Transcript for:
Conic Sections Overview

in this video we're going to focus on how to graph conic sections like ellipses parabas hyperbolas and also circles as well and in addition to that we're going to talk about how to look at equation um 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 uh graphing part let's say if we have this equation x^2 + y^2 = 9 and this is a circle but how can we graph this circle so the standard equation for a circle it's it's x - h^ 2 + y - k^ 2 is equal to R 2 the center of the circle is H comma K but there's no H and K in this uh problem so the center is based on the origin it's 0 0 now R 2 is basically 9 so if r s is equal to 9 if you take the square root of both sides um you'll see that the radius is three 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 uh that particular Circle or that equation let's try another example so let's say if we have this equation x - 3^ 2 + y - 4^ 2 actually let's say y + 4 2 = 16 so we need to find H and K so the center is basically if you set x - 3 equal to 0 and you solve for x you'll get three so H is three so basically you change a negative3 to a positive three and the same thing with the four you Chang a positive4 into a ne4 so the center is 34 and since R 2 is equal to 16 the radius of the circle is four 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 four now because the radius is four 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 + 2^ 2 + y - 3 2 is equal to 4 so the center is -2 comma 3 you change to positive2 into a -2 and the negative3 into a positive 3 and the radius is the square root of four so it's two and then plot the center so we need to go two to the left up three so it's over here and radius is two so we need to go down two and then up two two 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 on all sides let's start with this equation x^2 over 25 + y^2 over 49 and let's say that's equal to 1 now the general equation for an ellipse it's x - h^ 2 a^2 and then y - k^ 2 over b^2 is equal to 1 now the a s is not always under the X A2 could be under y - K we need to know to distinguish between a and b a s is the larger of the two so in this case um for this problem that we have this is actually a squ because 49 is larger than 25 25 is actually b^2 now we need to realize that the center is 0 comma 0 because if you see just X2 and y^2 if it's not like x -2 or y + 3 um then the center is 0 0 so it's based on the origin now since A2 is equal to 49 what this means is that a is equal to 7 and S is under uh y^2 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^2 is 25 so if you take the square root of 25 that's five and B is under X so we need to go five 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 um what is the uh 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 ma for major axis since it's equal to 2 a it's 14 the minor axis they're both n the length of the minor axis is twice the value of B so 2 * B is 10 by the way the end points of the major axis which is um 07 and 07 those end points known as vertices which is the plural form of vertex so some books will call it the major vertices and other textbooks will call the end points of the minor axis like the minor vertices if they don't use the word minor then the vertices correspond to the end point of the major axis now the only other thing we need to find is the folky the folai is it's along the major axis um but whenever you have a graph that opens up the folai is somewhere in the middle so this should be a folky somewhere between along this major axis between 0 and 7even to find it you need to use this equation c^2 is equal to a 2 not plus but minus b^2 for our hyperbola it's a squ plus b square kind of like the pythag theorem but for um for an ellipse it's a square minus b^ squ because um C is less than a and I'll explain why shortly but let's solve for c^ S we already know the value of a squ and this is b^2 so it's 20 not 25 but 49 - 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 root4 andun 6 so it's 2un 6 so the folky the coordinates of the folky x is zero Y is C so it's plus or minus 2un 6 as a decimal 2un 6 is about 4.89 so just under five so the folai 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 the that's why the folky um that's why C is less than a and so therefore it's c^2 = A2 minus B squ 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 c is the distance between the center of the ellipse and the folai so if the origin is not Z 0 then the coordinates of the folai will be affected by the way sometimes you may get a question that ask you for the X and Y intercepts for this particular example the Y intercepts are basically the end points of the major axis which is 07 and 07 so those are the Y intercepts the X intercepts are the end points of the minor axis so in this case it's a plus or minus 5 comma 0 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 - 2^ 2 / 16 plus y + 3^ 2 / 9 and let's say that's equal to 1 now the first thing you need to do is find the center so since we have x minus two we need to reverse it so it's gonna be plus two and for y + 3 change it to 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 squ and what's b s a s is the larger of the two this is a squ and this is b s so if a squ 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 is 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 and B is three go up three units and then down three units so it's not very difficult to graph an ellipse it's fine in the other details about it that 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 xaxis 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 uh horizontal so the major axis is at Y3 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 xal 2 the length of the major axis is equal to 2 a twice the value of a and since a is four it's about 8 units so this is POS 6 this is -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 is6 here and Y is zero at this point so you can see uh this is a distance of six and and here we have a distance of eight now if you want to find the end points of the major axis since you already graphed it you can see that it's going to be -23 and 63 but if you wish to find it from the center the end points 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 -3 so keep in mind H is 2 K is3 so basically what this means is that we go four to the right along the x axis and four to the left from the center and so we get these two points 2 + 4 is 6 and 2 - 4 is uh -2 which correlates to this point this is uh 6-3 and -23 so that's the vertices or the end points of the major axis now if we wish to find a minor vertices or the end points into the minor um 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 -3 plus or minus the value of B which is three so the two answers that we get is 2 comma -3 + 3 is 0 so that's basically this point right here two 0 and-3 minus 3 is -6 so we get this point to6 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 two 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 zero for y and then algebraically solve for x if you need to find a y intercept plug in zero for x and then solve for y now the last thing we need to do is find the folky the folky 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^2 is a 2 minus b^ 2 which is um a square is 16 B2 is 9 so that's s so if you take the square root of both sides C is radical 7 so to write the coordinates of the folky 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 um plus or minus C comma K so let's see if I could put that here so the folai is going to be H which is 2 plus or minus C which is radical 7 comma -3 if you want you can convert that to um a decimal value but the FOC should be somewhere around here and somewhere 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 + 1 s over 9 plus y - 2^ 2 over uh 25 and let's say that's equal to one so first let's find a center it's a -1 comma 2 which is right here and a s actually that's b squ b squ is the smaller of the two b square is 9 a squ is 25 since 25 is larger than 9 so that means a is five and B is three so a is under the Y so we need to go up five units 1 two 3 four five and then down five 3 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 autter points so that's our ellipse and if you want to find a folky solve for C so since c^2 is a 2us b^ 2 we know a s is 25 b^2 is 9 25 - 9 is 16 the square root of 16 is four so C is four so you need to go up four units which is over here and down four units from the center so that's the coordinates of the folai but to write it this time it's not h plus orus C it's h comma K plus orus C because the major axis is vertical the folky the center and the major vertices they all pass through the major axis so therefore the folky is H which is 1 comma K which is 2 plus or minus the value of C which is four so 2 + 4 is 6 that gives us this point it's at16 and then the other one 2 - 4 is -2 which is over here it's at 1 comma well actually this one1 comma -2 so those are the coordinates of the fog guy so now let's find the end points of the vertices or the end points of the major axis so those two points you can probably see it what it is graphically um 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 -1 2 plus or minus the value of a which was 5 so 2 + 5 is 7 and 2 - 5 is3 so these are the major vertices which are the end points 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 10 that's the length of the major axis the length of the Min 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 zero for one I mean z 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 + 1 2 -2 2 is 4 and this term times 9 and 25 the 25s cancel so you just get nine and then here 1 * 9 * 25 the value of 94 that's $225 so we know it's 225 so 4 * 9 is 36 and 225 - 36 that's that's 289 I believe and then divide both sides by 25 so x + 1^ 2 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 + 1 is equal to plus or minus 17 over the < TK of 25 which is 5 so X is-1 plus to minus 17 over 5 now if you go back to the example that we uh 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 zero for X solve for y and just basically follow the example that we just did but now let's move on to hyperbolas so let's say if we have x^2 over 9- y^2 over 4 = 1 the general equation of hyperbola it's x - h^ 2 over a^ 2 - y - k^ 2 over B ^2 = 1 now a is not the larger of this two this time a squ is simply whichever comes first you can also have the other form of this equation which is y - k^ 2 a 2 - x - h^2 over B2 = 1 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^ S now we need to realize is that for our hyperbola if x² comes comes uh first it opens uh left and right if y^2 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 three we need to go three to the left and three to the right and b^ 2 is four 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 asmp tootes make sure that these two lines crosses the uh make sure it forms like a diagonal uh through the rectangle it has to pass through the center as well now these two points are where the graph is actually going to uh 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 follow follows the ASM toote it touches the vertex and then it follows the other ASM toote so that's how you graph a hyperbola these are called the uh the vertices those are the vertex of the uh hyperbola so the vertices are plus or minus 3 comma 0 the folky is somewhere here it it's always where the graph opens towards so if the graph opens this way the folai is like towards it if it opens this way the folai is over here now to find it it's a little different instead of it's c^2 equals a s not minus but plus b^2 for an ellipse it's a squ minus b 2 this time as you can see the value of C is larger than the value of a so that's why c^ s is a 2 plus b s Square so it's going to be 9 + 4 that's 13 so C is the squ < TK of 13 so the folky therefore is plus or minus radical 13 comma 0 now the only other thing that we need to do for this type of graph is we need to find the asmp tootes the equation for the asmp tootes it's y - K is equal to now it can be B over a or A over B plus or minus and then x - H here's how you tell so what we have is the ASM toote 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 X is M which represents the 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 VI over run so for this actual equation we know that H and K are zero so the right the ASM toote 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 uh two b^ squ is four so B is two and a if a s is 9 a is 3 so it's Y is equal to plus orus 2/3 x that's the those are the asmp tootes for this particular hyperbola so let's try another example so let's say if we have uh x + 1^ 2ar over 25 - y - 2^ 2 over 16 so a s 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^2 so we know that a is five and B is 4 and the center is -1 pos2 just don't forget to change the signs to find a folky it's c^2 is equal to a 2 + B2 so a s is 25 B2 is uh 16 so that's 41 so C is the square root of 41 so let's plot the center first it's at -1 pos2 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 -6 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 -2 and then draw the rectangle and then draw the ASM tootes 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 uh 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 uh h plus or minus a comma K so H is uh the x coordinate of the center and a is 5 and K is 2 so -1 + 5 that's 4 and-1 - 5 is -6 so this point as we can see it's um it's 4 comma 2 here this is two and this is four so that's four comma 2 and this one it's at -6 comma 2 which is what we have so those are the vertices of the graph now to find the folky it's also along um the transverse axis is horizontal so it's going to be H but plus or minus C comma K so we got to find the value of C which we already know it to beunk 41 so it's -1 plus or minus < tk41 comma 2 now we can't add 1 plusun 41 so we're just going to leave it like this so we know the folky is somewhere over here let's and it's somewhere on the outside it's past um the vertices now the only other thing we need to find is um the asmp tootes which is y - K is equal to plus or minus now rise starting from here we need to rise B units because B is uh vertical a is horizontal so the run is still a So it's b over a rise over run and then x - H so that's Yus K is uh -2 actually no K is uh K is plus two so it's y - 2 is equal to plus or minus B over a this is B this that's a so 4 over 5 xus H so H is um negative 1 so negative negative 1 you basically get what's here so x + 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 ASM tootes now the reason why we have plus and minus is because there's two of them if it's plus four of five the slope is positive so it's going up minus 4 five the slope is negative so it's going down the negative 1 is four the one that decreases as you go from left to right the positive one is for the one that goes up as you 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 - 2^ 2 actually not let's start with Y this time so y - 2^ 2 over 4 - x + 3^ 2 over 9 and that's equal to one so a squ is whatever comes first and this is b^ s so a therefore is the square otk of 4 which is two B is the squ OT of 9 which is three and to find C c^2 is a 2 + b^ 2 so 4 + 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 -3 and the Y value + 2 so it's at -3 comma 2 now since a is two 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 that's a find we know B is3 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 draw the ASM tootes 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 folai is somewhere over here it's up and down rather than left and right and to find the coordinates of the folky 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 and right so the folai is um is the center -3 comma 2 but plus or minus C or plus or minus root3 so that's the coordinates of the folai and for the vertices which are basically these two points it's h comma K plus or minus a so it's -3 and K is 2 so 2 + 2 is 4 and 2 - 2 is zero so those are the major vertices now the last thing we need to find is the ASM toote which is y - K is equal to plus or minus okay so we went up a units the rise is a the run is B and since the slope is rise over run It's A over B and then xus H so y - K is basically exactly what you see here so it's y - 2 and a a is 2 B is 3 so 2/3 x - H is basically x + 3 and that's how you find the equation of the ASM tootes 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 squ and if it opens downward y isal tox2 if it opens to the right X is equal to y^2 and when it opens to the left X is equal to y^2 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 p the folky is p units um above the the the vertex so wherever the graph is that's where the focus so the focus is over here the dric is over here in this case the coordinates of the focus is um it's p comma 0 but for this example is 0 comma P because you want P units up here you want pits to the right for this one the directrix 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 0ga 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 derric if you have a horizontal line it's y equals if you have a vertical line it's x equals here it's X x = postive p and the focus is actually P comma 0 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 - k^ 2 is equal to 4 p x - H so basically this is the equation that's like y^2 is equal to X or x = y 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 - h^2 is equal to 4 p y - K so this is basically y = x^2 which can open up or down it opens upward if p is positive and if p is negative it opens uh downward all right so let's try some practice problems let's start with with the basics let's say that y^2 is equal to 8X so that means there's no h in K so the center is0 0 or at least we really don't have a center we should really call it the vertex that's a z z so this is in the form y - k^ 2 is = to 4 p x - H so notice that 8 is equal to 4 P so set 4 P equal to 8 if you divide both sides by four 8 ID 4 is 2 so p is two and then we can graph it now 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^2 equal to X or X = y^2 we know it's going to open to the right or to the left and since p is positive two we know the graph opens to the right which means the the retrix is over here and P is two so the folky is over here so the fola it's um it's p comma 0 or in this case 2A 0 since p is 2 and the directrix is X is equal to -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 derric 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 for Y which one is easier if you have y^2 it's easier if you plug in values for y rather than x let's plug in one for y if you plug in one um actually one is going to be too small let's solve for x y^2 / 8 is equal to X let's say if we plug in two for y That's 2^ 2 is 4 4 8 is a half if you plug in -2 for y when you square it it's still going to be positive4 so you're going to get the same x value let's try a bigger number like four plus or minus 4 because it's going to give us the same x value if you plug in four 4 S is 16 16 ID 8 is 2 the last point to plug in would be like eight for uh y 8 squ is 64 and 64 divid 8 is 8 but we may not need to go that far so when we plug in two for y x was about 0.5 so that means it's somewhere over here but also for -2 we get positive point5 for X let me use a different color the key is to choose the right points you want to choose points that you can get uh 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 4 and 2a4 so we have a very wide graph now let's say if we were to extend this graph so this would be three four five six 78 and let's go up along the Y AIS so 5 6 7 8 and I'm out of space over here so this is somewhere over here it's at 88 and 8 negative 8 it's 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 in point U 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 U pick the right ones and eliminate the wrong ones let's try another example so let's say if we have y - 2^ 2 is equal to 4 x - 3 so our our general equation is y^2 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 directrix is over here so the center is 3 for X positive2 for y so we need to go three units to the right up two units so that's the I keep saying Center but I meant the the ver the vertex so now when need to find P whatever this number is set it equal to 4 P 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+ P comma K or 4 + 1 I mean 3 + 1 which is four and K is two so that's the coordinates of the F the focus for the directrix we need to go one unit backwards and then we can draw a vertical line so the directrix is that X is equal to positive2 so now let's see if we can find some points to graph it so we have a four in front let's say if we plug in let's make a table we know the center is three or the vertex is three uh two let's plug in four for x and let's see what we get for y if we plug in four for x it's going to be 4 * 4 - 3 is 1 so we get that and if we square root both sides y - 2 is equal to two well we get plus or minus two so y - 2 is equal to 2 and Y - 2 is equal to -2 so Y is equal to 4 and here if we um add two Y is equal to Zer so that means when X X is 4 y can be zero and it could be four so 40 is basically over here and 44 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 five for X it's going to be 2 * 4 which is 8 and I can't take the square root of eight so the next value I'm going to choose for X is going to be seven 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 four so y - 2 is equal to plus or- 4 so y - 2 is 4 that means if you add two you get six and Y - 2 is equal to4 as well if you add two you get um -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 -2 so that should be that's enough points to get a 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 + 1^ 2 let's say that's equal to um 2 * y - 3 so here we have the example where x^2 is equal to y or Y is equal x^2 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 a center it's -1 comma 3 so one to the left up three and I actually mean the vertex I keep calling it The Center so now we need to find the value of P so -2 is equal to 4 P so P if you divide by 4 that's Nega a half 2 over 4 is 12 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 a half and up by a half so if you go down by a half that's the focus but it's too small to draw the dric we know it's uh 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 is1 k is three but we need to go down by half so that's we need to subtract three by 1/2 so three is basically 6 over two minus a half that's 5 over two as you can see the focus is at 2.5 it's 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 3 +2 or 6 over2 + 1 over 2 so it's Y is equal to 7 /2 that's the directrix 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 um 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 need to plug in some values you can choose to 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 it's 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 at 7 over2 and the graph the 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 three for y let's say if I choose four 4 - 3 is 1 1 * -2 is -2 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 Nega 1 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 -2 it's going to give me a perfect square so I want to choose 1 for y because 1 - 3 is -2 and -2 * this -2 in front is four and I can take the square root of four so if x + 1^ 2 is equal to 4 x + 1 is therefore equal to plus or minus 2 so it's equal to 2 and -2 if x + 1 = 2 If I subtract both sides by 1 I get X = to 1 and for the other one if I subtract 1 from -2 -2 - 1 that's three now I need to choose another point where I can get another perfect square perfect squares are like four 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 Yus 3 equal to half of n which is 4.5 but negative 4.5 so if I add three to both sides Y is going to be equal to 1.5 I'm going to choose that for my yvalue so if I plug in 1.5 I get -2 * 1.5 - 3 which is -2 * 4.5 that's 9 so therefore x + 1^ 2 is equal to 9 and if you square root both sides x + 1 is equal to 3 which means 3 - 1 is 2 and x + 1 is = to -33 - 1 is4 so that should be enough points for this particular example so when X is 1 Y is 1 so that's over here and when X is -3 Y is 1 so we can see the Symmetry about uh this line here when X is 2 Y is 1.5 which is about there somewhere and when X is -4 it's over here so this is enough points to graphic so now let's say if you're given these four equations x^2 + 6 x - 4 y + 1 is equal to Z and 4x^2 - 9 y^2 - 16 x + 54 y let's say that's equal to zero as well and then this one 2x^2 and the last one so given these four equations which one is a circle which one one is an ellipse which one is a parabola and which one is hyperbola now the first thing I would look for is the parabola if you have all four options the parabola is the one that has an X squ but not a y^ s or it has a y s but not an X squ so here we see you know I meant this to be a y squ by the way so here we have an X squ and A y^ S so that's not a problem X2 y^2 not a problem X2 y^2 this is the the first one is the only one that doesn't have an X2 and a y^2 so this is the parabola that's how you can distinguish it from the other three here's the X squ but there's no y squ or they could be a y squ but no X squ so now which one is the hyperbola that's the next thing you want to look for now the hyperbola has a positive X2 and a negative y^2 or vice versa the ellipse and a circle both X2 and Y squ are positive so both of these are positive both of those are positive but notice this we have positive x^2 negative y^2 the ellipse uh the hyperbola has the minus sign so whenever you see that a positive X stive y^2 or vice versa it's a hyperbola now to distinguish the circle from the ellipse look at the coefficients of X2 and Y squ 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^2 + 8x + 2 y^ 2 + 4 y - 6 equal to Z 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 um the GCF of the coefficient so between two and eight we can take out a two and so we're left with x^2 + 8x and leave a space and between the next two terms we can take out a two as well so we have y^2 + 2 y leave a space and then the -6 let's move it to the other side now we need to complete the square half oh wait this is should be um this should be a four not an eight because that took away two so you want to take half of four half of four is two and then Square it half of two is one and then Square that's how you complete the square but now notice that we added two squar * 2 to the left side 2^ 2 is 4 * 2 is 8 so we got to add 8 to the other side as well 1 2 * 2 is two so let's add two 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 + 2 and then squared to factor the next term it's going to be whatever this variable is y Plus 1^ 2ar 8 + 2 is 10 10 + 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 two so our final answer is x + 2^ 2 + y + 1^ 2 the twos cancel and then 16 ID 2 is 8 so now it's in standard form the center of the circle is -21 and the radius is the square < TK 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^2 plus 25 y^ 2 - 24x + 100 y + 36 is equal to Zer so first let's move the x's and the Y's together so 4x^2 - 24x + 25 y^2 + 100 Y and let's move the 36 to the other side so let's factor out of four from the first two terms we want X2 to have a coefficient of one so it's x^2 and then -4x / 4 that's -6x and let's leave a space let's factor out 25 100 y / 25 is 4 y leave a space and then let's complete the square so half of 6 is three ignore the negative sign and then Square it half of 4 is 2 so that's 2^ 2 so 3^ 2 is 9 * 4 that's 36 so we got to add 36 to this side 2^ 2 is 4 * 25 is 100 so let's add 100 to that side so now let's factor it's going to be four whatever this variable is x minus this number squared and then plus 20 5 this variable y plus this number 2 squared - 36 and 36 they cancel so we're just left with 100 now for ellipses and hyperbolas this number has to be a one so divide by what you see there so we're going to divide both sides by 100 so we have x - 3^ 2 so 4 over 100 divide it backwards 100 ID 4 is 25 but since you divide it backwards 25 goes in the bottom 100 ID 25 is 4 and 100 divid 100 is 1 so now we have the equation of the ellipse in standard form as we can see the center of the ellipse is 3 -2 and a is the larger of the two a squ is 25 so that means a is five b^2 is four so B is 2 and keep in mind four and uh for an ellipse it's c^2 = a 2us B2 if you want to find the fol guy so now let's try this one the hyperbola 4x^2 - 9 y^2 - 16x + 54 y - 101 = 0 so let's move the x's and y's together so we have 4x^2 - 16x and then - 9 y^ 2 + 54 Y and let's move the 101 to the other side so now let's factor out a four so 4x^2 ID 4 is x^2 -6x ID 4 is -4x leave a space next let's factor out 9 54 /9 is -6 and now let's complete this Square half of four is two so let's add 2 squar half of six is three so we're going to add 3 squ now 2^ 2 is 4 * 4 we need to add 16 to this side 3 2 is 9 * 9 so that's 81 so let's add 81 to that side so now let's factor it's going to be 4 * xus 2^ 2 and then - 9 * y - 3 squar so 101 minus 81 that's 20 20 + 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 / 4 is 9 and 36 / 9 is 4 and this we get one so the center for the hyperbola it's 2 comma 3 whatever is positive that's a s and therefore this is b^2 now the last one this is a parabola and we want to put it in a form where it's like this x - h s is equal to 4 p y - 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^2 + 6 x is equal to -4 y becomes POS 4 Y when you move it from left to right and a positive one becomes negative 1 so now let's complete the square half of six is three and so we need to add three squar to the left and whatever you add to the left side add to the right side so three square is not so to factor the left side it's going to be X plus this number before you square it and then over here we have 4 y- 1 + 9 is 8 so all we got to do now is factor out of four from the right side so it's 4 y + 2 and now the parabola is in standard form the vertex is -3 comma -2 and this number is equal to 4 P so p is 1 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 - h^2 + y - k^ 2 which equals R2 and the center is h k so once you have the center let's say it's let's say this is H this is K you need to go R units to the right R units to the left up r down R and then graph the circle now for an ellipse if you have x - h^ 2 over a 2 + y - k^ 2 over B ^2 = 1 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're going to have a graph that looks like this so if this is the center this is therefore a that's negative a this is B this is negative B just remember these are the end points of the major axis so those are the vertices and these are the end points of the minor axis and your folky 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 folky 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^2 is equal to a 2us B2 for an ellipse and if you have the other version where you have x - h^ 2 but let's say B is under X and Y - k^ 2 where a is under y this ellipse it's 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 your folai 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 Tel now keep in mind the length of the major axis is 2 a the length of the minor axis is 2 B so to review for a hyperbola it can be x - h^2 over a^2 minus y - k^2 over b^2 equal 1 or it can be y - k 2 over a 2 and x - h^ 2 over b^ 2 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 squ to find a f guy is a s + b s rather than a s minus b^2 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 um the ASM tootes is y - K which is equal to rise over run rise is B run is a so plus or minus b/ a x - 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 orus a and the asmp tootes will be Yus K is equal to this time A over B so plus or minus A over B x - H and that's really what you need for a hyperbola those are the main equations oh I forgot the folky the folky is going to be h k Plus plus or minus C for this example and for this one it's going to be h h plus or minus C comma K so just so you know the folky is where it opens towards so the folai 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^2 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 + P now for the other one it's x - h^ 2 is equal to 4 p y - K so for y = x^2 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 + p and for this example it's h comma K minus P the directrix is horizontal so it's Y is equal to k+ p and here um it's Y is equal to uh K minus P so that's all the equations that you need um for conic sections so basically that's the end of this video so thanks for watching and have a great day