 Parabola we all know how a parabola is formed by intersection of a plane with a conic but again as I said it is just for you to know why it is called a conic there is no you know relation of that particular diagram with what we are going to study actually so what is the locus definition of a parabola so what's the locus definition just like we had a locus definition for a circle right we'll have a locus definition of a parabola as well so parabola is basically defined as locus of or path traced by a moving point path traced by a moving point on a plane on a plane such that the distance of the point point from a fixed point is equal to the distance or its distance from a fixed line okay so as per this definition let's say there is a fixed point I'll call it as s and let's say there is a fixed line let's say I call it as d equal to 0 okay now there is a point P this is a moving point okay it is moving in such a way that it is maintaining the same distance from this point as from this line in other words SP should be equal to PM in other words SP should be equal so if this point moves in such a way that these two distances are equal okay so let's say it comes here okay then again the perpendicular drop is same as this distance okay it may not be appearing like that but let me draw a fish diagram okay let's say it comes here okay let's say it comes here let's say it comes here okay let's say it comes here so whatever point I'm choosing okay these points are such points such that if you drop a perpendicular from this point onto the line and take the distance of that point from the S point they will all be both the distances will be equal okay so the path traced by this fellow P will be something like a curve which is looking like this okay this is basically your parabola okay and this fixed point that you have that fixed point is called the focus this point is called the focus okay and this line that you have that line is called the direct tricks okay now many people ask me sir where was the focus and direct tricks for a circle because circle is also a part of the conic right so if parabola shows a focus show should circle show right now remember the circles focus is actually as it at its center okay and its direct tricks is actually at infinity now this is something which I will talk about when I do ellipse with you every conic has actually two four psi and two directresses every conic has two four psi and two directresses so for a circle the two four psi coincide at the center and two directresses are located at plus minus infinity that's why we never draw it okay now when you start stretching the circle a bit it takes the form of an ellipse ellipse has got the two four psi separated out you can say s1 s2 from the center and these two directresses they come at finite distances they come at finite distances okay now let's say if you further separate out let's say further start pulling this then what will happen one of the four psi and one of the direct tricks will disappear to infinity and you will left you will be left with something like this structure so you'll see one focus and one direct tricks only this will become a parabola okay now the other other arm and other focus and the directors which went to infinity will come again back from negative side of infinity and this will give you a hyperbola okay and slowly when they coalesce they will actually become a pair of straight lines okay so this is a transition normally that these conic sections actually face anyways we'll talk about it when we are discussing it in more detail as of now this is just a small story I wanted to tell you just to relate all the conic sections with each other now let's talk about let's talk about some terminologies some terms that we are going to use for a parabola by the way just to tell you the practical use of parabola the biggest practical use of a parabola is in parabolic reflectors I'm sure you would have seen the headlight of your cars right I don't know whether it is seen right now but in corolla cars you see that they would be a structure like this you can see from the side okay so those are basically parabolic reflectors okay and what is the purpose of this parabolic reflectors so they are basically two light bulbs okay one is kept at the focus if you switch on the light bulb which is on the focus after hitting the parabolic reflector it will become parallel to the road actually okay actually it becomes parallel to the axis of the parabola this is called the axis we'll talk about it okay and if you switch the other one it will get reflected and will show you some nearby object on the road okay so parabolic reflectors are used in the headlamps of the car it is also used in the field of surgery where let's say the doctors will normally let's say they want to destroy a stone in the kidney okay so they'll put the stone at the focus and they will pass laser beam laser beam after hitting the parabolic reflector will pass through the stone and it'll destroy the stone okay a concave convex mirrors are spherical mirrors my dear they're not parabolic mirrors they are part of a sphere concave and convex mirrors come from part of a sphere they don't come from a part of a parabola okay anyways so let us look into the terms that we are going to use so please be aware of these terms we are going to use it time and again so let me again draw the direct text and we'll do one thing I'll draw one parabola and keep in case I keep requiring it I'll draw and parabola and keep sir how is this a parabola I will make it a parabola don't worry okay so for the purpose of understanding think it's a parabola okay so direct text is here and let's say your focus is here okay let's say this is your focus right now all of you please understand few terms that we are going to use you already know that this is called focus this is called focus why it is called focus because if you pass a ray which is parallel to the axis of the parabola now I'll draw the axis also axis is a line which is passing through the focus of the parabola and perpendicular to the direct text in other words axis is such a line which divides the parabola into two symmetrical halves okay this is called the axis of the parabola by the way I'll do one thing I'll take an image of this and keep because I may need this time and again come down yeah normally I keep using the same figures I don't want to draw it over and over again but doesn't a spherical mirror give a parallel yes but they have spherical aberrations that's why they have been discontinued kinshuk if you see the ambassador cars which are given to your dad by the army okay they will have spherical mirrors okay but they have been discontinued because of the spherical aberrations in them so parabolic reflectors have now replace spherical mirrors okay so those have those were the shortcomings of those ambassador lights which we used to have people having ambassador cars I mean I used to own one when I was very small my dad used to have one he bought it in 1987 or something and the mirrors used to be like full round and all okay so those used to have a lot of issues anyways so this is called the focus this is called the directrix this is called the axis the point where the axis meets the parabola that is called the vertex this point is called the vertex okay now remember vertex is the midpoint of n and s n is the point where the axis is hitting the directrix okay so this is your directrix so this point n s v is the midpoint of it okay so I'll write it down here vertex is is the midpoint of mid point of n and s why n why is the midpoint because you know the definition itself says any point on the parabola should be equidistant from the line and from this point so it has to be midpoint in this case it has to be midpoint okay and also remember axis is perpendicular to the directrix always it is true for any parabola axis is perpendicular to the directrix okay next important term that we need to understand is chord okay so what is the chord what is the chord the chord is basically a line which connects any two points on the parabola so let's say I connect these two points okay this line would be this line segment will be called as a chord of the parabola okay and there's something called focal chord so let's say I join a I join our two points which passes through the focus okay then this line will be called as or this part will be called as the focal chord okay then there's something called double ordinate double ordinate is a line connecting any two points on the parabola which basically makes it perpendicular to the axis this will be called as double ordinate why the name double ordinate is because the end points of this double ordinate will have the same value of y but of course opposite sign would be there in this case that is why it is called double ordinate don't go by the literal meaning that the ordinate will be two times so the end points of the double ordinate the y coordinate in this case will be same but of opposite sign is it fine if a double ordinate happens to pass through if a double ordinate happens to pass through the focus then that particular chord or that particular double ordinate will be called as the lattice rectum lattice rectum lattice means side rectum means line so as you can see it is drawn at the side okay lattice rectum is basically the shortest focal chord lattice rectum is the shortest focal chord of the parabola okay so again I'll repeat all the terms once again this fixed point is your focus this line is your directrix a line which is passing through the focus and perpendicular to the directrix or you can say it is symmetrically dividing the parabola into two equal halves that is called the axis any line connecting any two points or any chord connecting any two points is called the chord okay I use the word chord for explaining a chord so any line segment connecting any two points on the parabola is the chord if a line is perpendicular to the axis and connecting two points on the parabola that is called a double ordinate if the double ordinate happens to pass through the focus it is called the lattice rectum okay and any chord which is passing through the focus is called the focal chord so far these terms should be well known to you okay any questions so far okay now what are we studying in this chapter we are studying coordinate geometry so where is coordinate coming in here I was only talking about the definition I was talking about the critical point so let us introduce coordinate geometry so coordinate geometry means equations points okay so I'll be starting with equations of standard parabola parabola or parabolas both are accepted words for multiple parabola okay so standard parabolas or standard parabola so let me start with one of the standard forms there are actually four standard parabola so we'll begin with one of the forms for that I will pull out my diagram which I had okay now in order to get the equation first of all I need to fix up my coordinate axes I need to choose an origin right so normally for the purpose of getting a standard parabola we choose the vertex as the origin that means I'm going to align my x and y axes in such a way that my x-axis is along the axis y-axis is perpendicular to the axis and passing through the vertex so I choose vertex to be my origin so in standard parabola your vertex is chosen as the origin okay so let us now give some coordinates so the first thing that I would choose is the coordinate of the focus since I my focus always lies on the axis which happens to be the x-axis in this case let me choose the coordinate of the focus to be a comma zero I remember here a represents the distance of the focus from the vertex and since it is a distance we always keep it as a positive quantity so always a will be positive this is something which is very very important many people while solving the question wrongly take a values as a negative value no a will always be positive so if you're let's say your parabola was having its focus on the negative x-axis you will write a minus a there okay because minus a a being positive will only give you a negative x-axis correct so remember this fact a will always be positive always do not make any mistake about it I'm putting question marks here so that you never make a mistake okay now if this is origin and this is a comma zero it is very obvious that this point would be minus a comma zero and if your direct x is parallel to the y-axis because it is perpendicular to the axis and you have chosen x-axis as your axis its equation has to be x equal to minus a okay now in order to derive the equation of the parabola we will choose a moving point H comma K over here okay so H comma K locus will give me the equation of the parabola so I will use the fact that this distance let me call it as P P M should be equal to SP or SP should be equal to PM so what is SP SP is under root of H minus a the whole square K minus zero the whole square what is PM PM is nothing but it will be if you see the distance here would be H plus a mod okay so equation of a line is known point is known the distance here will be H plus a mod okay now having got SP equal to PM relation let's square both the sites if you square both the sites you will end up getting H minus a the whole square plus K square is H plus a the whole square let's try to simplify this so K square is equal to H plus a the whole square minus H minus a the whole square which gives you H plus a plus H minus a and H plus a minus H plus a okay so this clearly gives you K square is equal to 2 H times 2 a or K square is equal to 4 a H or you can say generalizing it you can say Y square is equal to 4 a X right so for such a parabola for such a standard form of a parabola your equation of this parabola will become Y square is equal to 4 a X okay so please note for such a case your vertex will be at origin your focus will be at a comma zero your directrix will be X equal to minus a your axis equation will be Y equal to zero okay what else you need I think this is more than enough now I would request you all to give me the length of the lattice rectum so if I draw the lattice rectum for such a standard case can you tell me the length of the lattice rectum in terms of a what is the length of this lattice rectum L R length I know most of you have done this in school instantaneously you will say 4 a am I right yeah I know you have done it so let's do it from as if we have not learned this before so see when you're talking about oh so sorry yeah when you're talking about the length of the lattice rectum let's say I call this point as point L1 and I point call this point as L2 so both L1 and L2 will have the X coordinate as a only right let's say Y coordinate is K for the time being so a comma K should satisfy this curve so you can say K square is equal to 4 a a so K square is going to come out to be 4 a square so K will be equal to plus minus 2 a now remember since a is a positive quantity to signify your top value or you can say the top coordinate you'll have to use plus 2 a and here you have to use minus 2 a so the length here will be the distance between a comma 2 a and a comma minus 2 a which happens to be 4 a okay so please make a note of this that is another thing that we need to keep in mind length of lattice rectum length of lattice rectum is 4 a units is 4 a units you don't have to write a mod over here because a is only a positive quantity so it'll be 4 a units now before we go for a break I'll quickly tell you what about the other forms of the other standard forms of the parabola so this is just one of the standard forms remember there are four standard forms this is just one of the standard form anything that you would like to copy from here please do so because I'm going to change the slide to the next okay so other standard forms will be the one where your parabola is opening oh let me just choose the same colors which I have used because I mean it's a crooked one I believe this is your direct text this is called the previous one is called the right word opening parabola I'll name it I'll name it I think I did not name it so this is our standard form and we will refer to this standard form as a right word opening parabola what we'll call a right opening or right word opening parabola okay the next one which we are doing that will be called as left word opening parabola yeah this will be called as the left word opening parabola okay so here your focus is at minus a comma zero directrix is x equal to a vertex is still at origin okay this is your y-axis this is your x-axis for such a case the equation will become y square is equal to minus 4x we don't have to derive it separately because it is just like taking the reflection of the right word opening parabola about the y-axis now all of you have undergone a rigorous crash course with us not crash course bridge course with us so I've been habituated to saying crash course because your seniors are undergoing crash course so you have done a bridge course with us where you have learned that if a curve gets reflected about the y-axis we actually change the sign of x so instead of okay let me write in a more proper way is it off for a x now you will write for a minus x so as a result this becomes y square is equal to minus 4x okay so few important points to be noted so please note that for such a case your vertex will be at origin your focus will be at minus a comma zero directrix equation I've already written x equal to a okay our length of the lattice rectum and length of lr will be again for a units no change okay so this is your left word opening parabola coming to a right word sorry upward opening let me choose this color so this is a upward opening parabola so for upward opening parabola the focus is at 0 comma a this is y equal to minus a vertex is still at origin this is your x-axis this is your y-axis okay now if I if somebody says derive the equation of this parabola you don't have to again reinvent the wheel you just realize the fact that your right word opening parabola if you reflect it about a line y equal to x so let's say if you reflect it about a line y equal to x it will actually become an upward opening parabola see okay so this parabola is actually the reflection of reflection of a right word opening parabola right word opening parabola about about y equal to x line okay and again if you recall your bridge course days we had done this concept that if you're reflecting a curve about y equal to x line just change the position of x and y okay just change the position of x and y so in y square is equal to 4ax if you change the position of x and y it becomes x square is equal to 4 a y okay so in this case please note this is first of all called upward opening parabola upward opening parabola critical points here your vertex again at origin okay focus 0 comma a directrix equation y equal to minus a length of latter sector will still remain for a units because on reflection about y-axis or x axis or whatever you're doing it doesn't change the length of the latter so this is your third standard form okay now going to the fourth standard form in the fourth standard form we are going to reflect the parabola downwards so I'll take the last one okay and directrix will be here okay so this is your x-axis y-axis focus will be 0 comma minus a y equal to a will be the directrix 0 comma 0 will be the vertex so for such cases just have to do one thing change your y with a minus y because we know that when the graph is affected about x-axis the sign of y gets changes so the sign of y gets changed and it becomes x square is equal to minus 4 a y okay so this is our downward opening parabola this is a downward opening parabola so please make a note the critical points over here the vertex will be again at origin no change focus as I've already drawn 0 comma minus a directrix will become y equal to a length of latter sector is again for a units as of now we'll take a small break on the other side of the break we'll talk about some few basic questions and then we'll move on to the generalized version of a parabola alright I hope everybody is back so before going for a break we had done the four standard cases of a parabola correct guys when you look at these equations for a very very long time even in school and now you get a feeling that parabola equations always look like y square is equal to 4 a x or plus minus or x square is equal to 4 a y plus minus now don't get me wrong over here these are standard forms only your parabola equation can become as complicated as your general form of a conic so a parabola equation can look as ugly as this are you getting my point right so in order to make you realize this I will take a small question to begin with my question here is find the equation of a parabola find the equation of a parabola of a parabola whose focus is at whose focus is at one minus one comma minus two and direct tricks is and direct tricks is x minus 2 y plus 3 equal to 0 everybody please solve this question and then after solving this question we will see how complicated or how we can say lengthy and ugly an equation of a parabola can look like because many people we just see parabola as that y square is equal to 4 a x or y x square is equal to 4 a y no those are of course the equation but those are very very simplistic cases they're very standard cases right and the reason why they are standard is because they're very simple okay but you can have an ugly looking expression as well so first solve this question and then we'll talk about it just type done if you're done follow the locus definition here kinshuk what are the locus definition of a parabola parabola is a locus of a point which moves in such a way that its distance from this fixed point should be same as the distance from this line isn't it follow that just say done once you're done okay let me sketch it also to a certain extent this line is nothing but x by 2 plus 3 by 2 x by 2 plus 3 by 2 line will look like this okay and its focus is here minus 1 comma minus 2 so parabola is of this nature I'm just doing a rough sketch this will become the axis of the parabola this is your vertex okay now what is the locus definition locus definition is if I take any point h comma k on the parabola if I take any point h comma k on the parabola its distance from the focus should be same as its distance from the directorics isn't it this distance and this distance should be same and this line is already x minus 2 y plus 3 equal to 0 so distance of the point p from s will be h plus 1 the whole square k plus 2 the whole square under root this should be the distance of h comma k from this particular line which is h minus 2 k plus 3 mod by under root of 5 okay so what I have done I have just used the fact that sp is equal to p m okay square it simplified so I'll directly do a lot of calculations here so I'll take 5 on the other side and I'll square the whole term so it'll become 5 times h plus 1 the whole square k plus 2 the whole square on this side you will have h minus 2 k plus 3 the whole square if you collect your h square terms you'll have 4 h square if you collect your k square terms you will have plus k square if you collect your x y terms x y terms will come from here so it'll give you 4 h k terms and this will give you a 10 h from here minus 6s so this will be 4 h and y will be this is 4 k 20 20 plus 12 32 k and constant terms would be 5 25 25 minus 9 which is 16 sorry 16 okay simplified I mean you don't have to follow me for simplifying it right so now if you generalize this if you generalize it you will see that this is how a equation of a parabola can look like okay that means your equation of a parabola will have all the characteristics that a phonic can have a x square b y square 2 h x y 2 g x 2 f y plus c equal to 0 so don't be under the impression that parabola is always like y square is equal to 4 a x or y square is equal to minus 4 a x or x square is equal to 4 a y or x square is equal to minus 4 a y okay so such complicated questions or such complicated parabola I should say not questions question was not complicated such complicated equation of the parabola can also come normally such kind of parabola we call as oblique parabola okay it's an oblique parabola that means it is not rightward opening or leftward opening or upward opening or downward opening it's something oblique that means at a certain angle it is okay for school exams I don't think so you will be tested on this okay in J also not many questions will be asked okay but whatever is going to be asked we're going to we are going to take that up once we we are done with our you know school exams any question anybody okay now we'll come back to our standard cases so some problems based on standard cases so problems on standard parabola so let me start with some questions on standard parabola very simple question I'll begin with find the coordinates of the foresight equation of the directorics and length of the latter sector for the following four cases okay we'll quickly do it just to understand that all of you are good with your basics okay and will not waste much time first one let us start y square is equal to 8x 8x you can write 4 into 2 into x correct so your a is actually 2 over here right now let's figure out what's the coordinate of the foresight in fact there will be only one focus so foresight is a misnomer here so what will be focus let's write it down what will be the directorics equation let's write it down and what is the length of the latter sector let's write it down yeah so all of you please write down on your screen or write down on your chat box focus is 2 comma 0 very good jumbo directorics x equal to minus 2 brilliant length of the latter sector 8 units awesome okay second one x square is equal to 6 y 6 y you can write it as 4 into 1.5 4 into 1.5 y okay so here a is 1.5 so tell me focus focus yes absolutely yes focus is 0 comma 1.5 very good directorics y equal to minus 1.5 very good LR 6 units very good very good so all of you are comfortable with your basics so I don't have to worry much about it third one y square is equal to minus 12 x so minus 12 x you will write it as minus 4 into 3 into x so what is a here what is a here a is 3 never say minus 3 okay never say minus 3 yes so what is the focus here it's a left foot opening so focus is minus 3 comma 0 very good directorics x equal to a latter sector 12 units very good next one fourth one x square is equal to minus 16 y minus 16 y I will write it like minus 4 into 4 into y so a here a here is going to be 4 yes so where is focus where is focus it's a downward opening parabola 0 comma minus 4 very good that equation of the directorics y equal to 4 length of the sector 16 units very good so everybody is pretty fine with it okay alright now we will move towards the generalized form of a parabola also called the shifted form of the parabola okay so what we did was those cases where your vertex was at the origin okay but in these cases your vertex will not be at the origin it will be shifted somewhere in space but but its axis will still be parallel to the x axis or y axis as discussed in the standard cases okay so right now I have basically shown a parabola where the vertex has now gone to alpha comma beta and let's say the distance of the focus from the vertex is still a okay and the axis of the parabola is parallel to the x axis can you tell me what will be the equation for such a parabola all of you first write it down on your notebook what do you think will be the equation of a parabola whose distance between the vertex and the focus is a and the vertex has gone to a coordinate or a place called alpha comma beta this was your origin this is your y axis this is your x axis so how will you write this equation so you will say it's a simple it is just a case where your standard form of a parabola got shifted shifted alpha units right okay and beta units up okay when a curve gets shifted to the right and up what do you do with your x and the y values when a curve shifts to the right we normally replace our x with x minus alpha and we replace our y with y minus beta I hope you are all aware of this yes or no okay now Shradha is making a mistake Shradha you curve got shifted right and up not the origin origin will go in the reverse direction guys let me tell you the inside story is the curve doesn't shift what shifts origin shifts if you recall your shifting of origin concept you would recall that you would recall that when the origin shifts to H comma K when the origin shifts to H comma K what do we do we replace our x with capital X plus H and we replace our y with capital Y plus K do you remember this or not you remember this shifting of origin concept so now many people will start asking sir why didn't you write then y plus beta and X plus alpha guys when the vertex has gone to alpha comma beta it means your origin has shifted to minus alpha minus beta see originally it was like this right now how should you shift your origin how should you shift your origin so that your curve starts looking like this you will say you have to take your origin to minus alpha minus beta right so your origin comes to this position so if your origin comes to this position only then your curve starts looking like this isn't it so your X will get changed with X plus H and H here is minus alpha and your Y will get replaced with Y plus K and Y is minus beta don't confuse okay don't confuse this is the concept of shifting of origin okay any questions here any questions here similarly if your curve was your y square is equal x square is equal to 4 a y and you are shifting your origin or you are shifting your vertex to any position alpha comma beta you just have to do a simple act take the standard form change your X with X minus alpha change your Y with Y minus beta that's it okay so let me begin with a simple question for you all find vertex focus equation of directrix length of lattice rectum okay equation of axis for this parabola I just give you one of them first to begin with y plus 1 the whole square is equal to let's say 8 X minus 2 okay so first of all think very carefully and let's answer these five things vertex focus equation of directrix length of lattice rectum equation of axis so all of you first think carefully and then we'll solve this question and if somebody is done just type that if you feel you want to draw a diagram to solve it you are most welcome to do it arch it is done very good arch it that was fast this type done lucita if you're done just say done okay we'll discuss the answers I don't want you to take the effort of writing everything okay lucita is done to the is done arch it is done Shadda is done okay so when you look at this it basically gives you an idea that y square is equal to 8 X was shifted to right and one down okay so one down and two right will bring the vertex at this position so this was your parabola this was your parabola okay so this is your parabola okay this will be your vertex this will be your vertex this will be your axis okay and focus will be somewhere over here this will be your focus okay and of course lattice equation directrix would be somewhere over here I'm not exactly commenting on its equation right now but I'm just drawing one okay anyways so first of all vertex is clear to everybody vertex should be at 2 comma minus 1 2 comma minus 1 will be the vertex okay second thing that you would observe here is that in this case of a parabola your a value was actually a 2 correct so if you compare this with 4 a your a value was actually a 2 correct so that means you have to move 2 units on this side to get the to get the focus correct so this is 4 comma minus 1 okay so vertex is 2 comma minus 1 focus is 4 comma minus 1 correct what about equation of the directrix directrix would be a line parallel to the x axis y axis and it would be passing through this point now what will be this point this point will be 0 comma minus 1 correct yes or no have I made any mistake okay so when I say 0 comma minus 1 it will actually become your it will actually become your y axis only correct so your directrix will actually be aligned along the y axis only okay so directrix equation will be x equal to 0 correct next what was required length of lattice sector length of lattice sector will be 4 a only 4 into 2 which is 8 units length of the lattice doesn't change on shifting the parabola whether you shift it down right up left whatever length of the lattice system will be fixed fixed to 8 units only in this case next is equation of the axis this is a line which is basically y equal to minus 1 line okay now how many of you got all these five things correct vertex 2 comma minus 1 focus 4 comma minus 1 equation of the directrix x equal to 0 length of the lattice sector 8 units equation of the axis y equal to minus 1 absolutely very good very good those who got it correct okay now there are many students who ask me is there any shortcut to find it I don't want to draw a diagram and all okay because people think drawing a diagram is a bonus work right we should avoid doing it no drawing diagram is sometimes the only way to solve a question many a times okay especially in exams like J so never refrain yourself from drawing a sketch anyway since you have asked for a method without a diagram let's discuss it so here what do we do I'll take the same question again y plus 1 the whole square is equal to 8 x plus 2 so what do we do here we do a role change what role change we do instead of y plus 1 we'll put a y okay and instead of x plus 2 we'll put a capital X okay capital Y capital X thereby converting this equation to y square is equal to 8 x or in a more granular way y square is equal to 4 into 2 x okay now recall everything that you know about this standard form of a parabola where was where was the vertex of this parabola the vertex was actually supposed to be at 0 comma 0 right 0 comma 0 for my square is equal to 8x vertex is 0 0 right now all you need to do is for the case which was given to you all you need to do is put x as 0 because here this is 0 put y as 0 because this y is also 0 and just change the values of capital X or you can say expressions for capital X and capital Y like this solve for X and you will end up getting your vertex coordinates so this is your answer for the vertex okay as you can see here this is what you had opted okay now focus focus for this parabola was a comma 0 that is 2 comma 0 right okay so just do a role change just call X as 2 y as 0 and put your X as X plus 2 okay and put your Y as I hope I have got I haven't done any mistake in copying a yo Ramakrishna this was X minus 2 sorry yeah so X will become a 4 and capital Y was Y plus 1 so Y will become a minus 1 okay so 4 comma minus 1 will become your focus as you can see over here we have already got that okay next is directrix equation so for this the directrix equation is going to be X equal to minus a now you change your change your X with X minus 2 and this will automatically reduce to X equal to 0 okay so all you're doing is a role change so whatever you knew for this you write it down and just do a role change of capital X in terms of small x and capital Y in terms of small y that's it next was equation of the axis for this case the equation of the axis was your X axis X axis is capital Y equal to 0 so just write y plus 1 equal to 0 which is nothing but y equal to minus 1 okay length of the latter sector will be the same and in this case you don't have to draw any diagram to solve the question make sense you can solve this question just by doing a role change if you have understood this concept let me give you few problems based on this so that you are you know accustomed to it this is the maximum your school teachers can ask you okay so let's have another example again find number one I just write down the questions what we need vertex focus equation of directrix equation of axis length of ladder system okay and let me just take a snapshot in case I need it again and again okay and the parabola is y minus 3 the whole square is equal to X plus 2 with a minus sign over it let's say yeah everybody please solve it only after solving the five of them say done done kinshok is done very good with two done anybody else done chakda done very good okay so first compare with which standard case will you compare this this standard case was shifted to get this parabola you will say y square is equal to minus 4 a xr am I right am I right how did I know minus 4x because of the presence of this minus sign okay so let us do a comparison so on comparing your capital Y is small y plus 3 capital X is small x plus 2 and a my dear is one fourth I hope everybody is fine with this okay now recall everything that you know about this parabola recall everything so for this parabola where was the vertex recall it where was the focus recall it what was the equation of the directrix recall it what was the equation of axis recall it okay what was the length of the lattice system recall it let's do that so what takes for this guy used to be zero zero focus used to be minus a zero directrix used to be x equal to a okay let me better write capital X okay so that yeah correct equation of the axis used to be capital Y equal to zero and length of lattice sector used to be four a units okay now on the very same figure you just have to do a role change what role change instead of zero zero right capital X equal to zero capital Y equal to zero change your capital X to X plus two came your capital Y to Y plus three so X becomes minus two Y becomes minus three so minus two minus three will be your vertex as everybody got this give yourself a pat on your back if you've got this very good next minus a comma zero write it like this X is minus a Y is zero and do a role change X plus two is minus a minus a is minus one fourth and Y plus three is zero so this means your coordinate will be I'm directly writing it minus nine by four minus three as everybody got this directrix already I've written capital X equal to a that means X plus two is one fourth so directrix equation is X equal to minus seven by four if I'm not mistaken or you can also write it four X plus seven equal to zero has everybody got this check check check check check okay capital Y equal to zero means Y plus three is equal to zero so that means this is this itself is the equation so check it out length of the lattice rectum is one unit okay does it give you a fair bit of idea how to work with these cases okay got all of them correct very good very good okay now normally the question setter will not be a kind enough person to give you a equation like this what he will do is he will open this up he will open the brackets which you call as simplification but he will do the complication of it okay and it is our duty to figure it in this way and then figure out the answers so what we will do is we'll take some questions when the question setter has actually aggravated that situation by opening the brackets okay all right so let's have this question first of all this represents a parabola okay find its vertex equation of axis equation of lattice rectum also this fellow has asked very good coordinates of the focus equation of the directrix extremities of the lattice rectum length of the lattice rectum in fact in short it has asked full bio data of parabola okay now before you start solving a question one very small thing I would like to you know inform you many people ask me sir is this a case of a shifted parabola or is this a case of a oblique parabola let me tell this to you very straightforward in the beginning of the process itself oblique parabola will always contain x y terms does this equation contain any x y term no so it is not a case of an oblique parabola so when there is no x y term then it is either a case of shifted parabola or standard parabola but it is not a standard one because had it been a standard one it would have been like y square is equal to some constant into x isn't it but it has two y also so it's a case of shifted parabola okay so let nobody tell you what case of a parabola it is you can yourself recognize you can yourself make out now solve this question first everybody give a honest attempt then we'll discuss it done sir okay anybody else done sir done okay all right so here one more step you need to do and that step is completing the square okay so first of all let me write it like this so normally I separate out the y terms and x terms of course constant I keep with the x terms now remember since this question shows you a quadratic in y there has to be a completion of square in terms of y hadn't it been a quadratic in x you would have completed a square in x so that would have been a case of shifting of an upward or a downward opening parabola okay so in this case since you are since it is very clear that it is a quadratic in y it has to be either in the form of shifting of right word opening parabola or a left word opening parabola so that will come to know when we once we complete the square so add a one I hope you all are good at the art of computing the square okay I will do I'll go one more step here I'll do four in into one by four times this this is y square is equal to four a x so here is a role change your y role is being paid by y plus one x role is being played by x minus four and your a role is being played by one fourth okay now vertex where is the vertex of this guy zero zero correct so I will write y plus one equal sorry x minus four equal to zero okay and y plus one equal to zero so it is four comma minus one how many of you got this right four comma minus one very good excellent next equation of the axis equation of the axis for this parabola is capital y equal to zero isn't it the x axis so capital y equal to zero so let me write vertex over here axis equation is capital y equal to zero means y plus one equal to zero okay vertex axis next is equation of the latter now equation of the latter sector is given by capital x equal to a okay if you recall your standard case your latter sector used to be like this so this is small x equal to a so here it will become capital x equal to it so equation of lr will be capital x which is x minus four equal to a in other words it is four x minus 17 equal to zero how many of you got this correct very good coordinates of the focus coordinates of the focus the focus of this parabola is capital a comma zero so x is capital x is your a y should be zero that means you're saying x minus four is one fourth and y plus one is equal to zero so the coordinates will be 17 by four comma minus one that will be your focus okay equation of the directrix equation of the directrix for this parabola was capital x equal to minus a so your x minus four will be minus one by four that will give you four x minus 15 equal to zero so this will be the equation of the directrix this for your focus let me bubble it up finally extremities of the latter spectrum now extremities of the latter spectrum how will you find it out now remember your parabola is a parabola which has been shifted in such a way that the vertex has gone to four comma minus one so four comma minus one is here okay or you can do one one simple activity you can simply put you can simply put in the equation x equal to 17 by 4 okay so when you put excess 17 by 4 in the equation of the parabola let's see what values of y it throws out so y square minus 2 y i hope it is minus or plus or plus plus minus 17 by 4 plus 5 okay so what does this give you this gives you y plus 1 the whole square is equal to 17 by 4 minus 4 which is 1 by 4 okay so automatically from here y plus 1 comes out to be plus minus half so your y can be either minus half or it could be minus 3 by 2 so your ends of the latter spectrum will be 17 by 4 comma minus half and 17 by 4 comma minus 3 by 2 make sense Shraddha why why it will be 4x plus 50 may know the reason for that is this correct directrix those who have solved it yes what is the problem Shraddha you're saying 4x plus 15 15 y 15 will go okay next is the length of the latter system length of the latter system is going to be one unit make sense now you know how to deal with these situations also okay now before we take on more problems one small thing I would like to cover with you and that is the parametric form of a parabola just like we did the parametric form for the case of a circle we'll also do a parametric form for the case of a parabola and then we'll take a few questions and wind this topic up and there's nothing more that is required for your school more than this but again just like circle parabola is also a big chapter okay there are a lot of tangent properties and normal properties that will come up so parametric form of a parabola so I will start with a very simple case let's start with this guy y square is equal to 4x okay now though there can be millions and trillions and zillions of parametric form possible the one that we prefer using for such a parabola is x equal to at square and y is equal to 280 right I don't know who came up with this but the person who came up with this very beautifully you know represented this and you can see that if you try to eliminate the t from x and y you will end up getting you'll end up getting this equation so t here is a parameter okay this is one of the widely used parametric form for solving j level questions so please keep this in mind okay no problem Shraddha no problem but my question to you is you suggest me in a similar fashion what should I have the parametric form for the second standard case so can you suggest me a parametric form for this guy take a clue from this very good very good so for such case you can take x equal to minus at square and y is equal to 280 excellent very good okay right now for the others I'll give you one more opportunity tell me for x square is equal to 4a y this is simple I mean yes absolutely so you'll say sir why don't you change the position of x and y so x will become 280 y will become 80 square okay is it fine any questions okay and finally the fourth one which is not a surprise to anybody if I have to write for x square is equal to minus 4a y I will write the parametric form as x equal to 280 and y is equal to minus 80 square make sense okay now having known this one small question I would like to you to answer this out suggest a parametric form find a parametric form for this parabola y minus 1 the whole square is equal to minus x plus 2 suggest a parametric form for this parabola please give me your response on the chat box I want to see everybody's response very good kinshuk kinshuk your y is not here okay very good okay so here you will see that it resembles this format correct so so you will write x plus 2 as minus a t square that means your x is minus 2 minus 1 fourth t square and your y which is y minus 1 you will write it as 280 that means y is nothing but 1 plus t by 2 so you can suggest a parametric form for this equation to be like this I am writing it in one point minus 2 minus 4 t square and y is equal to 1 plus t by 2 okay so this is a parametric form make sense any questions any concerns please do highlight please ask me okay so for the remaining 15 minutes of the class we'll do some problem solving okay so we'll make ourselves comfortable with this concept okay so let's have this find the equation of a parabola whose vertex is at 2 comma minus 3 and focus is at 0 comma 5 just let me know done with your done I will be happy to see that you have tried it out diagram is nothing can replace a beautiful almost accurate diagram focus is 0 5 okay so focus is somewhere over here okay vertex is at 2 comma minus 3 2 comma minus 3 is here okay they're very far apart so let me just take 2 comma minus 3 here okay 2 comma minus 3 so how will your parabola look like it will look like this great fat one actually kinshukh let me remind you that this is an oblique parabola okay this is an oblique parabola here you can see your axis is tilted okay it is not a shifted case of a parabola like what you are thinking of okay it's an oblique parabola correct so for oblique parabola we need to know the directrix how will I get the directrix who will tell me how will I get the directrix this is the vertex this is the focus yeah absolutely first of all it is perpendicular to the axis so I know the slope because I know your slope of sv correct so what is the slope of sv first of all slope of sv will be y1 minus y2 y1 minus y2 by x1 minus x2 that's minus 4 so slope of this guy is plus 4 correct so slope of the directrix will be plus 4 correct and this point how will I find out remember v is the midpoint of snn okay midpoint now you do use distance also midpoint use curve isn't it so this guy will become if I'm not wrong this will become four and this will become minus 11 okay ha ha sorry thank you Dave thank you you saved my lot of effort yeah yeah sorry I realized that thanks Dave and Ritu for correcting me okay now directrix equation is what y minus y1 is slope times x minus x1 correct that means 4 y plus 44 is x minus 4 so x minus 4 y minus 48 equal to 0 okay now I will use the locus definition now I'll say that any point on this particular parabola the distance from this directrix should be same as the distance from the focus even though that diagram wise it doesn't look like that but you can consider it to be like this and save your time just call it as x and y no need to waste your time writing it as h and k and then writing it in terms of x and y it was just for your initial understanding that I wrote like this so brevity is the key okay now don't waste your time simplifying it I hope you can simplify it from here on my idea was just for you to tell how to proceed in these kind of questions remember not all the parabola are your shifted parabola not all the parabola are your standard parabola okay Dave we should be the midpoint of s and some point over here so what is this point midpoint formula use Karloff you know take it as some p comma q so p plus 0 by 2 should be 2 so p will be 4 correct q plus 5 by 2 should be minus 3 so q should be minus 11 okay will you be able to proceed from here on any questions any doubts any kesta please please let me know okay all good all right so we'll move on we'll take few more questions this is something which we have already done focus vertex questions okay let's do this there are three questions I can say the three sub parts are like three questions find the equation of a parabola whose vertex is at 1 comma 0 focus is at 3 comma 0 first one let us do the first one then we'll come to the second and the third first one done make the diagram diagram will speak volumes about it yes vertex is at 1 comma 0 my dear friend focus is at 3 comma 0 my dear friend that means your parabola will look like this my dear friend that means it is our right word opening parabola which has been shifted one unit to the right isn't it okay a value is 2 so had it the had the vertex been at origin it would have been y square is equal to 8x but actually what had happened the parabola has moved one unit to the right so what you will do this is your answer Dave shifting is not clear my dear friend what is happening is it fine second question please try it out focus 2 comma 0 and one extremity of its lattice rectum is 2 comma 2 I am not solving it diagram to be there okay one extremity is here 2 comma 2 which clearly indicates that your parabola should have been like this down down down down down down down this is your lattice rectum yes okay very good very good kinshok very good they uh were above anybody else who would like to participate okay so guys this guy is 2a so you know a is going to be one isn't it this is half the lattice rectum full lattice rectum is 4a half of it is 2 correct so your vertex will be when you go back one unit over here so this guy is 1 comma 0 correct a is 1 vertex is at 1 comma 0 so your equation will be like this y square is equal to 4x minus 1 am I right am I right any questions here the above is absolutely correct okay kinshok where you went wrong third one third one focus 0 comma minus 3 directrix is y equal to 3 this looks like the standard form only isn't it this is a standard form only okay so focus focus focus focus focus is here 0 comma minus 3 directrix is y equal to 3 so it is top here so your parabola will look like this okay so it is x square is equal to minus 4a a y a is 3 my dear this is 3 a is 3 so x square is equal to minus 12y I hope everybody has got this right at least okay next question last question let's do a application application a beam is supported at its end by two supports which are 12 meter apart since the load is concentrated at the center there is a deflection of 3 centimeter at the center and the deflective beam is in the shape of a parabola how far from the center is the deflection 1 centimeter one and a half minutes draw it out think it out and then we'll solve it okay let's draw the diagram so see what has happened there is a beam which was kept supported on two supports okay and these two supports were 12 meters apart and because the beam was heavy at the center there is a deflection of 3 centimeter over here watch out the units my dear friends watch out the units one unit is meter another unit is centimeter okay now they're asking you how far from the center is the deflection 1 centimeter that means how far is this distance what is d they're asking us now in order to find it I will first position my coordinate axes in this way I'll keep my x axis like this and I'll keep my y axis like this okay in other words the lowest point of my beam I will take it as the origin now if you know the parabola which is positioned in this way we know it has to be x square is equal to 4 a y but what is a how do I figure that out not to worry if you take this point let me call this point as a p we know that this point is 6 comma 3 by 100 am I right so this equation must be satisfied by 6 comma 3 by 100 so 36 is equal to 4 a into 3 by 100 in other words in other words a will become 300 correct me if I'm wrong okay that means your parabola is x square is equal to 1200 y okay now I want to know I want to know the location of this coordinate this is d comma correct me if I'm wrong this is d comma 2 by 100 isn't it because it is d units to the right and since it is one down from the x axis it is 2 up actually doesn't look like from the diagram so it is 2 up so if I put my d comma 2 by 100 in this particular expression I should be able to get my d so d square is equal to 1200 into 2 by 100 that means d is equal to a root of 24 that is 2 root 6 meters okay so from the center if you go to root 6 to the right or to the left you will end up getting the the deflection as 1 centimeter is this correct wherever a small mistake you did you thought that 1 centimeter is the y coordinate but 1 centimeter is not 2 centimeters the y coordinate is it fine any questions any concerns okay so we'll give this topic a rest here and the next class would be on