing anything I think that's it I think we covered everything there so now let's move on to our next topic so let's focus on the parabola the general equation for a parabola well before I do that let's go over the different types this Parabola is y is equal to x squared and if it opens downward Y is equal to negative x squared if it opens to the right X is equal to Y squared and when it opens to the left X is equal to negative y squared now somewhere over here this is the directrix and the distance between the vertex and the directrix is p so in this case it's Y is equal to negative p and the foci is p units above the the vertex so wherever the graph is that's where the focus so the focus is over here that the Matrix is over here in this case the coordinates of the focus is um it's it's p comma zero but for this example is zero comma p because you want P units up here you want p is to the right for this one the Matrix is X is equal to negative P rather than uh y and over here it's Y is equal to positive p and since the focus is below um the origin the coordinates is zero negative p so you have to see you have to know the general shape of the graph to see where um the focus is going to be if you have to add P or subtract p so keep in mind for the directrix if you have a horizontal line it's y equals if you have a vertical line it's x equals here it's x equals positive p and the focus is actually negative P comma zero this time that's if the parabola is based on the origin now let's say if it's moved away or transformed or shifted away from the origin these are the equations that you're going to need y minus K squared is equal to 4p x minus h so basically this is the equation that's like y squared is equal to X or x equals y squared so this can open to the right or to the left it opens to the right if p is positive or greater than zero if p is negative it opens towards the left now the next equation that you need x minus H squared is equal to 4p y minus k so this is basically y equals x squared which can open up or down it opens upward if p is positive and if p is negative it opens downward all right so let's try some practice problems let's start with the basics let's say that y squared is equal to 8x so that means there's no H and K so the center is zero zero or at least we really don't have a center we should really call it the vertex that's a zero zero so this is in the form y minus K squared is equal to 4p x minus H so notice that 8 is equal to 4p so set 4p equal to 8. if you divide both sides by 4 8 divided by 4 is 2 so p is 2. and then we can graph it now it really depends on what you want to do here I'm going to draw a smaller graph but more like a rough sketch since we have y squared equal to X or x equals y squared we know it's going to open to the right or to the left and since p is positive 2 we know the graph opens to the right which means that the Matrix is over here and p is 2 so the foci is over here so the foci it's um it's p comma zero or in this case two comma zero since p is two and the directrix is X is equal to negative 2 because you got to go P units to the left so the first thing we should do is plot the vertex then go P units to the right and then P units to the left so there's your directrix and here's the focus but now how can we get how can we like plot this graph accurately what you could do is you can plug in some points now it's easier if you plug in let's see we could make a table if you want should we plug in values for x or 4y which one is easier if you have y squared it's easier if you plug in values for y rather than x let's plug in one for y if you plug in 1 um actually one is going to be too small let's solve for x y squared divided by 8 is equal to X let's say if we plug in 2 for y that's 2 squared is 4 4 over 8 is a half if you plug in negative 2 for Y when you square it it's still going to be positive four so you're going to get the same x value let's try a bigger number like 4. plus or minus 4 because it's going to give us the same x value if you plug in 4 4 squared is 16 16 divided by 8 is 2. the last point to plug in would be like 8 for y 8 squared is 64. and 64 divided by 8 is 8. but we may not need to go that far so when we plug in 2 for y X was about 0.5 so that means it's somewhere over here but also for negative 2 we get positive 0.5 for x let me use a different color the key is to choose the right points we want to choose points that you can get whole numbers which four was the best example when Y is 4 X is 2. so that means it's over here that's 2 comma four and two comma negative 4. so we have a very wide graph and let's say if we were to extend this graph so this would be three four five six seven eight now let's go up along the y axis so 5 6 7 8 and I'm out of space over here so this is somewhere over here it's at 8 8 and 8 negative eight it's as well I'm out of space there so this graph looks something like this so if you want to get a very accurate graph if that's important make a table plot some points if you just want a rough sketch you just need to know what the direction is because if your test is multiple choice you don't need a table you can pick the right ones and eliminate the wrong ones let's try another example so let's say if we have y minus 2 squared is equal to 4 x minus 3. so our general equation is y squared is equal to X so this one's going to open to the right again and we know the focus is going to be over here the Matrix is over here so the center is 3 for X positive 2 for y so we need to go three units to the right up two units so that's the I keep saying the center but I'm at the the vertex so now we need to find p whatever this number is set it equal to 4p so therefore p is one and since it opens to the right we need to go one unit to the right that gives us the focus so the focus is basically h plus P comma k or four plus one I mean three plus one which is four and K is 2. so that's the coordinates of the folk the focus for the directrix we need to go one unit backwards and then we can draw a vertical line so the Matrix is that X is equal to positive two so now let's see if we can find some points to graph in so we have a 4 in front let's say if we plug in let's make a table we know the center is 3 or the vertex is 3 2. let's plug in 4 for x and let's see what we get for y if we plug in 4 fractions it's going to be 4 times 4 minus 3 is 1. so we get that and if we square root both sides y minus 2 is equal to 2. well we get plus or minus 2. so y minus 2 is equal to 2 and Y minus 2 is equal to negative 2. so Y is equal to four and here if we add to Y is equal to zero so that means when X is 4 y can be zero and it could be four so four zero is basically over here and 4 4 is over there in the last example we plugged in for y and solve for x but in this example I chose to solve for um y instead now if I choose 5 for x it's going to be 2 times 4 which is 8 and I can't take the square root of 8. so the next value I'm going to choose for x is going to be 7. because 7 minus 3 is 4 which is 16. and we can take the square root of 16 because that's going to give me an integer which is 4. so y minus 2 is equal to plus or minus 4. so y minus 2 is 4. that means if you add 2 you get six and Y minus 2 is equal to negative 4 as well if you add 2 you get um negative 2. so I kind of picked the points differently from the last example so we need to go to seven and at seven it's going to go up to six which is over here and it's going to go down to negative 2. so that should be that's enough points to get a rough sketch for this graph and that's how it looks like so our last example for today let's try this One X plus one squared let's say that's equal to um two times y minus 3. so here we have the example where x squared is equal to y or Y is equal to x squared and let's actually add a negative sign so it's going to open downward so the focus should be somewhere over here and the directrix is horizontal this time so let's find the center it's negative one comma three so one to the left up three and actually mean the vertex I keep calling the center so now we need to find the value of P so negative two is equal to 4p so P if you divide by 4 that's negative a half two over four is one half so this is the center we need to go half units well it opens the graph opens downward so we need to go down by half and up by half so if you go down by half that's the focus but it's too small to draw the directrix we note this line here so the focus is going to be uh H comma k but since we need to go down it's minus p so H is negative 1 K is three but we need to go down by half so that's we need to subtract three by one half so three is basically six over two minus a half that's five over two as you can see the focus is at 2.5 is between two and three so I'm going to draw a little F here for the focus the directrix it's going to be uh Y is equal to since it's horizontal it's going to be K plus P we need to go up P units so that's three plus one half or six over two plus one over two so it's Y is equal to seven over two that's the Matrix okay so we have the general shape of the graph but the hard part is choosing the values for X and Y that we should pick so if it's not in standard form it's easier for you to plug in well I guess you have to pick which way it's easier for you um but we'll need to plug in some values you can choose the plug-in values from for x and solve for y or you can plug in values for y and solve for x it really depends on whichever you think is easier what I'm going to do is I'm going to plug in a value for y and solve for x now my directrix is that 7 over 2. and the graph to Center the y-coordinate is at uh three so the graph doesn't go past three it doesn't go past the vertex if it opens downward so I can't choose any value that's greater than 3 for y let's say if I choose four four minus 3 is 1 1 times negative two is negative two and you can't square root a negative number so I have to choose something that's less than three I already have the value for three when Y is 3 x is negative one that's the vertex so I need to choose something less than three but I want to choose a number where when I multiply by negative 2 it's going to give me a perfect square so I want to choose one for y because 1 minus 3 is negative 2 and negative 2 times this negative 2 in front is 4. and I can take the square root of 4. so if X Plus 1 squared is equal to four X Plus 1 is therefore equal to plus or minus two so it's equal to two and negative 2. If X Plus 1 equals 2 If I subtract both sides by 1 I get x equals to 1. and for the other one If I subtract 1 from negative 2 negative 2 minus 1 that's negative 3. now I need to choose another point where I can get another perfect square perfect squares are like 4 9 and 16. I already use four I don't want to use nine because I need a fraction to get that well actually let me use nine let's see I need to set y minus 3 equal to half of 9 which is 4.5 but negative 4.5 so if I add 3 to both sides Y is going to be equal to negative 1.5 I'm going to choose that for my y value so if I plug in negative 1.5 I get negative 2 times negative 1.5 minus 3 which is negative 2 times negative 4.5 that's a 9. so therefore X Plus 1 squared is equal to 9. and if you square root both sides X Plus 1 is equal to 3 which means three minus 1 is 2 and X plus one is equal to negative three negative three minus one is negative four so that should be enough points for this particular example so when X is one y is one so that's over here and when X is negative three Y is one so we can see the symmetry about this line here when X is 2 Y is negative 1.5 which is about there somewhere and when X is negative four it's over here so this is enough points to graph it so now let's say if you're given these four equations x squared plus 6x minus 4y plus one is equal to zero and 4x squared minus 9y squared minus 16x Plus 54y let's say that's equal to zero as well and then this one 2x squared and the last one so given these four equations which one is a circle which one is an ellipse which one is the parabola and which one is hyperbola now the first thing I would look for is the problem if you have all four options the parabola is the one that has an x squared but not a y squared or it has a y squared but not an x squared so here we see you know I meant this to be a y squared by the way so here we have an x squared and a y squared so that's not a problem x squared y squared not a problem x squared y squared this is the the first one is the only one that doesn't have an x squared and a y squared so this is the parabola that's how you can distinguish it from the other three here's the x squared but there's no y squared or they could be a y squared but no x squared so now which one is the hyperbola that's the next thing you want to look for now the hyperbola has a positive x squared and a negative y squared or vice versa the ellipse and a circle both x squared and Y squares are positive so both of these are positive both of those are positive but notice this we have positive x squared negative y squared the ellipse the hyperbola has the minus sign so whenever you see that a positive x squared negative y squared or vice versa it's hyperbola now to distinguish the circle from the ellipse look at the coefficients of x squared and Y squared a circle is perfectly even so if the coefficients are the same it's a circle if the coefficients are different it's an ellipse and that's how you can distinguish these four conic sections if it's not in standard form so now we're going to talk about how to put these equations in standard form so let's start with a circle so we had 2x squared plus 8x plus 2i squared Plus 4y minus 6 is equal to zero so the first thing you want to make sure is you want to group The x's and the Y's together which we already have that so for the first two terms let's factor out the GCF of the coefficient so between two and eight we can take out a two and so we're left with x squared plus eight X and leave a space and between the next two terms we can take out a two as well so we have y squared plus two y leave a space and then the negative six let's move it to the other side now we need to complete the square half oh wait this should be um this should be a 4. not an 8 because that took away two so you want to take half of four half of 4 is 2 and then Square it half of 2 is 1 and then squared that's how you complete the square but now notice that we added 2 squared times 2 to the left side 2 squared is 4 times 2 is 8. so we got to add 8 to the other side as well 1 squared times two is two so let's add 2 as well so now we can Factor it's going to be X plus whatever that sign is and then this number before you square it X plus 2 and then squared to factor the next term it's going to be whatever this variable is y Plus 1 squared 8 plus 2 is 10 10 plus 6 that's 16. so now when you have a circle you got to get rid of these coefficients so we're going to divide everything by 2. so our final answer is X plus 2 squared plus y plus 1 squared the twos cancel and then 16 divided by 2 is 8. so now it's in standard form the center of the circle is negative 2 1 and the radius is the square root of 8. so that's how you can write um the equation of a circle in standard form and then if you want to you can graph it if if needed so now let's move on to the ellipse so we had 4x squared plus 25y squared minus 24x Plus 100y plus 36 is equal to zero so first let's move the x's and the Y's together so 4x squared minus 24x plus 25y squared plus 100y and let's move to 36 to the other side so let's factor out a 4 from the first two terms we want x squared to have a coefficient of one so it's x squared and then negative 24x divided by 4 that's negative 6X and let's leave a space let's factor out 25. 100y divided by 25 is 4y leave a space and then let's complete the square so half of 6 is 3 ignore the negative sign and then Square it half of 4 is 2 so that's 2 squared so 3 squared is 9 times 4 that's 36 so we got to add 36 to this side 2 squared is 4 times 25 is 100 so let's add 100 to that side so now let's Factor it's going to be 4 whatever this variable is X minus this number squared and then plus 25 this variable y Plus this number to squared negative 36 and 36 they cancel so we're just left with a hundred now for ellipses and hyperbolas this number has to be a one so divide by what you see then so we're going to divide both sides by a hundred so we have x minus 3 squared so 4 over 100 divided backwards 100 divided by 4 is 25 but since you divide it backwards 25 goes to the bottom 100 divided by 25 is 4. and 100 divided by 100 is 1. so now we have the equation of the ellipse in standard form as we can see the center of the Ellipsis to read negative two and a is the larger of the two a squared is 25 so that means a is 5. B squared is 4 so B is 2. and keep in mind for uh for a in ellipse it's c squared equals a squared minus B squared if you want to find the foci so now let's try this one the hyperbola 4x squared minus 9y squared minus 16x plus 54y minus 101 equals 0. so let's move the x's and y's together so we have 4x squared minus 16x and then minus 9y squared Plus 54y and let's move the 101 to the other side so now let's factor out a 4. so 4x squared divided by 4 is x squared negative 16x divided by 4 is negative 4X leave a space next let's factor out negative nine 54 divided by negative 9 is negative 6. and now let's complete this Square half of 4 is 2 so let's add 2 squared half of six is three so we're going to add 3 squared now 2 squared is 4 times 4 we need to add 16 to this side 3 squared is 9 times negative 9 so that's negative 81. so let's add negative 81 to that side so now let's factor it's going to be 4 times x minus 2 squared and then minus 9 times y minus 3 squared so 101 minus 81 that's 20 20 plus 16 is 36. now we need the 36 to be a one for a hyperbola so let's divide every term by 36. so if we divide it backwards 36 divided by 4 is 9. and 36 divided by 9 is 4. and this we get 1. so the center for the hyperbola it's 2 comma 3. whatever is positive that's a squared and therefore this is B squared now for the last one this is a parabola and we want to put it in a form where it's like this x minus H squared is equal to 4p Y minus k so all the X's we want to move to the left side and everything else let's move it to the right side so x squared plus 6X is equal to negative 4y becomes positive 4y when you move it from left to right and a positive one becomes negative one so now let's complete the square half of 6 is 3. and so we need to add 3 squared to the left and whatever you add to the left side add to the right side so 3 squared is 9. so to factor left side it's going to be X Plus this number before you square it and then over here we have 4y negative 1 plus 9 is 8. so all we got to do now is factor out a 4 from the right side so it's 4 y Plus 2. and now the parabola is in standard form the vertex is negative three comma negative two and this number is equal to 4p so p is one and that's basically it so let's review the general equations for every conic section so starting with a circle keep in mind that your general equation is x minus H squared plus y minus K squared which equals r squared and the center is h k so once you have this Center let's say it's say this is H this is K we need to go R units to the right R units to the left of r down r and then graph the circle now for an ellipse if you have x minus H squared over a squared plus y minus K squared over B squared equals one so remember a is the bigger one a is greater than b so what's going to happen is let's assume the center is the origin you can have a graph that looks like this so this is the center this is therefore a that's negative a this is B this is negative B just remember these are the endpoints of the major axis so those are the vertices and these are the endpoints of the minor axis and your full guy is along the major axis which is somewhere over here so your vertices for this particular example it's uh h plus or minus a comma k and for your Foci it's h plus or minus C comma k and the last thing that you need to know is to find C it's c squared is equal to a squared minus B squared for an ellipse and if you have the other version where you have x minus H squared but let's say B is under X and Y minus K squared where a is under y this ellipse is going to be elongated it's like a vertical ellipse it's going to look like this so this is a this is negative a this is negative B that's B so now euphokai to find it it's going to be H comma K plus or minus C so you add and subtract C to the y coordinate instead of the X and your vertices are H comma K plus or minus a so the major axis is vertical the minor axis is horizontal now keep in mind the length of the major axis is 2A the length of the minor axis is 2b so to review for a hyperbola it can be x minus H squared over a squared minus y minus K squared over B squared equals one or it can be y minus K squared over a squared and x minus H squared over B squared so keep in mind for a hyperbola a always come in front so B could be bigger than a or a could be bigger than b it doesn't matter and c squared to find the full guy is a squared plus b squared rather than a squared minus B squared so for this example it's going to open this way so let's say if we drew the Box so this is a that's negative a this is a here this is B and that's negative B so for this particular example the asymptotes is y minus K which is equal to rise over one Rises B run as a so plus or minus B over a x minus h and the vertices for this equation it's um h plus or minus a comma k now for this one the graph opens this way I'm not going to draw the Box though but you know how how it is you know how to do it but this time the vertices is going to be H comma K plus or minus a and the asymptotes will be y minus K is equal to this time A over B so plus or minus A over B x minus h and that's really what you need for a hyperbola those are the main equations oh I forgot the foci the foci is going to be h k plus or minus C for this example and for this one it's going to be h plus or minus C comma k so just so you know the foci is where it opens towards so the four Chi is over here and in this example it's over here and lastly what we have is the parabola so for this one here we have X is equal to Y squared so it can open to the right or it can open to the left so your focus is always on the interior towards where it opens and your directrix is here so for this example we know that the vertex is H comma k so the focus is p units from the vertex so it's h plus P comma K and for this example the focus is H minus P comma k the directrix is X is equal to H minus p and here it's X is equal to h plus p now for the other one it's x minus H squared is equal to four p y minus k so for y equals x squared it can open up or it can open down here p is positive p is negative for this one when it opens to the right p is positive when it opens to the left p is negative so the focus is somewhere over here and the directrix is horizontal this time so the focus is going to be H comma K plus p and for this example it's h comma K minus p the directrix is horizontal so it's Y is equal to K plus p and here um it's Y is equal to K minus p so that's all the equations that you need for conic sections so basically that's the end of this video so thanks for watching and have a great day

in this video we&#39;re going to focus on how to graph conic sections like ellipses parabolas hyperbolas and also circles as well and in addition to that we&#39;re going to talk about how to look at equation and to tell if it&#39;s going to be a circle ellipse Parabola and hyperbola and how to put it in standard form so let&#39;s focus on the graphing part let&#39;s say if we have this equation squared plus y squared equals 9. and this is a circle but how can we graph this circle so the standard equation for a circle is it&#39;s x minus H squared plus y minus K squared is equal to r squared the center of the circle is H comma k but there&#39;s no H and K in this problem so the center is based on the origin it&#39;s zero zero now r squared is basically 9. so if r squared is equal to 9 if you take the square root of both sides um you&#39;ll see that the radius is 3. so starting from the origin which is 0 0 you need to go three to the right three to the left up three and then down three and then draw a circle connecting those points so that&#39;s how you can graph that particular Circle or that equation let&#39;s try another example so let&#39;s say if we have this equation x minus 3 squared plus y minus 4 squared actually let&#39;s say y plus 4 squared equals 16. so we need to find H and K so the center is basically if you set x minus 3 equal to zero and you solve for x you&#39;ll get three so H is 3. so basically you change the negative 3 to a positive three and the same thing with the four you change the positive 4 into a negative 4. so the center is three negative four and since r squared is equal to 16 the radius of the circle is 4. and then we just have to graph it so we need to go three units to the right and up four I mean down four units so here&#39;s the center it&#39;s at 3 negative 4. now because the radius is 4 and we need to go four units to the left which will give us this point four units to the right and then up four units and then down four units and so that&#39;s the circle let&#39;s try one more example so let&#39;s say if we have this equation x plus two squared plus y minus 3 squared is equal to four so the center is negative two comma three you change the positive two into a negative two and the negative three into a positive three and the radius is the square root of four so it&#39;s two and then plot the center so we need to go 2 to the left of three so it&#39;s over here and in radius is two so we need to go down two and then up to 2 to the right and then two to the left and then connect those four points and so that&#39;s how you can graph circles so now let&#39;s move on to an ellipse an ellipse looks similar to a circle but it&#39;s uneven it&#39;s not equal in all sides let&#39;s start with this equation x squared over 25 plus y squared over 49 and let&#39;s say that&#39;s equal to 1. now the general equation for an ellipse it&#39;s x minus H squared a squared and then y minus K squared over B squared is equal to 1. now the a squared is not always under the X a squared could be under y minus k we need to know to distinguish between A and B A Squared is the larger of the two so in this case um for this problem that we have this is actually a squared because 49 is larger than 25. 25 is actually B squared now we need to realize that the center is zero comma zero because if you see just x squared and Y squared if it&#39;s not like x minus 2 or Y plus 3 then the center is 0 0. so it&#39;s based on the origin now since a squared is equal to 49 what this means is that a is equal to 7. and 7 is under y squared so from the origin we need to go seven units up and seven units down the way we plot is very similar to a circle but um there&#39;s no radius A and B are different if a and b were the same then it would be a circle now B is uh B squared is 25. so if you take the square root of 25 that&#39;s 5 and B is under X so we need to go 5 units to the right and five units to the left and then we just connect those points so that&#39;s how you can graph an ellipse now you need to know that this axis right here is called the major axis but it&#39;s vertical so it&#39;s the major vertical axis this line here because it&#39;s shorter it is known as the minor axis but it&#39;s horizontal so it&#39;s the minor horizontal axis and this one is the major vertical axis so let&#39;s say if you get a question and they ask you for what is the major vertical axis length how long is it the major vertical axis it&#39;s always equal to twice the value of a and so in this case I&#39;m going to write M A for major axis since it&#39;s equal to 2A it&#39;s 14. the minor axis they&#39;re both n the left of the minor axis is twice the value of B so 2 times V is 10. by the way the endpoints of the major axis which is um zero seven and zero negative seven those endpoints are known as vertices which is the plural form of vertex so some books will call it the major vertices and other textbooks we&#39;ll call the endpoints of the minor axis like the minor vertices if they don&#39;t use the word minor then the vertices corresponds to the endpoint of the major axis now the only other thing we need to find is the foci the foci is it&#39;s along the major axis but whenever you have a graph that opens up the foci is somewhere in the middle so this should be a full guy somewhere between along this major axis between 0 and 7. to find it you need to use this equation c squared is equal to a squared not plus but minus B squared for a hyperbola it&#39;s a squared plus b squared kind of like the Pythagorean theorem but for um for an ellipse it&#39;s a squared minus B squared because um C is less than a and I&#39;ll explain why shortly but let&#39;s solve for c squared we already know the value of a squared and this is B squared so it&#39;s 20 not 25 but 49 minus 25. and that&#39;s 24 but if you take the square root of both sides the square root of 24 we can reduce that to root 4 and root 6. so it&#39;s 2 root 6. so the foci the coordinates of the foci x is 0 Y is C so it&#39;s plus or minus 2 root 6. as a decimal two root 6 is about 4.89 so just under 5. so the forecast should be somewhere over here and somewhere over here so it&#39;s inside imagine this ellipse has like a half Parabola it&#39;s inside of it so that&#39;s why the foci that&#39;s why C is less than a and so therefore it&#39;s c squared equals a squared minus B squared because a is bigger than C as you can see this is the value of a which is longer and this is the value of C is the distance between the center of the ellipse and the foci so if the origin is not 0 0 then the coordinates of the foci will be affected by the way sometimes you may get a question to ask you for the X and Y intercepts for this particular example the y-intercepts are basically the endpoints of the major axis which is 0 7 and 0 negative seven so those are the Y intercepts the x-intercepts are the endpoints of the minor axis so in this case it&#39;s a plus or minus five comma zero that&#39;s a nice way to write it if you don&#39;t want to write it twice but let&#39;s try another example so let&#39;s say if we have this problem x minus 2 squared divided by 16. Plus y plus 3 squared divided by 9. and let&#39;s say that&#39;s equal to one now the first thing you need to do is find the center so since we have x minus 2 we need to reverse it so it&#39;s gonna be plus 2 and for y plus 3 change it to the minus 3. so we need to travel two units to the right and down three so that&#39;s where the center is but keep in mind the center is not actually on the graph so now we need to know what&#39;s a squared and what&#39;s B squared a squared is the larger of the two this is a squared and this is B squared so if a squared is 16 that means a is the square root of that which is four and the square root of 9 is 3. so notice that a is under um the it&#39;s under X so you need to travel four units along the x-axis so four units to the right and four units to the left and B is under the the Y variable so in V is three go up three units and then down three units so it&#39;s not very difficult to graph an ellipse it&#39;s finding the other details about it that you have to be careful so let&#39;s identify what&#39;s the major axis and what&#39;s the minor axis the major axis is it vertical or horizontal the major axis is wherever a is a is along the x-axis so the longer side is the major axis the major axis is that is basically the x coordinate of this actually the y-coordinate because it&#39;s horizontal so the major axis is that y equals negative three that&#39;s the equation for it the minor axis is wherever B is this is the minor axis because B was over here so the minor axis if you want to write that as an equation it&#39;s x equals 2. the length of the major axis is equal to 2A twice the value of a and since a is four it&#39;s about eight units so this is positive six this is negative 2. they differ by eight units and the length of the minor axis that&#39;s twice the value of B and so B is three so it&#39;s six this this this is negative six here and Y is zero at this point so you can see this is a distance of six and here we have a distance of eight now if you want to find the endpoints of the major axis since you already graphed it you can see that it&#39;s going to be negative 2 3. and 6 3. but if you wish to find it from the center the endpoints of the major axis is basically the vertices and it&#39;s for this particular example is h plus or minus a comma k because a is along the uh the major axis which is uh it&#39;s horizontal so we add or and subtract a to the x coordinate of the center so basically we do 2 plus or minus 4 comma negative 3. so keep in mind H is 2 K is negative 3. so basically what this means is that we go four to the right along the x axis and 4 to the left from the center and so we get these two points two plus four is six and two minus 4 is uh negative two which correlates to this point this is 6 negative 3 and negative 2 negative 3. so that&#39;s the vertices or the endpoints of the major axis now if we wish to find the minor vertices or the endpoints of the minor axis which is right here we can use this equation so I&#39;m going to put MV for like minor vertices so it&#39;s going to be H comma K plus or minus B because we added and subtracted B to the y coordinate of the center so it&#39;s 2 comma the y coordinate which is negative 3 plus or minus the value of B which is three so the two answers that we get is a two comma negative three plus three is zero so that&#39;s basically this point right here two zero and negative three minus 3 is negative six so we get this point two negative six so these are our minor vertices now if you need to find the intercepts as we can see we have one x intercept only in this graph which is 2 0. and there is a y-intercept which is here and here now we don&#39;t know exactly what those points are just by looking at this graph because this is a rough sketch but for any equation if you have to find the x-intercept plug in 0 for y and then algebraically solve for x if you need to find a y intercept plug in 0 for x and then solve for y now the last thing we need to do is find the foci the focus should be somewhere along the major axis now those points aren&#39;t precise but it&#39;s a rough estimate the first thing you need to do is find the value of C so c squared is a squared minus V squared which is um a squared is 16 B squared is nine so that&#39;s seven so if you take the square root of both sides C is radical seven so to write the coordinates of the foci since it&#39;s along the since the major axis is horizontal you need to add and subtract C to the x coordinate of the center so the equation for this particular graph it&#39;s going to be h plus or minus C comma k so let&#39;s see if I could put that here so the foci is going to be H which is 2 plus or minus C which is radical seven comma negative 3. if you want you can convert that to a decimal value but the full class should be somewhere around here and some way there as you can see it&#39;s inside of the ellipse so for the sake of practice let&#39;s try one more example let&#39;s say if we have a X Plus 1 squared over nine Plus y minus 2 squared over uh 25 and let&#39;s say that&#39;s equal to 1. so first let&#39;s find a center it&#39;s a negative one comma two which is right here and a squared actually that&#39;s a b squared B squared is a smaller of the two B squared is nine a squared is 25 since 25 is larger than 9. so that means a is 5 and B is 3. so a is under the Y so we need to go up five units one two three four five and then down five three four five B is associated with X so we need to go three units to the right and then three units to the left and then connect the the auto points so that&#39;s our ellipse and if you want to find a foci solve for C so since c squared is a squared minus V squared we know a squared is 25 B squared is 9. 25 minus 9 is 16. the square root of 16 is 4 so C is 4. so we need to go up four units which is over here and down four units from the center so that&#39;s the coordinates of the foci but to write it this time it&#39;s not h plus or minus C it&#39;s h comma K plus minus C because the major axis it&#39;s vertical the foci the center and the major vertices they all pass through the major axis so therefore the foci is H which is negative one comma K which is 2 plus or minus the value of C which is four so two plus four is six that gives us this point it&#39;s at negative 1 6. and then the other one two minus 4 is negative 2 which is over here it&#39;s at negative one comma well actually this one negative 1 comma negative two so those are the coordinates of the foci so now let&#39;s find the endpoints of the vertices or the endpoints of the major axis so those two points you can probably see it what it is graphically but I&#39;m going to do it this way it&#39;s h comma K plus or minus a since the major axis is vertical so it&#39;s negative one two plus or minus the value of a which was five so two plus five is seven and two minus 5 is negative 3. so these are the major vertices which are the endpoints of the major axis now keep in mind the length of the major axis is two times the value of a so in this problem it&#39;s ten that&#39;s the length of the major axis the life of the minor axis is two times the value of B which is six so let&#39;s try an example where we have to find the X and Y intercepts just for the sake of practice so let&#39;s say if you want to find the x-intercept plug in 0 for 1 I mean zero for y and then solve for x so what I would do to get rid of all the fractions is I would multiply by 9 and 25. so if you take this term and you multiply by 9 and 25 the nines cancel and so you get 25 X Plus 1 squared negative 2 squared is 4. and this term times 9 and 25 the 25s cancel so you just get 9. and then here 1 times 9 times 25 the value of nine quarters that&#39;s two dollars and 25 cents so we know it&#39;s 225. so 4 times 9 is 36. and 225 minus 36. that&#39;s 289 I believe and then divide both sides by 25. so X Plus 1 squared is 289 over 25. and then take the square root of both sides the square root of 289 I believe that&#39;s 17. and so you get X Plus 1 is equal to plus or minus 17 over the square root of 25 which is 5. so X is negative 1 plus the minus 17 over 5. now if you go back to the example that we had notice that we have two x-intercepts because the ellipse it touches the x-axis twice if you want to find a y intercept pretty much plug in 0 for X solve for y and just basically follow the example that we just did but now let&#39;s move on to hyperbolus so let&#39;s say if we have x squared over nine minus y squared over four equal one the general equation of the hyperbola it&#39;s x minus H squared over a squared minus y minus K squared over B squared equals one now a is not the larger of this 2 this time a squared is simply whichever comes first you can also have the other form of this equation which is y minus K squared over a squared minus x minus H squared over B squared equals one so whichever one is positive is associated with a whichever one is in front of the negative sign or whichever one is negative um has the B squared now we need to realize that for a hyperbola if x squared comes first it opens uh left and right if y squared comes first it opens up and down but we&#39;ll go over a few examples so let&#39;s solve this one the center is 0 0 because there&#39;s no H and K for this example so that&#39;s where the uh that&#39;s the midpoint of the two vertices since a is 3 we need to go three to the left and 3 to the right and B squared isn&#39;t 4 so that means B is two so we need to go up two and down two and then make a box in this case like a rectangle and then draw two lines these are the asymptotes make sure that these two lines crosses the uh make sure it forms like a diagonal through the rectangle it has to pass through the center as well now these two points are where the graph is actually going to touch it&#39;s going to meet at those two points it doesn&#39;t touch these two points here those are imaginary points so this graph is going to look like this it follows the asymptote it touches the vertex and then it follows the other asymptote so that&#39;s how you graph a hyperbola these are called the uh the vertices those are the vertex of the hyperbola so the vertices are plus or minus 3 comma zero the foci is somewhere here it it&#39;s always where the graph opens towards so if the graph opens this way the foci is like towards it if it opens this way the foci is over here now to find it it&#39;s a little different instead of it&#39;s c squared equals a squared not minus but plus b squared for an ellipse it&#39;s a squared minus B squared this time as you can see the value of C is larger than the value of a so that&#39;s why c squared is a squared plus b squared so it&#39;s going to be 9 plus 4. that&#39;s 13 so C is the square root of 13. so the foci therefore is plus or minus radical 13 comma zero now the only other thing that we need to do for this type of graph is we need to find the asymptotes the equation for the asymptotes it&#39;s y minus K is equal to now it can be B over a or A over B plus or minus and then x minus h here&#39;s how you tell so what we have is the asymptote is basically a line and whenever you have a linear equation in slope intercept form it&#39;s MX plus b notice that the letter in front of X is M which represents a slope so B over a is the slope of the line which is rise over run so from the center to this point you need to go up B units that&#39;s your rise and then you run a units so it&#39;s a rise over run B over a for this particular example when it opens this way it&#39;s actually going to be A over B but you can tell by just doing Vise over run so for this actual equation we know that H and K are zero so to write the asymptotes it&#39;s going to be just y there&#39;s no k is equal to plus or minus the value of B which is um B is a two B squared is four so B is 2. and a if a squared is nine a is three so it&#39;s Y is equal to plus or minus two thirds X that&#39;s the those are the asymptotes for this particular hyperbola so let&#39;s try another example so let&#39;s say if we have X plus one squared over 25 minus y minus 2 squared over 16. so a squared is the first number it&#39;s not the bigger or the smaller one whichever comes first whichever is positive and the second one is usually negative so that&#39;s B squared so we know that a is 5 and B is 4. and the center is negative one positive two just don&#39;t forget to change the signs to find a full guy it&#39;s c squared is equal to a squared plus b squared so a squared is 25 B squared is 16. so that&#39;s 41 so C is the square root of 41. so let&#39;s plot the center first it&#39;s at negative 1 positive two so it&#39;s over here now a is under the X variable so we need to go five units to the left and five units to the right so this coordinate is at negative six this is four and B is four so we need to go up four units so that&#39;s at positive six and down four units to negative two and then draw the rectangle and then draw the asymptotes now this is the value of a same as here so a is horizontal this time just like last time by the way this is called the transverse axis and the graph is going to touch at those two points so it should look something like this so the vertices which are these two points but let me write the general equation it&#39;s a h plus or minus a comma k so H is uh the x coordinate of the center and a is five and K is 2. so negative one plus five that&#39;s four and negative 1 minus five is negative six so this point as we can see it&#39;s um it&#39;s four comma two here this is two and this is four so that&#39;s four comma two and this one it&#39;s at negative six comma two which is what we have so those are the vertices of the graph now to find the foci it&#39;s also along the transverse axis is horizontal so it&#39;s going to be H but plus or minus C comma K so we&#39;ve got to find the value of C which we already know it to be root 41 so it&#39;s negative 1 plus or minus root 41 comma 2. now we can&#39;t add negative 1 plus root 41 so we&#39;re just going to leave it like this so we know the foci is somewhere over here let&#39;s and it&#39;s somewhere on the outside it&#39;s past the vertices now the only other thing we need to find is um the asymptotes which is y minus K is equal to plus or minus now rise starting from here we need to rise B units because B is a vertical a is horizontal so the run is still a so it&#39;s a b over a rise over run and then x minus h so that&#39;s y minus K is negative two action okay is a case Plus 2. so it&#39;s a y minus 2 is equal to the plus or minus B over a this is B this that&#39;s a so four over five x minus h so H is um negative one so negative negative one you basically get what&#39;s here so X plus one so notice that what you see here is what&#39;s here and what you see there is the same as that but this is the equation of the asymptotes now the reason why we have plus and minus is because there&#39;s two of them if it&#39;s plus four over five the slope is positive so it&#39;s going up minus four over five the slope is negative so it&#39;s going down the negative one is four the one that decreases as you go from left to right the positive one is for the one that goes up has to go from left to right now for the sake of practice we&#39;re going to try one more example let&#39;s try this one X minus 2 squared actually not let&#39;s start with Y this time so y minus 2 squared over 4 minus X plus 3 squared over nine and that&#39;s equal to 1. so a squared is whatever comes first and this is B squared so a therefore is the square root of 4 which is two B is the square root of nine which is three and to find c c squared is a squared plus b squared so 4 plus 9 which is 13. so therefore C is root 13. the center is uh let&#39;s look at the x value first so it&#39;s negative 3 and the Y Value Plus 2. so it&#39;s at negative three comma two now since a is 2 but it&#39;s under the Y we need to go up two units and then down to units so wherever um the points that a give you is the vertices so that&#39;s where the graph is going to be so the transverse axis this time is actually vertical now to find we know B has to be so we need to go three units to the right three units to the left and let&#39;s draw our rectangular box next let&#39;s try the asymptotes so keep in mind the major the vertices are here so this graph is going to open like this it opens up and down because Y is in front this time so the foci is somewhere over here it&#39;s up and down rather than left and right and to find the coordinates of the foci it&#39;s we need to add and subtract C to the y-coordinate of the center because we went up and down and Y is associated with up and down X is associated with left and right so the foci is the center negative three comma two but plus or minus C or plus or minus root 13. so that&#39;s the coordinates of the foci and for the vertices which are basically these two points it&#39;s h comma K plus or minus a so it&#39;s negative three and K is 2 so 2 plus 2 is 4. and 2 minus 2 is 0. so those are the major vertices now the last thing we need to find is the asymptote which is y minus K is equal to plus or minus okay so we went up a units the Rises a the run is B and since the slope is rise over run It&#39;s A over B and then x minus H so y minus K is basically exactly what you see here so it&#39;s y minus 2. and a is 2 B is 3 so 2 3. x minus H is basically X plus three and that&#39;s how you find the equation of the asymptotes and I believe we&#39;re not missing anything I think that&#39;s it I think we covered everything there so now let&#39;s move on to our next topic so let&#39;s focus on the parabola the general equation for a parabola well before I do that let&#39;s go over the different types this Parabola is y is equal to x squared and if it opens downward Y is equal to negative x squared if it opens to the right X is equal to Y squared and when it opens to the left X is equal to negative y squared now somewhere over here this is the directrix and the distance between the vertex and the directrix is p so in this case it&#39;s Y is equal to negative p and the foci is p units above the the vertex so wherever the graph is that&#39;s where the focus so the focus is over here that the Matrix is over here in this case the coordinates of the focus is um it&#39;s it&#39;s p comma zero but for this example is zero comma p because you want P units up here you want p is to the right for this one the Matrix is X is equal to negative P rather than uh y and over here it&#39;s Y is equal to positive p and since the focus is below um the origin the coordinates is zero negative p so you have to see you have to know the general shape of the graph to see where um the focus is going to be if you have to add P or subtract p so keep in mind for the directrix if you have a horizontal line it&#39;s y equals if you have a vertical line it&#39;s x equals here it&#39;s x equals positive p and the focus is actually negative P comma zero this time that&#39;s if the parabola is based on the origin now let&#39;s say if it&#39;s moved away or transformed or shifted away from the origin these are the equations that you&#39;re going to need y minus K squared is equal to 4p x minus h so basically this is the equation that&#39;s like y squared is equal to X or x equals y squared so this can open to the right or to the left it opens to the right if p is positive or greater than zero if p is negative it opens towards the left now the next equation that you need x minus H squared is equal to 4p y minus k so this is basically y equals x squared which can open up or down it opens upward if p is positive and if p is negative it opens downward all right so let&#39;s try some practice problems let&#39;s start with the basics let&#39;s say that y squared is equal to 8x so that means there&#39;s no H and K so the center is zero zero or at least we really don&#39;t have a center we should really call it the vertex that&#39;s a zero zero so this is in the form y minus K squared is equal to 4p x minus H so notice that 8 is equal to 4p so set 4p equal to 8. if you divide both sides by 4 8 divided by 4 is 2 so p is 2. and then we can graph it now it really depends on what you want to do here I&#39;m going to draw a smaller graph but more like a rough sketch since we have y squared equal to X or x equals y squared we know it&#39;s going to open to the right or to the left and since p is positive 2 we know the graph opens to the right which means that the Matrix is over here and p is 2 so the foci is over here so the foci it&#39;s um it&#39;s p comma zero or in this case two comma zero since p is two and the directrix is X is equal to negative 2 because you got to go P units to the left so the first thing we should do is plot the vertex then go P units to the right and then P units to the left so there&#39;s your directrix and here&#39;s the focus but now how can we get how can we like plot this graph accurately what you could do is you can plug in some points now it&#39;s easier if you plug in let&#39;s see we could make a table if you want should we plug in values for x or 4y which one is easier if you have y squared it&#39;s easier if you plug in values for y rather than x let&#39;s plug in one for y if you plug in 1 um actually one is going to be too small let&#39;s solve for x y squared divided by 8 is equal to X let&#39;s say if we plug in 2 for y that&#39;s 2 squared is 4 4 over 8 is a half if you plug in negative 2 for Y when you square it it&#39;s still going to be positive four so you&#39;re going to get the same x value let&#39;s try a bigger number like 4. plus or minus 4 because it&#39;s going to give us the same x value if you plug in 4 4 squared is 16 16 divided by 8 is 2. the last point to plug in would be like 8 for y 8 squared is 64. and 64 divided by 8 is 8. but we may not need to go that far so when we plug in 2 for y X was about 0.5 so that means it&#39;s somewhere over here but also for negative 2 we get positive 0.5 for x let me use a different color the key is to choose the right points we want to choose points that you can get whole numbers which four was the best example when Y is 4 X is 2. so that means it&#39;s over here that&#39;s 2 comma four and two comma negative 4. so we have a very wide graph and let&#39;s say if we were to extend this graph so this would be three four five six seven eight now let&#39;s go up along the y axis so 5 6 7 8 and I&#39;m out of space over here so this is somewhere over here it&#39;s at 8 8 and 8 negative eight it&#39;s as well I&#39;m out of space there so this graph looks something like this so if you want to get a very accurate graph if that&#39;s important make a table plot some points if you just want a rough sketch you just need to know what the direction is because if your test is multiple choice you don&#39;t need a table you can pick the right ones and eliminate the wrong ones let&#39;s try another example so let&#39;s say if we have y minus 2 squared is equal to 4 x minus 3. so our general equation is y squared is equal to X so this one&#39;s going to open to the right again and we know the focus is going to be over here the Matrix is over here so the center is 3 for X positive 2 for y so we need to go three units to the right up two units so that&#39;s the I keep saying the center but I&#39;m at the the vertex so now we need to find p whatever this number is set it equal to 4p so therefore p is one and since it opens to the right we need to go one unit to the right that gives us the focus so the focus is basically h plus P comma k or four plus one I mean three plus one which is four and K is 2. so that&#39;s the coordinates of the folk the focus for the directrix we need to go one unit backwards and then we can draw a vertical line so the Matrix is that X is equal to positive two so now let&#39;s see if we can find some points to graph in so we have a 4 in front let&#39;s say if we plug in let&#39;s make a table we know the center is 3 or the vertex is 3 2. let&#39;s plug in 4 for x and let&#39;s see what we get for y if we plug in 4 fractions it&#39;s going to be 4 times 4 minus 3 is 1. so we get that and if we square root both sides y minus 2 is equal to 2. well we get plus or minus 2. so y minus 2 is equal to 2 and Y minus 2 is equal to negative 2. so Y is equal to four and here if we add to Y is equal to zero so that means when X is 4 y can be zero and it could be four so four zero is basically over here and 4 4 is over there in the last example we plugged in for y and solve for x but in this example I chose to solve for um y instead now if I choose 5 for x it&#39;s going to be 2 times 4 which is 8 and I can&#39;t take the square root of 8. so the next value I&#39;m going to choose for x is going to be 7. because 7 minus 3 is 4 which is 16. and we can take the square root of 16 because that&#39;s going to give me an integer which is 4. so y minus 2 is equal to plus or minus 4. so y minus 2 is 4. that means if you add 2 you get six and Y minus 2 is equal to negative 4 as well if you add 2 you get um negative 2. so I kind of picked the points differently from the last example so we need to go to seven and at seven it&#39;s going to go up to six which is over here and it&#39;s going to go down to negative 2. so that should be that&#39;s enough points to get a rough sketch for this graph and that&#39;s how it looks like so our last example for today let&#39;s try this One X plus one squared let&#39;s say that&#39;s equal to um two times y minus 3. so here we have the example where x squared is equal to y or Y is equal to x squared and let&#39;s actually add a negative sign so it&#39;s going to open downward so the focus should be somewhere over here and the directrix is horizontal this time so let&#39;s find the center it&#39;s negative one comma three so one to the left up three and actually mean the vertex I keep calling the center so now we need to find the value of P so negative two is equal to 4p so P if you divide by 4 that&#39;s negative a half two over four is one half so this is the center we need to go half units well it opens the graph opens downward so we need to go down by half and up by half so if you go down by half that&#39;s the focus but it&#39;s too small to draw the directrix we note this line here so the focus is going to be uh H comma k but since we need to go down it&#39;s minus p so H is negative 1 K is three but we need to go down by half so that&#39;s we need to subtract three by one half so three is basically six over two minus a half that&#39;s five over two as you can see the focus is at 2.5 is between two and three so I&#39;m going to draw a little F here for the focus the directrix it&#39;s going to be uh Y is equal to since it&#39;s horizontal it&#39;s going to be K plus P we need to go up P units so that&#39;s three plus one half or six over two plus one over two so it&#39;s Y is equal to seven over two that&#39;s the Matrix okay so we have the general shape of the graph but the hard part is choosing the values for X and Y that we should pick so if it&#39;s not in standard form it&#39;s easier for you to plug in well I guess you have to pick which way it&#39;s easier for you um but we&#39;ll need to plug in some values you can choose the plug-in values from for x and solve for y or you can plug in values for y and solve for x it really depends on whichever you think is easier what I&#39;m going to do is I&#39;m going to plug in a value for y and solve for x now my directrix is that 7 over 2. and the graph to Center the y-coordinate is at uh three so the graph doesn&#39;t go past three it doesn&#39;t go past the vertex if it opens downward so I can&#39;t choose any value that&#39;s greater than 3 for y let&#39;s say if I choose four four minus 3 is 1 1 times negative two is negative two and you can&#39;t square root a negative number so I have to choose something that&#39;s less than three I already have the value for three when Y is 3 x is negative one that&#39;s the vertex so I need to choose something less than three but I want to choose a number where when I multiply by negative 2 it&#39;s going to give me a perfect square so I want to choose one for y because 1 minus 3 is negative 2 and negative 2 times this negative 2 in front is 4. and I can take the square root of 4. so if X Plus 1 squared is equal to four X Plus 1 is therefore equal to plus or minus two so it&#39;s equal to two and negative 2. If X Plus 1 equals 2 If I subtract both sides by 1 I get x equals to 1. and for the other one If I subtract 1 from negative 2 negative 2 minus 1 that&#39;s negative 3. now I need to choose another point where I can get another perfect square perfect squares are like 4 9 and 16. I already use four I don&#39;t want to use nine because I need a fraction to get that well actually let me use nine let&#39;s see I need to set y minus 3 equal to half of 9 which is 4.5 but negative 4.5 so if I add 3 to both sides Y is going to be equal to negative 1.5 I&#39;m going to choose that for my y value so if I plug in negative 1.5 I get negative 2 times negative 1.5 minus 3 which is negative 2 times negative 4.5 that&#39;s a 9. so therefore X Plus 1 squared is equal to 9. and if you square root both sides X Plus 1 is equal to 3 which means three minus 1 is 2 and X plus one is equal to negative three negative three minus one is negative four so that should be enough points for this particular example so when X is one y is one so that&#39;s over here and when X is negative three Y is one so we can see the symmetry about this line here when X is 2 Y is negative 1.5 which is about there somewhere and when X is negative four it&#39;s over here so this is enough points to graph it so now let&#39;s say if you&#39;re given these four equations x squared plus 6x minus 4y plus one is equal to zero and 4x squared minus 9y squared minus 16x Plus 54y let&#39;s say that&#39;s equal to zero as well and then this one 2x squared and the last one so given these four equations which one is a circle which one is an ellipse which one is the parabola and which one is hyperbola now the first thing I would look for is the problem if you have all four options the parabola is the one that has an x squared but not a y squared or it has a y squared but not an x squared so here we see you know I meant this to be a y squared by the way so here we have an x squared and a y squared so that&#39;s not a problem x squared y squared not a problem x squared y squared this is the the first one is the only one that doesn&#39;t have an x squared and a y squared so this is the parabola that&#39;s how you can distinguish it from the other three here&#39;s the x squared but there&#39;s no y squared or they could be a y squared but no x squared so now which one is the hyperbola that&#39;s the next thing you want to look for now the hyperbola has a positive x squared and a negative y squared or vice versa the ellipse and a circle both x squared and Y squares are positive so both of these are positive both of those are positive but notice this we have positive x squared negative y squared the ellipse the hyperbola has the minus sign so whenever you see that a positive x squared negative y squared or vice versa it&#39;s hyperbola now to distinguish the circle from the ellipse look at the coefficients of x squared and Y squared a circle is perfectly even so if the coefficients are the same it&#39;s a circle if the coefficients are different it&#39;s an ellipse and that&#39;s how you can distinguish these four conic sections if it&#39;s not in standard form so now we&#39;re going to talk about how to put these equations in standard form so let&#39;s start with a circle so we had 2x squared plus 8x plus 2i squared Plus 4y minus 6 is equal to zero so the first thing you want to make sure is you want to group The x&#39;s and the Y&#39;s together which we already have that so for the first two terms let&#39;s factor out the GCF of the coefficient so between two and eight we can take out a two and so we&#39;re left with x squared plus eight X and leave a space and between the next two terms we can take out a two as well so we have y squared plus two y leave a space and then the negative six let&#39;s move it to the other side now we need to complete the square half oh wait this should be um this should be a 4. not an 8 because that took away two so you want to take half of four half of 4 is 2 and then Square it half of 2 is 1 and then squared that&#39;s how you complete the square but now notice that we added 2 squared times 2 to the left side 2 squared is 4 times 2 is 8. so we got to add 8 to the other side as well 1 squared times two is two so let&#39;s add 2 as well so now we can Factor it&#39;s going to be X plus whatever that sign is and then this number before you square it X plus 2 and then squared to factor the next term it&#39;s going to be whatever this variable is y Plus 1 squared 8 plus 2 is 10 10 plus 6 that&#39;s 16. so now when you have a circle you got to get rid of these coefficients so we&#39;re going to divide everything by 2. so our final answer is X plus 2 squared plus y plus 1 squared the twos cancel and then 16 divided by 2 is 8. so now it&#39;s in standard form the center of the circle is negative 2 1 and the radius is the square root of 8. so that&#39;s how you can write um the equation of a circle in standard form and then if you want to you can graph it if if needed so now let&#39;s move on to the ellipse so we had 4x squared plus 25y squared minus 24x Plus 100y plus 36 is equal to zero so first let&#39;s move the x&#39;s and the Y&#39;s together so 4x squared minus 24x plus 25y squared plus 100y and let&#39;s move to 36 to the other side so let&#39;s factor out a 4 from the first two terms we want x squared to have a coefficient of one so it&#39;s x squared and then negative 24x divided by 4 that&#39;s negative 6X and let&#39;s leave a space let&#39;s factor out 25. 100y divided by 25 is 4y leave a space and then let&#39;s complete the square so half of 6 is 3 ignore the negative sign and then Square it half of 4 is 2 so that&#39;s 2 squared so 3 squared is 9 times 4 that&#39;s 36 so we got to add 36 to this side 2 squared is 4 times 25 is 100 so let&#39;s add 100 to that side so now let&#39;s Factor it&#39;s going to be 4 whatever this variable is X minus this number squared and then plus 25 this variable y Plus this number to squared negative 36 and 36 they cancel so we&#39;re just left with a hundred now for ellipses and hyperbolas this number has to be a one so divide by what you see then so we&#39;re going to divide both sides by a hundred so we have x minus 3 squared so 4 over 100 divided backwards 100 divided by 4 is 25 but since you divide it backwards 25 goes to the bottom 100 divided by 25 is 4. and 100 divided by 100 is 1. so now we have the equation of the ellipse in standard form as we can see the center of the Ellipsis to read negative two and a is the larger of the two a squared is 25 so that means a is 5. B squared is 4 so B is 2. and keep in mind for uh for a in ellipse it&#39;s c squared equals a squared minus B squared if you want to find the foci so now let&#39;s try this one the hyperbola 4x squared minus 9y squared minus 16x plus 54y minus 101 equals 0. so let&#39;s move the x&#39;s and y&#39;s together so we have 4x squared minus 16x and then minus 9y squared Plus 54y and let&#39;s move the 101 to the other side so now let&#39;s factor out a 4. so 4x squared divided by 4 is x squared negative 16x divided by 4 is negative 4X leave a space next let&#39;s factor out negative nine 54 divided by negative 9 is negative 6. and now let&#39;s complete this Square half of 4 is 2 so let&#39;s add 2 squared half of six is three so we&#39;re going to add 3 squared now 2 squared is 4 times 4 we need to add 16 to this side 3 squared is 9 times negative 9 so that&#39;s negative 81. so let&#39;s add negative 81 to that side so now let&#39;s factor it&#39;s going to be 4 times x minus 2 squared and then minus 9 times y minus 3 squared so 101 minus 81 that&#39;s 20 20 plus 16 is 36. now we need the 36 to be a one for a hyperbola so let&#39;s divide every term by 36. so if we divide it backwards 36 divided by 4 is 9. and 36 divided by 9 is 4. and this we get 1. so the center for the hyperbola it&#39;s 2 comma 3. whatever is positive that&#39;s a squared and therefore this is B squared now for the last one this is a parabola and we want to put it in a form where it&#39;s like this x minus H squared is equal to 4p Y minus k so all the X&#39;s we want to move to the left side and everything else let&#39;s move it to the right side so x squared plus 6X is equal to negative 4y becomes positive 4y when you move it from left to right and a positive one becomes negative one so now let&#39;s complete the square half of 6 is 3. and so we need to add 3 squared to the left and whatever you add to the left side add to the right side so 3 squared is 9. so to factor left side it&#39;s going to be X Plus this number before you square it and then over here we have 4y negative 1 plus 9 is 8. so all we got to do now is factor out a 4 from the right side so it&#39;s 4 y Plus 2. and now the parabola is in standard form the vertex is negative three comma negative two and this number is equal to 4p so p is one and that&#39;s basically it so let&#39;s review the general equations for every conic section so starting with a circle keep in mind that your general equation is x minus H squared plus y minus K squared which equals r squared and the center is h k so once you have this Center let&#39;s say it&#39;s say this is H this is K we need to go R units to the right R units to the left of r down r and then graph the circle now for an ellipse if you have x minus H squared over a squared plus y minus K squared over B squared equals one so remember a is the bigger one a is greater than b so what&#39;s going to happen is let&#39;s assume the center is the origin you can have a graph that looks like this so this is the center this is therefore a that&#39;s negative a this is B this is negative B just remember these are the endpoints of the major axis so those are the vertices and these are the endpoints of the minor axis and your full guy is along the major axis which is somewhere over here so your vertices for this particular example it&#39;s uh h plus or minus a comma k and for your Foci it&#39;s h plus or minus C comma k and the last thing that you need to know is to find C it&#39;s c squared is equal to a squared minus B squared for an ellipse and if you have the other version where you have x minus H squared but let&#39;s say B is under X and Y minus K squared where a is under y this ellipse is going to be elongated it&#39;s like a vertical ellipse it&#39;s going to look like this so this is a this is negative a this is negative B that&#39;s B so now euphokai to find it it&#39;s going to be H comma K plus or minus C so you add and subtract C to the y coordinate instead of the X and your vertices are H comma K plus or minus a so the major axis is vertical the minor axis is horizontal now keep in mind the length of the major axis is 2A the length of the minor axis is 2b so to review for a hyperbola it can be x minus H squared over a squared minus y minus K squared over B squared equals one or it can be y minus K squared over a squared and x minus H squared over B squared so keep in mind for a hyperbola a always come in front so B could be bigger than a or a could be bigger than b it doesn&#39;t matter and c squared to find the full guy is a squared plus b squared rather than a squared minus B squared so for this example it&#39;s going to open this way so let&#39;s say if we drew the Box so this is a that&#39;s negative a this is a here this is B and that&#39;s negative B so for this particular example the asymptotes is y minus K which is equal to rise over one Rises B run as a so plus or minus B over a x minus h and the vertices for this equation it&#39;s um h plus or minus a comma k now for this one the graph opens this way I&#39;m not going to draw the Box though but you know how how it is you know how to do it but this time the vertices is going to be H comma K plus or minus a and the asymptotes will be y minus K is equal to this time A over B so plus or minus A over B x minus h and that&#39;s really what you need for a hyperbola those are the main equations oh I forgot the foci the foci is going to be h k plus or minus C for this example and for this one it&#39;s going to be h plus or minus C comma k so just so you know the foci is where it opens towards so the four Chi is over here and in this example it&#39;s over here and lastly what we have is the parabola so for this one here we have X is equal to Y squared so it can open to the right or it can open to the left so your focus is always on the interior towards where it opens and your directrix is here so for this example we know that the vertex is H comma k so the focus is p units from the vertex so it&#39;s h plus P comma K and for this example the focus is H minus P comma k the directrix is X is equal to H minus p and here it&#39;s X is equal to h plus p now for the other one it&#39;s x minus H squared is equal to four p y minus k so for y equals x squared it can open up or it can open down here p is positive p is negative for this one when it opens to the right p is positive when it opens to the left p is negative so the focus is somewhere over here and the directrix is horizontal this time so the focus is going to be H comma K plus p and for this example it&#39;s h comma K minus p the directrix is horizontal so it&#39;s Y is equal to K plus p and here um it&#39;s Y is equal to K minus p so that&#39;s all the equations that you need for conic sections so basically that&#39;s the end of this video so thanks for watching and have a great day

Transcript for:Graphing Conic Sections: Circles, Ellipses, Parabolas, and Hyperbolas

Transcript for:
Graphing Conic Sections: Circles, Ellipses, Parabolas, and Hyperbolas