start with ellipse okay ellipse is basically a conic which is generated when a right circular double cone when a right circular double cone is cut by a plane okay this is a right circular double cone this is the axis of the cone okay and we all know this angle is called the this angle is called the semi vertical angle okay this angle is called the semi vertical angle so if you have a plane which is cutting this right circular cone let me make it in white yeah so if you have a plane which is cutting this right circular cone at such an angle which is more than alpha but lesser than 90 degree so as you can see this angle beta this angle beta is more than alpha but lesser than 90 degree then the cross section area that you would be getting here that would be of a ellipse okay so this would be a this would be a ellipse okay so this is just to tell you this is just to relate why ellipse is a conic now let us see the locus definition of an ellipse by the way the locus definition that we had for parabola and that we'll be having for ellipse and that we will take up for hyperbola they're all same they just differ in one aspect we'll be discussing that aspect right now so please note down ellipse is the locus of a point or it's a path it's a path traced by a point traced by a point by the way this is locus definition one there's a locus definition two also we will talk about it later on in our discussion this is locus definition one okay so it's a path traced by a point a point moving in the plane moving in the plane in such a way in such a way that the ratio of its distance distance from a fixed point from a fixed point to that from a fixed line that from a fixed line is a constant is a constant where this constant has a value between zero and one not including zero not including one okay so let me repeat this definition once again and I'll also point out what are the what is that critical point and what is that line and what is this constant actually called so it's a path traced by a point moving in a plane yes ellipse is a two-dimensional object okay so it has to be moving in a plane it cannot move in a three-dimensional plane so it is moving in a plane in such a way that the ratio of its distance from a fixed point okay now this fixed point as we already know it's called the focus okay and that from a fixed line that we already know is called the directrix directrix okay don't call it as dielectric it's a directrix is a constant and this constant is called the eccentricity okay denoted by symbol e so your e is a value which is between zero and one okay if you recall for a parabola it was equal to one for a circle it is going to be zero so if somebody asks you casually also what is the eccentricity value for a circle for a circle it is zero for a parabola it is one for ellipse it is somewhere between zero to one okay so ellipse is basically a intermediate position between a circle and a parabola correct so circle right ellipse e is zero if you extend it e starts becoming more and more and more so for any e between zero and one not including zero and one it will be an ellipse by the way if you stretch it so much that one of the arms of the ellipse goes to infinity you'll be left with the parabola okay so that time e becomes one and when the other arm comes back from infinity like this it becomes a hyperbola so for hyperbola e becomes greater than one we will talk about that in our next session okay any questions here so this is one of the locus definition there is one more locus definition which is going to come your way very soon maybe towards the last part of this discussion so please note this down and let me just show you here with a quick diagram so let's say this is your directrix and this is your focus so let's say b and s okay and this is the point let's say p which is a moving point okay this is a moving point please note this point s is fixed okay this is a fixed point and this line is also fixed there's no movement in this line in the point s so this point s moves in such a way that the ratio of its distance from the point s to the line is a constant okay where this constant is between 0 to 1 so it could be a value like 0.5 0.3 but for the entire ellipse it will be fixed value right it cannot change at every instant it would be some fixed value between 0 to 1 okay so for a particular ellipse that value will be fixed okay it could be any number between 0 to 1 so when this point p moves it is going to trace a path which is going to look like this let me use a different color yeah it's going to look like somewhat like this okay so anywhere it moves anywhere it moves okay let's say I keep it here okay this distance let's say p is here divided by this distance divided by this distance okay that will also be the same value e okay so anywhere it goes on the ellipse it's going to satisfy it now it has been figured out that maintaining the same e value okay for the same set of points there could be another position of focus and directrix so let me make that with a different color let's make it with a blue color so there could be another focus and there could be another directrix so unlike a parabola unlike a parabola an ellipse has a pair of focus and directrix that means there are there is one focus one directrix which is this one okay which I have shown over here with s and d and there could be another focus and another directrix corresponding to it such that the very same point maintains the same ratio that means if I take if I take s dash p by p m dash that ratio will also be e okay so unlike a case of unlike the case of a parabola an ellipse has got two fosai and two corresponding directruses or you can say a pair of fosai and directruses okay now ideally speaking I mean if I just generalize the situation there is always two fosai and two directruses so as to be you know very generic how see in a circle okay circle also has two fosai but those two fosai overlap with the center that's why you don't see it okay and the two directruses of a circle are at infinity one is at minus infinity other is at plus infinity that is why you can't see the directruses also and we don't even draw it okay now when the stretching starts when you start stretching a circle to make it an ellipse the two fosai get separated from the center and the and the directruses corresponding to those two fosai which were initially at infinity and minus infinity they come at a finite distance okay and when you stretch it so much that one of the one of the branch or one of the you can say part of the ellipse breaks and goes to infinity then what happens one focus and its corresponding directrix goes off to infinity okay that is why in a parabola we only see one focus and one directrix but in reality actually there are two fosai and two directruses corresponding to it but one pair is at infinity that is why you are not able to see it you are not able to see it is it fine any questions you can say so yes yeah so ellipse is the general case of that absolutely right so circle and parabola both are extreme cases ellipse you can say is a generic case right is it fine now I would like to demonstrate this on the geojibra tool as well just to show that how on every point on the ellipse this ratio is maintained as a fixed value so let us see that so right now I'm going to write an ellipse equation most of you would be aware if you have done it in school but don't be scared that oh how come sir has written this equation we are not we are clueless about this equation don't worry I'll be you know deriving such equations a lot like these in today's session okay so if you see on your screen there is a there is a ellipse let me just make it slightly bigger okay now thanks to the geojibra software that we can write down directrix equation for this conic as well I mean they can show the directruses as you can see these two lines are the directruses okay and we can also write down or we can also show the focus of this conic okay as you can see a and b are the two fosai now what I'm going to do is I'm going to pick up a generic point on the ellipse any point let's say I pick up a point c over here right and now I'm going to find out what is the distance of c from a as you can see it is h value and h value is 3.68 as you can see on the left side of the two okay now I'm going to drop a perpendicular from c onto this line okay and I'm also going to find out what is the perpendicular distance of c from the directrix corresponding to a and as you can see it has named it as j j is 4.93 now what I'm going to do I'm going to take the ratio of h by j h divided by j okay as you can see it is a number which is 0.75 which actually is the eccentricity now I'll make this point die dance around the ellipse and please keep your eyes glued to j does this value a and you'll see that it is not going to change you see that c is moving c is moving as you can see c is moving c is taking a tour around this ellipse and you can see this h by j is a number which is 0.75 it is not budging from there correct and not only that for the very same point now I'm going to take its distance from the other focus let's say this guy k okay and I'm also going to I'm also going to take this distance c e it'll call it l yeah okay now let's see k by l okay so k divided by l as you can see it is also 0.75 so b and a the values that you see on the left hand side both are 0.75 and keep your eyes glued to them they're not going to change as the point moves do you see that 0.75 0.75 both have the values fixed is it right so this is what the locus definition okay of the ellipses but this is one of the locus definition there's one more locus definition which will come your way little later in the session okay any questions with respect to this demonstration please let me know all good okay great now let's learn certain terms certain terms associated with the parabola so we had many terms learned in our sorry ellipse parabola chapter so some terms with respect to parabola also sorry ellipse also we are going to learn okay so terms associated with associated with ellipse okay by the way give me some examples where you have seen elliptical curves where have you seen ellipse I'm sure you're going you you have seen ellipses at so many places right Kepler's law exactly okay so motion of celestial bodies in fact the Somersault model also says that the electron moves around the nucleus forming an elliptical path okay whereas any other examples okay many of the bridges and arcs they are made up of elliptical they are of elliptical path actually yeah the base there's a you know baseball what do you call that not it's not a baseball it's actually you can say baseball also but American football they call it that ball is also somewhat elliptical in shape anyways let's look at certain terms so as you already seen that these you know these points are called the foci and these are the directices corresponding to those points here all I would like to repeat here s1 and d1 are associated to each other that means if you're referring to any points distance from s1 you should take the point distance from d1 then only that ratio will be e okay please do not interchange it don't do like let's say if I take a point p here don't say that p s1 or s1 p divided by the perpendicular distance of p from d2 that is e don't say that that is will be going to be wrong so these two you know pairs they are linked to each other so this focus and this directrix are married to each other similarly this focus and this directrix is married to each other so if you're referring many times in the question also they will say this is a focus and this is the directrix so they mean to say this is the directrix associated with that focus okay so we already know that these two are called the focus so this is focus one focus two there's nothing like focus one focus two this is this two foresight and these two are directices you have already you know seen them during the locus definition okay apart from that please note that the line segment which connects the two focus okay this line segment this line in fact this line is called the major axis okay this line is called as the major axis we'll discuss the reason why they are called major axis okay and the midpoint of the two foresight is called the center of the ellipse this is called the center of the ellipse by the way the major axis is always perpendicular to the two directrises okay and it connects the two foresight and the center a line which is perpendicular to the major axis and passing through the center okay this line is called as the minor axis this line is called as the minor axis a line segment connecting any two points let's say I connect any two points that is called a chord this is called a chord okay if the chord happens to pass through the focus then it will be called as a focal chord so let's say I you know make one of the focal chords so this is a focal chord this is a focal chord a line which is or you can say a line connecting or a line segment connecting two points on the ellipse and perpendicular to the major axis that is called a double ordinate so let me show a double ordinate so this is an example of this is an example of a double ordinate okay this is a double ordinate this line is called the double ordinate you also had a double ordinate in a parabola so nothing new here only thing new here here is that in place of an axis now there are two axes okay one is the major axis and one is the minor axis that is the only change of course I forgot to tell you one thing that the major axis cuts the ellipse at two points which are called the vertices so these two points are called the vertices note that these two points let me show you here these two points they are not called the vertices okay they are not called the vertices so these two points are called the vertices and last but not the least a double ordinate which passes through the focus a double ordinate which passes through the focus or a focal chord which is also a double ordinate that is called the lattice rectum okay so in case of an ellipse there are two lateral recta the plural for lattice rectum is lateral recta okay so there are two lateral recta for ellipse so in a parabola there was only one lateral rectum but here there will be two lateral recta of course both the lateral recta would be of equal size okay same size we will discuss about the length of it also in some time so let's quickly recap what are the terms associated as I already told you there are two four psi and two corresponding directories then a line connecting the two four psi along with the you know center that line is called the major axis major axis is perpendicular to both the directories a line passing through the center and perpendicular to the major axis is called the minor axis a line segment connecting any two points on the ellipse is called a chord if the chord passes through the focus it will be called as a focal chord a line or you can say a chord which is perpendicular to the major axis that is called double ordinate and if a double ordinate passes through the focus it is called lattice rectum so there are two lateral recta in case of an ellipse any doubt any questions to respect to this okay so having learned this you know names and terms let us now move towards standard forms of an ellipse or standard forms of ellipses so what are standard forms first let me you know clarify this we had already done standard form in case of a circle so in case of a circle the standard forms are those forms where the center of the circle is passing through or center of the circle is at origin okay that was a standard form in case of a parabola the standard forms where where the vertices are at origin where the vertices vertex of the parabola is at origin and not only that the axis is along one of the coordinate axes so either along x axis or either along the y axis so those were called as the standard forms in case of a parabola so in case of an ellipse the standard forms are those forms please note down where the center is at origin the major minor axis major and minor axis minor axes are along the coordinate axes are around the coordinate axes okay so those cases will be called as the standard forms so let us take one of the cases like this so first let me quickly draw an ellipse by the way of all the conics ellipse is much easier to draw you just have to take a you know circle and drag it that's it okay now in order to make a standard form the center has to be at the origin so what i will do is the center i will keep it as the origin okay and as i already told you the major and the minor axes are along the coordinate axes so this is let's say your x axis okay and this is let's say your y axis x axis y axis this becomes your center origin okay now in order to derive the equations we have to give some dimensions to it so what do i do here is the first thing that i would say is let's say this is a comma zero now again here note that a represents the semi major axis length semi major axis length and that's why it is always kept as a positive quantity okay just like in case of a parabola a was the distance of the vertex to the focus right and it was kept positive in the same way here a represents the semi major axis length so it is always kept as a positive quantity so if this becomes a comma zero okay let me name it vertex v1 then the vertex v2 will automatically become minus a comma zero okay now in order to satisfy our locus condition in order to utilize utilize will be the right word here in order to utilize our locus definition we need to have a focus and we need to place a directrix okay so what i'll do is i'll call the this focus as c comma zero okay let me call it as s1 so i will have to call this focus as minus c comma zero let's say s2 and let me make a directrix corresponding to s1 over here let's say this is your directrix okay d1 and this is your directrix d2 this is your directrix d2 now i will again choose some coordinates over here i'll call this point as d comma zero and i will call this point as minus d comma zero let's name them also so that we can refer to them also this is minus d comma zero okay let's call this as n1 let's call this as n2 okay now as you can see i have overloaded this with lot of variables a c d and there's already a e also sitting what is e that in the locus definition there was an eccentricity also right so if i take any point let's say i take a point p okay then the distance of p from s1 divided by distance of p from the directrix corresponding to that that is e so too many you know i can say variables or alphabets have been introduced so what i'll do here for our benefit we will reduce the number of variables so what i will do i will instead of c and d i will try to use a and e only so i will make some arrangements to replace c and d in terms of a and d okay so lesser variables you know hassle free right so there's not much of a problem so we will write c and d in terms of a and e so we will only deal with any no c no d okay but how should i remove it so in order to remove this i'm going to use again the locus definition let me raise this this is not not used to me right now okay we'll come back to it so we'll use the locus definition applied to v1 okay so v1 is also a point this point is also a part of the ellipse so it must satisfy this locus definition okay so every point on the ellipse must meet this locus definition isn't it so can i say the distance of v1 from s1 divided by the distance of v1 from the directrix which is v1 n1 this should also be e everybody agrees with it everyone here in the session do you all agree with the statement of mine correct great now can i also say that the distance of v1 from s2 okay so same guy v1 from s2 divided by distance of v1 from the other directrix which happens to be v1 n2 that will also be this is my second equation correct all of you are convinced of this okay so this equation let me now use my actual values v1 s1 is nothing but a minus c and v1 n1 is nothing but d minus a so this is equal to e which means a minus c is equal to a minus c is equal to d e minus a e okay similarly from this one i can say a plus c by d plus a is equal to e please check v1 s2 is a plus c and v1 n2 v1 n2 is a plus d so from origin to here it is a and from origin to this fellow is a d right a plus d so from here i get and i get the second equation a plus c is equal to d e plus a okay let's do one thing let's subtract let me call this as 3 let me call this as 4 let's do 4 minus 3 so when you do 4 minus 3 you get 2 c is equal to 2 a e that means c is equal to a e okay so as to say that i don't need a c anymore i can manage with a and e only okay so this focus is nothing but a e comma 0 and this focus is nothing but minus a e comma 0 so let me make that change right away a e comma 0 and this is going to be this is going to be minus a e comma 0 okay now we'll do one more thing we will add these two also so let us add 4 and 3 so when you add 4 and 3 you get 2 a is equal to 2 d e so from here you get d as a by e okay so now going forward i will not call this point as d comma 0 i would rather call it as a by e comma 0 and it brings in one more advantage to us since your directrix is parallel to the y axis it's parallel to the major axis remember both the directrists are parallel to the major minor axis and they both are perpendicular to the major axis again i'll repeat both the directrists are parallel to the minor axis and perpendicular to the major axis okay so this equation now becomes x equal to a by e is it fine and this equation becomes x equal to minus a by e and let me remove this d from here and write a by minus a by e okay so not only you know you should also note down that how has the focus or how is the focus related to major axis length or you can say semi major axis length and eccentricity how are the directrists equation given to you everything that i have written in this particular you know screen you should form an image of it in your mind okay because you're going to refer to it in our subsequent discussions okay so again i'll repeat the two vertices a comma 0 minus a comma 0 the two foci a comma 0 minus a comma 0 the two directrists x equal to a by e x equal to minus a by e center is at origin okay so this is one of the standard cases of the ellipse now since we have found out the the coordinates of the foci foci in fact one of the focus and its corresponding directrists equation is also known let us now apply the actual locus definition which is this fellow let us apply this fellow to get the equation of the ellipse standard ellipse so now i'm going to take a point generic point okay and i'm going to use that this distance by this distance should be equal to e so now let us write that down s 1 p by p m 1 is equal to e s 1 p can be now written as e times p m 1 and what is s 1 p it is under root of h minus a e the whole square k minus 0 k minus 0 the whole square is equal to e times now p m length p m length is the distance of h comma k from this so it will be mod of h minus a by e by under root of 1 square isn't it i hope everybody remembers the distance of a point from a line anybody has any doubt till this step any question till this step do let me know because now i'm going to simplify the questions all fine great so let's simplify this let's square both the sides so h minus a e whole square plus k square is equal to now here this e could have been introduced inside also because e is a positive quantity so if you introduce that e inside it becomes e h and this loses a e from the base and this anyways a 1 so that becomes e h minus a the whole square so on squaring this will give you something like this by the way nps raja jhe negar have you completed this chapter in school was the ellipse taken in school or they have they're not yet started done also okay then then i can afford to go a little faster also because yeah so now take this term take this term to the other side okay and take this term to the other side oh wonderful you've completed everything also okay now divide both sides divide both sides by a square 1 minus e square so when you do that you end up getting h square by a square plus k square by a square 1 minus e square equal to 1 now here we need to generalize it so if you generalize it okay replace your h with x k with y if my voice is breaking do let me know okay because i think the internet is not that great today now this fellow this fellow actually looks a little ugly to remember so what do we normally do we call this term as a b square okay so what do we do we start calling this term as a b square so we'll say let this be b square okay now many people ask me sir you are calling it as a square of something that means you're sure it is a positive term is it a positive term i'm asking you are you sure that a square 1 minus e square will be positive there's an e in the right side of where did that go in the simplification okay that e got absorbed here harshetan didn't i tell you e is a positive term so i'm introducing it within the mod got it yes so most of you are in the absolutely see everybody e is a term less than 1 correct so you can say e square is also less than 1 so 1 minus e square is positive and if you're multiplying it with a square which is anyways a positive so this will also be a positive so it is not wrong to call that as a square of some number okay so in light of that this entire expression now changes to now let me use a different development x square by a square plus y square by b square equal to 1 okay now one more question i would like to ask you here look at this relation and tell me which is more a is more or b is more that also please tell me since you have told me so much let's also figure this out which is more here a is more or b is more and why i know most of you will say a is more why any reason to support it i mean mathematically what supports the fact that a is more than b okay so we all know that e square e square is positive number right yes or no e square is a positive number so 1 minus e square will be lesser than 1 because if you're subtracting a positive quantity from 1 of course it is lesser than 1 but it is still positive okay so so 1 minus e square will be lesser than 1 okay if you multiply a square throughout right so a square 1 minus e square will be lesser than a square and this term is anyways a b square and since a and b both are positive quantity here we will say b should be less than a okay so in our equation that i have written over here i would like to add one more thing that in this equation your b is less than a or a is more than b are you getting my point so this page is completely filled up i'm going to do some analysis related to this standard form in our next page okay so if you want to copy anything down from here or note anything down from here please do you know notice down all right so i'll again repeat the same thing this expression is what we got in the previous page where we realized that b square is a square 1 minus e square okay and of course by the virtue of this fact here b is or a is more than b okay so let me draw that ellipse once again and we will do some small analysis over here the first thing that we are going to discuss over here is the length of the major axis okay so please everybody note this down the distance from one vertex to the other that is called the length of the major axis the many people are surprised this is sir isn't a isn't a major axis a line so how are you talking about length of a line because line isn't infinitely extending geometrical figure so this is how it is conventionally you know defined so if somebody says what is the length of the major axis basically is asking you the distance between the two vertices which in our case is going to be now again i'm using the word in our case in this case it's going to be to a that means if i change the case this length will also change so that is something to be you know later on for discussion we will take that up but as of now for this ellipse please note down the major axis length is to a here and because because a is more than b in fact let me complete this i'll i'll i'll talk about that this length that is the part of the minor axis which is cut a bit by the which is cut by the ellipse that part is called the length of minor axis okay now as per our equation what is the coordinates of b1 and b2 that means what will be the what will be the points where the y-axis cuts this ellipse so for that you put y as 0 you would realize that your x values come out to be in fact sorry x as 0 you're finding the points b1 b2 so you have to put x as 0 because it is on the y-axis so when you do that you end up getting y as plus minus b which signifies the fact that this point is 0 b and this point is 0 minus b which you have already learned in your school also so the length b1 b2 is going to be 2b units now since since 2a is more than 2b that is why this particular axis was called as the major axis okay and this axis was actually called as the minor axis because of that so there was a reason for naming them major and minor okay next thing that I would like you to notice down here is that the eccentricity that you would get in terms of major and minor axis can be obtained from this equation itself a square is equal to sorry b square is equal to a square 1 minus e square so from there let me just write that down first so b square is equal to a square 1 minus e square so from here I can say b square by a square is 1 minus e square so e square is 1 minus b square by a square so e is under root of this okay now please make a note here this is a very you can say a generic case which is true for this standard ellipse but there is a universal formula for this universal formula for eccentricity is it's under root of 1 minus the semi minor axis length square upon the semi major axis length square please note this down this is true for any ellipse any ellipse that is why I've called it as a universal formula apply to any ellipse okay so eccentricity of an ellipse in terms of major and minor axis length is given by under root of 1 minus semi minor square by semi major square now this is such a beautiful formula that even if you try even if you try to misuse it you cannot because many people stop the positions of a and b by mistake okay but if you do that mistake the formula will automatically flash out that you are doing some error because 1 minus this quantity will become more than 1 in that case so under root of that will actually be like under root of a negative thing so even if you stop the position of a and b the formula is the formula will not let you do that okay so even if you want to make a mistake you cannot so please note this down please note this down okay the third thing that we are going to discuss here is the distance between the distance between the foci distance between the foci we have already discussed that will be 2 ae okay by the way there is a universal formula for this it is always eccentricity times the length of the major axis okay so the distance between the two foci is always for any ellipse is always eccentricity times the length of the major axis now let us go ahead and find the lens of the lateral recta lateral recta so let me just draw one of the lateral recta's one of the lateral recta okay so let's say l r what is this length l r so um may i ask you to spend some time and tell me what is the length l r everybody please do it on your respective notebooks and tell me the length on the chat box i know you've already done it in school but still a quick exercise for you all okay so in light of yeah so let's let's call this point as ae comma k because it is going to be lying on the lateral recta and the lateral recta is perpendicular to the x-axis in our case okay this happens to be the x-axis so it is going to be having the same x-coordinate throughout so if you put ae in our equation of the ellipse so in this equation i'm going to put x as put x as ae okay and y as k so when you do that this will become a square e square by a square plus k square by b square equal to 1 so k square by b square equal to 1 minus e square and we already know 1 minus e square from here 1 minus e square is b square upon a square right so i can write this as b square upon a square also so this gives me k square as b to the power 4 by a square which means k is plus minus b square by a in other words if i call this as b square by a the r coordinate will be ae comma minus b square by so this is b square by a so this point will be ae comma minus b square by okay so the length l r will be nothing but 2 b square by a units please note this down and i'll give you a universal formula also it is twice of semi minor square upon semi major this is a universal formula for length of the lateral recta so length of the lateral recta universal formula that means it's going to be applicable to any type of a ellipse clear everybody so i hope this analysis was you know worth noting down and we'll be using all these things in our problem solving so if you have any questions any concerns or is my voice breaking let me switch off my camera yeah if at all you have any concerns any doubt at this stage please do highlight it shall we move on to the second standard case so standard case two so this is a case as i told you the standard cases are those cases where the center is at origin and the major and the minor axis are oriented along the coordinate axis so in this case what i'm going to do i'm going to keep my major axis along the y axis okay so i'm going to keep the major axis along the y axis which means my force i should lie on the y axis so let me draw it let's use the different color okay so let's say this is my foresight s1 and s2 now what i'm going to do is i'm going to i'm going to give some coordinates to this major axis this is your minor axis i'm going to call this point as 0 comma b and this point as 0 comma minus b okay so this s1 becomes 0 comma b e and s2 becomes 0 comma minus b and let's let's make the two directories as the okay i'm making a small one not a big one okay and this direct x equation i will also mention next to it this is y equal to b by y equal to minus b by okay now again using the locus condition let's say i take any generic point h comma k so using the fact that s1 p by p m1 sorry for that crooked line let me just make it again so again using the fact that s1 p by p m1 is going to be e let us now derive the equation of this ellipse as well so s1 p is going to be e times p m1 so s1 p will be nothing but under root of h minus 0 the whole square k minus b e the whole square okay and p m1 will be mod of please note this on mod of k minus b by e so this length is nothing but k minus b by e mod yeah if you want you can write under root of 1 square also not a issue now what i'll do is i will introduce this e inside just like how i did it in the first standard case and i'm going to square both the sides so i'm just skipping some steps i don't need a bracket yeah so even introduce inside will become e k minus b and square so this will become x square k square minus or you can say plus b square e square minus 2 k b this also e square k square plus b square minus 2 k b 2 k b minus 2 k b will get cancelled off take this term to the left side and take this term to the right side so that is going to give you something like this now divide both sides by b square 1 minus e square so that's going to give you this let's generalize it by replacing our h with x and k with y so this becomes x square now everybody please note this down if you want me to scroll up or down please do let me know first note this down and then we will talk something related to this equation now let us say this point where it is cutting the x axis this is a comma 0 and this is minus a comma 0 so as this point satisfies a comma 0 or minus a comma 0 any of the two you can take they satisfy it so i can put x as a and y as 0 okay so that will give you something like this a square by b square 1 minus e square plus 0 equal to 1 which means i can write b square 1 minus e square as a a square as well so i'm going to give it a more better shape which will help you to remember this result okay now what do you see might surprise you you'll feel like your same equation back but yes there is a very very big change here in this case your a square is b square 1 minus e square please note this down and because of that a will be lesser than b so this is one of the vital differences between this form and the previous form so the equation looks exactly the same but in the first form a was more than b and the relationship between a and b was b square was a square 1 minus e square correct and in our case right now which you see on your screen here the relationship is a square is b square 1 minus e square in other words a is less than b in this case a is less than b in this case okay so we basically distinguish between the two ellipses by seeing their a and b values if a is more than b it is the case of the first standard form and just to name it mean i call it as a horizontal form because it is an egg flying horizontally okay and when you're b is more than a it is a vertical form of the ellipse or you can some people call it as an egg standing on the nose okay so a and b makes a difference here in the two forms however please let me tell you both of them have their center at origin only and in one case your major axis is along x axis in the other case your major axis is along the y axis is it fine any questions any concerns and also i would request you to note down the vertices coordinates the foci coordinates the equation of the directories everything over here from this diagram anywhere you want me to scroll up and down do let me know okay couple of analysis that i would like to do here number one the length of major axis in this case will be two b units okay the length of minor axis in this case will be two a units from this formula assenticity will come out to be under root of one minus a square by b square okay but remember the universal formula still remains the same the universal formula that is under root one minus semi minus square by semi major square that still holds true for this case as well okay and the length of the latter sector which you can always you know find it out in this case will become two a square by b units okay again the universal formula still remains true what are the universal formula twice the square of the semi minor by length of the semi major that still remains valid is it fine so please make a note of this and in the next slide i will do a comparison of the two standard cases so that you are very very clear when to use what when to use what i noted the zone should we go to the next slide then okay so let me do a comparative study of both the ellipses let me do a comparative study of both ellipses here is a division which i'm going to make so the first ellipse that we had taken today in this case your a was more than b okay and the second ellipse which are taken or the second standard form where your b is more than a so see the equations more many of the people say that the equations is same so how do we distinguish it so this the distinguishing factor here is your a and b magnitudes if a is more than b it is a case of a horizontal ellipse or case one of our standard form if b is more than a so these values will be given to you in the exam don't worry right now i'm writing a and b that doesn't mean the examiner will also give it to you in a and b of course they will provide some values for example they'll say x square by nine plus y square by four equal to one so you know a is three and b is two so a is more than b so you will be following everything which i write under the left column if your b is more than a let's say x square by four plus y square by nine is equal to one so there your a is two b is three so b is more than a there so you'll follow everything which is which is what i'm going to write another right column okay just a rough diagram for both of them so that you are clear about their images as well so in such cases your ellipse looks like egg kept on the base here it looks like an egg kept on the nose something like this so horizontal and vertical and let's also write down the distinct distinction between them with respect to their center eccentricity vertices fosai equation of directices directices length of the lattice system length of the major axis in fact equation of the major axis will also take up equation of major axis length of minor axis equation of minor axis i think full bio data we have created for this two standard cases of ellipse okay now this chart that i'm going to write over here this chart should be committed to your mind and of course you will be only able to commit this to your mind once you practice now all the questions given to you in the dbps which i'll be giving to you after the class so please note down center for both the ellipses is zero zero eccentricity for this ellipse is one minus b square by a square but for this ellipse it is one minus a square by b square as we have already discussed vertices a comma zero minus a comma zero here it is zero comma b and zero comma minus b fosai here is a e comma zero and minus a comma zero here it is zero comma b e and zero comma minus b e here directices are x equal to a by e and x equal to minus a by e here the directices are y equal to b by e and y equal to minus b by e latter-sector length here is two b square by a in this case it is two a square by b length of the major axis is two a here length of the major axis is two b length of the minor axis here is two b here it is two a equation of the major axis is your x axis equation of the major axis here is your y axis equation of the minor axis is your y axis an equation of the minor axis here is x axis. So please, you know, make a note of this chart. Is it fine? Any questions? Okay. Now let's take few questions right from your locus definition. Let's take few questions right from the locus definition. Okay, let's start with this question. Find the equation of an ellipse whose focus is minus one comma one. Ascenticity is half and directrix is x minus y plus three equal to zero. Give me a response in the chat box or are done is also good enough. This is just to check whether you have understood the locus definition. Done nearly done very good. So this is going to look like this. I'm just drawing the directrix first. So this is your directrix. Okay, minus one comma one is somewhere over here. So your ellipse is going to look like this. I'm just drawing a very small one. Yeah, so this is your directrix x minus y plus three equal to zero. Okay, and this is your minus one comma one. Now, in order to find out the equation of this ellipse, I'm going to again use my locus condition that the distance of P from S divided by distance of P from the directrix. So SP by PM will be equal to E will be equal to E in this case is half very good. Okay, let's check. So SP is under root of H plus one the whole square K minus one the whole square is equal to half PM. PM length is the distance of the point from the given directrix which is going to be H minus K plus three by root two. Am I right? Take this to root two on the other side. Take this to root two on the other side and square. Okay, let's square both the sides. So that is going to give us an eight H plus one the whole square K minus one the whole square. And on the right side, we'll end up getting H minus K plus three the whole square. Now, as you all know that I don't write all the terms I simplify focusing on what kind of a terms I will be getting. So I'll be getting a square term. So a square term will be contributed by the left side as eight at square right side as a square. So we'll end up getting a seven at square K square term. If I'm not mistaken that will also be seven K square only check it out. HK term will be contributed only by the right hand side term which is going to give you minus six HK but when it comes to the left hand side it will become six HK. Oh, sorry, two HK two H and H term here will be 16 H and from here you'll get six H. So you'll end up getting a 10 H K term will be minus 16 K and from here you'll end up getting minus six K. So that will be minus 10 K. Constants on the left will be 16 one plus one into eight, which is two into eight 16 and the right side will be nine. So that will leave you with a seven. Now you can generalize it. You can generalize it by replacing your H with an X and K with a Y. So that's going to give you a seven X square, seven Y square, two XY plus 10 X minus 10 Y plus seven equal to zero. So as you can see, the equation of the ellipse can become as ugly and complicated as what you see on your screen right now. So the ones which you saw, those were very special cases, those were standard cases. Please don't forget that. Why I'm emphasizing on this fact is because the moment I say an ellipse, the image that people get in their mind is X square by a square plus Y square by B square equal to one. That's a very special form. Please understand it. So an ellipse equation can have the thorough of other terms like XY term can also be there. X can be also there. Y can also be there. Are you getting my point? So don't be like so rigid in your understanding of ellipse that all the ellipses in this world are standard ellipse. Now, this is no mistake of yours because in CVSC we only talk about standard forms, right? There's no other form discussed in CVSC at all, but that's not going to be the case for competitive exams. So for competitive exams, standard forms will be very rare. Mostly we'll be dealing with non-standard forms. Okay. So please understand that you may end up getting a case like this also where your ellipse is an oblique one. Oblique one means making whose axes are making certain angle with the coordinate axes. Is this fine? Any question, any concerns with this problem itself? Okay. Should we go to another question? So we'll take this question. Find the equation of the ellipse referred to its axes as the axes of coordinates. That means basically they're saying that the axes of the ellipse are along the coordinate axes. Okay. Now, there are four sub-questions in it. I will not ask you to do everything. Maybe we can touch upon the first one. Okay. Let's take third one and fourth one. First one, I would need the answer. Table. Yeah, why not? I should not show. I'll show you the table after this. First one. Are you all ready? First one. Okay. So in the first one, in the first one, I'm assuming that the major axis is along the X axis and the minor axis is along the Y axis. So in other words, I'm assuming that my A is more than B. Okay. Now again, you can also assume B is more than A. And accordingly, the result is going to differ. Okay. So any of the two forms, if you give me, I'm happy with that. Now, first thing that I can use here is that the ellipse must pass through minus three comma one. So minus three comma one would satisfy this equation. So can I say nine by A squared plus one by B squared is equal to one. Okay. That's number one. And second thing I know that it's a centricity is under root of two by five. So on a centricity value is one minus B square by a square. So if I'm assuming A is more than B, I have to use one minus B square by a square under root, which means B square by a square is three by five. Correct me if I'm wrong. Which means B square is three a square by five. Let's call this as the second equation. Let's use this in the first one. So put two in one. So when you do that, you get two by nine by a square one by B square will be five by three a square equal to one, which means a square is going to be nine plus five by three, which is nothing but 32 by three. Okay. And B square will be 32 by five. Because B square is three by five a square. Yes or no. In other words, your equation of the ellipse will become x square by a square. Y square by B square is equal to one, which means three x square plus five y square is going to be 32. Is it fine. Any questions, anybody who got this answer September answers seems to be different than mine. You took the other case. Okay. All right. Next one. Third one, find the equation of an ellipse which passes through these two points. Okay. So, let me write it here. Third one, third one, if you're ready with your answer, please give it on the chat box. Your response on the chat box. I'm again assuming is more than. Do I need to show the question because I think this two points are required two comma two and one comma four. Again, same thing two comma two should satisfy it. So this is number one equation. One comma four should satisfy it. Okay, this is the second equation. Just subtract both of them. If you subtract both of them, we'll end up getting three by a square minus 12 by B square is equal to zero. In other words, in other words, B square is for a square. Okay. Put this in your, any one of the two, let's say I put it here. So four by a square and four by B square B square is again four by a square. So you end up getting five by a square is one. That means a square is five and B square is going to be 20. Right. So that leaves you with the equation as a square by five plus Y square by 20. What is wrong with this? Y square by 20 equal to one. If you want to get further simplified by writing it like this one and the same thing. Excellent enough. Very good. That's typical school level stuff. Okay, these are the questions which you can expect to come in your school exam. Next, our four psi is at plus minus two comma zero and lattice rectum is six. So fourth case, I will write it over here for psi plus minus two comma zero and lattice rectum length is six years. Yeah, please do it and give me a response in the chat box. Still doing. Okay, so this is a case where your A is two and two B square by A is six, which means B square is equal to three. Now what I'm going to do is I'm going to square the first equation. So from one I can say a square E square is four. Correct. Now, please understand here. E square is one minus B square by a square. So that means that means a square minus B square is given to you as four, which means a square minus three is four, which gives you a quadratic in a. I think this is easily factorizable as a minus four a plus one. Remember a can be four a cannot be minus one as I already told you. This is accepted. Okay, as I already told you a is the distance or is the length of the semi major axis cannot be negative. Okay, so if a is for so if a is for what is your B. B square is going to be 12. Correct. So you end up getting the equation as x square by a square y square by B square equal to one. Okay. Yes, you can easily write it as three x square plus four y square is equal to 48 as well. Is it like any questions, any questions. So for your school, this would be like a four marker, something like that. Now, Harshita wanted to copy something from the slide where I had given a comparison table that which part you wanted to copy second half second half is this one. Yeah. So based on this table also will take few questions. Let me take a look. Okay, let's take a very, very simple question. Find the length of the major axis, minor axis coordinates of foresight vertices, eccentricity of this ellipse. So they have asked you major, minor, foresight vertices, eccentricity and directories equation for this ellipse. Again, as a typical school level question. Let's do this. Please write all these answers in one go.  Okay, Vishal, very good. Where are the directories? I need the last line also find the equation of the directories. Okay, so Harshita has all together different set of answers. Okay. Okay. Nice. Anybody else. Okay, so some of you have responded. And few of your answers are common, but few of you are deviating from those answers as well. Okay, let's check who is right, who is wrong. We'll come to know in some time. So, first of all, this equation, let us write it in a standard form. Okay, actually, it is a standard form. Okay, so all of you appreciate that this is a case of a standard form. Okay, and looking at the values, you'll come to know that B is more than a. So in this case, your B is more than a. So it's a case of a vertical ellipse. Okay, so a rough diagram that I am going to show you. It's a case of a vertical ellipse like this. Okay, now that chart that I gave you. Okay, just close your eyes and remember that chart. Everything will be, you know, clear from there. So the first part length of major axis. That is going to be to be in this case. These remember to be two root three. So first answer is two root three. Okay. Length of minor axis to a in this case, it'll be two root two. Now, four side coordinates require ascenticity. So first we'll have to find ascenticity in this case. Ascenticity would be under root of one minus a square by B square, which is going to be one upon root three. Now comes the coordinate of the four side. Zero comma B E. Zero comma B E and zero comma minus B E. Sorry. Zero comma B zero comma minus B E. Yeah, correct. So zero comma one and zero comma minus one. Next vertices, vertices is zero comma B zero comma minus B. This is your vertices. Sorry. Yeah, vertices. Next is your direct cases. Direct cases will be X equal to no sorry Y equal to B by E. B by E will give you three. And Y equal to minus B by that will give you minus three. Okay. So this is the right answer to this question. Sutton has got everything perfectly correct. Well done. Why first minus root five and root three and all. We shall also some mistakes. I can see in your answer. Is it fine? Yes. Mainly the people who have, you know, got wrong answers is because it is as two and BS three, not as root two and root three. Is it fine? Any questions? Okay. Let's take some. Basic geometry related question. If the distance between the direct cases is twice the distance between the four side. Find the simplicity. Very simple question should not take you more than one minute. The distance between the direct cases is twice the distance between the four side. Find the assenticity. Okay. It's quick. This is not a difficult question. Okay. We shall. Oh, why three people, three different answers. How come? Okay. What are the distance between the direct cases to. A by right. This is twice the distance between the four side. That is to a. Okay. To a to a gone. So you end up getting one is equal to three square. So he's one by root three. Why so many different different answers I'm getting. Check. Check. Check. Okay. The next thing that we are going to talk about is the shifted forms. Of ellipses. Okay. Also called as the generalized form. Please don't confuse this. General form is something which is a x square plus b y square plus two h x y. Plus two g x plus two f five plus equal to zero, which we are not going to discuss at least for our school level. Of course, that is a part and parcel of your preparation. So for your day point of view or your competitive level point of view, we are going to talk about that. That is called the general form, not the generalized form. The general form is something which is a x square plus b y square plus two h x y plus two g x plus two f five plus two z. So we are going to talk about the general form, not the generalized form. The generalized form is the shifted form of the standard cases. Okay. So let me write down. In this form your center. Your center is going to be shifted. Okay. Center is going to be shifted to let's say alpha beta. Some point, alpha beta. But the. Major in the minor axis. Major. major slash minor axis will still remain parallel to the coordinate axis parallel to the coordinate axis. So let's begin with one of the standard cases that we have already seen x square by a square plus y square by b square equal to one. So this is one of the standard cases of an ellipse and I've assumed here is more than b. So that assumption doesn't make a difference to the concept. Now what I'm going to do with this ellipse, I'm going to shift this ellipse in such a way that I don't change its dimension at all but now its center has come to alpha comma beta. This is your new major axis and new major axis and the old major axis are still parallel and this is your new minor axis. Your new minor axis and your old minor axis are still parallel. So what will happen to the equation of this ellipse? Can anybody tell me? So first tell me where has the origin got shifted such that your ellipse, the yellow one becomes the blue one. How will you shift the origin so that the yellow ellipse becomes the blue one? Yes, last time some of you had done that mistake. Good that now you're not making that mistake. So remember the origin actually got shifted to minus alpha minus beta. Then only your yellow ellipse will appear to be the blue one. So in such cases the equation becomes x minus alpha the whole square by square y minus beta the whole square by b square equal to one. Please note the dimensions that is the length of the major axis, the length of the minor axis, the length of the lattice rectum or retro recta, they are not going to get changed because they are not going to be able to shift. Now we had already done a very extensive exercise on this when we were dealing with parabola. So the same concept is now going to be applied if you are asked a question on finding the center, length, foci, directices, equations. So role change mechanism that is what we had discussed in that role change method is what is again going to be applicable over here. So this doesn't require a separate discussion. So we will directly jump to a question. Let's take a shifted version of a standard case where a standard case has been shifted and let's write down the critical points and equations for the same. Oh, was it you don't remember? Okay, so should we go to a question now? All of you are ready? So all concepts that you have learned, all the methods that you have learned in a parabola would be applied to this case as well. Okay, let's do a question. Find number one, center, vertices, eccentricity, foci, equation of directices, length of ladder spectrum, length of length of in fact length and equation of major axis and length and equation of the minor axis. Yeah, four plus two the whole square by four plus y minus three the whole square by nine equal to one. Give me all the eight sub parts in one go. Okay, should we start the discussion or you want some more time? Almost done. Okay. Okay, so we are going to compare this equation with our standard case. Of course, I'm going to use capital X, okay, and capital Y.  A is two, B is three. Okay. Now just recall, just close your eyes and recall a case of a standard form of a ellipse where a is lesser than B or B is more than A. So as to say, so it's a case of a vertical ellipse, right? So in your mind, vertical ellipse, chart should be ready. Now ask yourself for a vertical ellipse, standard vertical ellipse, where was the center? 00, right? So instead of 00, just do one more extra labor here that put capital X as zero and capital Y as zero. Now just do a role change. So your X will become negative two, Y will become three. So negative two comma three will become your center, correct? Any questions? Next, second part, vertices. Vertices again, if you close your eyes, you'll realize that vertices used to be at zero comma B, zero comma minus B. So I will write it like this, zero comma B and zero comma minus B. Okay, I'll just do a role change. Zero, this also zero minus B. Okay.  Am I right? Those who are doing it, please also match your answer, right? So second part also done. Third part, ascenticity. Ascenticity has only to do with your A and B dimension. Ascenticity is not going to be affected by shifting off origin. So as per the dimensions, it will be under root of one minus A square by B square. Now even if you want to do B square by A square, automatically you'll come to know that there's an error happening. So you will, you can automatically know that error will happen in that case. So that's going to be root five by three. Am I right? Ascenticity is root five by three. If you do the other way around, it will give you one minus nine by four, which is going to be a negative quantity under root. Okay. Next, four side coordinate. So four side for this case used to be zero comma BE, zero comma BE, and zero comma minus BE. So zero BE, now remember BE will be root five. BE will be root five. Similarly, it will be X plus two equal to zero and Y minus three equal to minus root five. So your coordinate will be minus two comma three plus root five and minus two comma three minus root five. Now many people do this mistake and maybe you will also end up doing the same mistake sometime or the other if you don't pay attention right now. Many people think that, okay, after getting this coordinate, you just have to change the sign of the Y coordinate. Please note that three plus root five and three minus root five, they are not negatives of each other. Okay. So some people have this habit. Okay. Once one four side they will get and the second four side, they will change the sign of, you know, coordinate in this case, but that doesn't work like that in case of shifted forms. That works in case of a standard form, but not in a shifted form. So please don't do that mistake. Next, fifth one, equation of the directices. Equation of the directices for this case used to be Y equal to BE and Y equal to minus BE, isn't it? So your Y is Y minus three, B by E. B is a three. Three divided by this will be nine by root five. Oh yeah, nine by root five. And this will be Y minus three as minus nine by root five. So the two equations will become Y equal to three plus nine by root five and Y equal to three minus nine by root five. Here also don't make that mistake. Don't think like it's going to follow the same way as what the standard form followed. Here there's difference in the right side term. They're not negatives of each other. Please note that. So three plus nine by root five, three minus nine by root five, they are not negatives of each other. So length of the ladder system. Length of the ladder system in this case, I think it was the seventh one or something. No, sixth one. Yeah. So ladder system length in this case will be two A square by B. Right. So it is two times two square by three, which is eight by three units. Check it out. Okay. Length and equation of the major axis. So major axis length in this case is two B units. So length is two into three. So six units. And equation was what equation was major axis equation was capital X equal to zero. So in this case, this becomes your equation. Next, length of the minor axis will be two A units. In this case is two certainly four units and equation will be capital Y equal to zero, which is nothing but Y minus three equal to zero. How many of you got all these eight correct? No mistakes. Very good Satyam. Is it fine? Any questions, any concerns if you have? Do let me know. You got six correct. Too wrong. Very went wrong, I should be a major mind. Is it fine? Maybe we can take one more question. But this time, the question that I'll be taking is where the question setter has actually simplified the expression. So I would like you to attempt here question number four. Let's do four and five. So for the following ellipses, one, two and three are standard cases, which is very easy, but four and five are shifted cases. Many people ask me, sir, looking at the equation in this form, how do you know it is a shifted one and not a oblique one? It could be an ellipse like this also. How do I know it is not this case? Can anybody tell me that? I can surely say that these two are shifted cases. That means the center may not be at origin, but their major and minor axis are still parallel to the coordinate axis. Right, Harshita? Absolutely. So if it was an oblique case, you would definitely see an x, y term in the expression. Since both of them do not contain x, y term and they're not like your first, second and third case, the only option left is it is a shifted case. Okay, right. So for the fourth one, please do the needful. Very good, Vishal. Okay, let's discuss it. So here in this case, we need to complete the square. Okay, so as I already did in our case of a parabola, so we will complete the square by first, you know, separating them into their respective, you know, variable camps. So x cap, y cap and constant, I'll keep it separate. So here if I take five common, I get something like this. Okay, now in order to complete a square, I need to introduce a plus four over here. But because of this plus four, 20 has to be introduced on the right side. Okay, so now it becomes four x square, five times y minus two, the whole square is equal to 20 divided by 20. There you go. Yeah, this is what you see. Now, here a is root five, b is two, of course, a is more than b. Okay, so it's a case of a shifting of a horizontal type of an ellipse. And capital X role is being played by small x only and capital Y role is being played by y minus two. Okay, now let's see what does the question setter ask us. He's asking us first of all, the length of the major axis. So first part of the question, length of major axis that is going to be two cap, two a, that's nothing but two root five. Second length of minor axis that is going to be two into two, which is four units. Let's not forget writing units. Okay, so length of major axis, minor axis is clear. Coordinates of the four sides. So before coordinates of the four sides, you would require a centricity. So without a sense, it is a centricity, we cannot comment on the four side coordinates. So a centricity in this case will be one minus b square by a square, which is under root of one minus four by five, which is one upon root five. Next is four side coordinates. Remember, in such cases, four side coordinates will be a comma zero and minus a comma zero, a comma zero and minus a comma zero. So your capital X and small x are same a e a e will be one. Okay, and this will be y minus two equal to zero. So one comma two will be your one four side. Similarly, this will be minus one. Okay, so minus one comma two is the other four side. Next, they also ask us for the vertices. Okay, so vertices is vertices is let me write it down here. Vertices is a comma zero. Why not write on top because I am not having any space. Yeah, let's write here. So a comma zero, a comma zero, and minus a comma zero. Okay, these are your vertices. So a is root five. So you get root five comma two and minus root five comma two. These are your vertices. I hope all of you have got this right. Any questions, any concerns, please immediately highlight. Very good. So let's take one last one before we go for a break. Find the equation of the ellipse whose four psi are at two comma three and minus two comma three and whose semi minor axis is root five. Please give me a response for this on the chat box. Very good, Satyam. Okay, shall we discuss it out? See here, if you just make your coordinate axis, two comma three and two comma minus three. So two comma three is here. Okay, minus two comma three is here. Okay, so this is our four sides. So our ellipse is going to look like this. Okay, and we also know that the major axis is along the major axis length is root five. Okay, sorry, minor axis is root five, minor axis is root five. Now, all of you please pay attention. So this is a case where clearly a horizontal ellipse had been shifted. Right? So I can say this root five that has been given to you is actually your b value. Semi minor axis is going to be your b. And two a e will be the distance between the two four sides. So let's say I call it as S1, S2. So S1, S2 will be two a e. And that is clearly four units. So you have an information about a e value and b value. So from these two information, I need to get a lot of things. First of all, I need to get my a value. How will I get? Very simple. Square it. Correct. And in place of e square, right, one minus b square by a square that is equal to five, which means a square minus b square is given to you as four, which means a square minus five is four, which means a square is nine. So a square is nine, b is root five. Now I would clearly require the center of the ellipse as well. Center is, as you all know, is going to be the midpoint of S1, S2. So center will be nothing but minus two plus two by zero and three plus three by, sorry, minus two plus two by two and three plus three by two. Sorry for writing zero. Actually I calculated zero in my mind and I wrote that. Yeah. So it's at zero comma three. So in light of all these information, now you can frame that the equation of your desired ellipse will be X minus zero the whole square, which is as good as x square by a square y minus c the whole square by b square is equal to one. Absolutely right, Vishal. Harshita also. Satyam, did you simplify it?  Okay. Okay. Okay. Fine. So here we'll go for a break. On the other side of the break, we'll discuss a lot of other things like parametric equation of an ellipse. We'll also talk about the distance of a point from the focus, which is called the focal radii. And we will see an alternate definition of an ellipse as well. Okay. So let's take a break right now. So as of now, the time is six or 12 or sorry, six, 12, and we will have a 15 minutes break. Okay. See you at six, 27 sharp. So next thing that we are going to talk about ellipse is the parametric form of the ellipse. This is going to be especially very useful when we start taking the engineering content of this chapter. So parametric form is going to be used in solving a lot of questions. In fact, locus-based questions are based on parametric form. We will be learning later on the equation of tangents, normals, etc. So we prefer using the parametric forms of those because that really makes our life very simple. So we have already seen the parametric form for a circle. Okay. So even for the cases of standard ellipses, there are certain preferred parametric forms. So we will be introducing those parametric forms to you. So let me begin with our standard case x square by a square plus y square by b square equal to one. Now a could be more than b, b could be more than a, doesn't matter. This doesn't influence your parametric form. So this is actually the Cartesian form. We all know that. So this form is called the Cartesian form of the equation. So if you write the same equation like this, x is equal to a cos theta and y equal to b sin theta or vice versa, x equal to a sin theta, y equal to b cos theta. So as you all know that there is no fixed parametric form. So parametric forms, everybody can write their own parametric forms also. But this is the most preferred one where theta here is basically a parameter which is also called as the eccentric angle of that point. It is also known as, later on we will talk about it in more detail. This is also known as the eccentric angle of the point. Now what is this point? So basically when, what is the use of a parametric form? Parametric form is used to signify a point on that particular curve, isn't it? So let us say this is our standard ellipse. Okay, this is our coordinate axis. And if I take a point like this, okay, and I call it as a cos theta, comma b sin theta. That means I'm using these two as my x and y. Then please note theta here is called the eccentric angle of that point. That means every point on this particular ellipse will have different, different eccentric angles. For example, if you take this point, for this point, the eccentric angle will be pi by two, sorry, zero. Okay, because I will give you a comma zero, right? This point eccentric angle will be pi by two. So here your theta value is going to be zero. Here your theta angle is going to be, eccentric angle is going to be pi by two and so on. Okay, so theta is called the eccentric angle. Please note this down. Theta is called the eccentric angle. Now look at this figure and tell me, look at this figure and tell me which angle is theta here? Use a, b, p, c, etc. to tell me which angle is theta here actually. If I talk about this diagram, it is fine. Okay, we have understood this is called the eccentric angle of that point. But what actually is represented by theta? Which angle here is theta? Can anybody tell me that? Just write the name of the angle. A, C, B. A, C, B is 90 degrees, no? Vian, how can it be theta? This point p, there's a theta used, right? So which angle is theta here? That's what I'm asking you. Okay, now here, I mean, anybody who wants to take a guess also, please do so. I'm just asking your response. Need not be correct. Don't worry. So here many people believe that this angle is theta because for A it is zero and for B it is pi by two. So that makes people think that, oh, this is the theta angle. That is why Seth wrote for A it is zero and for B it is pi by two. Okay, now let me tell you this is not the theta value. Neither it is CPA. Now let me show you what is this theta over here. If somebody tells you that there is a point whose eccentric angle is theta on an ellipse. Okay, so let's say this is a point whose eccentric angle is theta. That means it is represented by this. Please note that this angle theta that you're seeing over here in the coordinates is basically obtained like this. So first make a circle with the major axis as the diameter. Okay, first make a circle with the major axis as the diameter. This circle is actually called the auxiliary circle. Okay, later on we will study more about auxiliary circle. So auxiliary circle is nothing but it's the locus of the foot of the perpendicular drawn from the focus on any tangent. Okay, to that ellipse. Again, I'll repeat it is the locus of the foot of the perpendicular dropped on any tangent to the ellipse. Okay, let me show that. It's better to show it on what is wrong with me. There was just here. Okay, so let's take a general form. Let's take a x square by nine y square by four equal to one. Okay, as you can see, this is our okay. Yeah. Now what I'm going to do is I'm going to draw a tangent at any orbit point. Let's say I take an orbit point here. Okay, and I'm going to sketch a tangent there. Okay, you see a tangent mean created. Now, let's figure out where is the focus for this. So from C, I'm going to drop a perpendicular onto the standard. So this is the foot of the perpendicular. As you can see D is the foot of the perpendicular. Okay, now all of you please pay attention. I'm not going to switch on the locus for this. So I'll be, I'll be moving my point D and just see what is the path. Sorry, I'll be moving a point A and just see the path space by D. Just see the path space by D that thick black line. Do you see that? So that basically forms, it's going very fast in this zone. Yeah, yeah. So it's basically forms a circle whose diameter is the length of the major axis. This circle is called the auxiliary circle. What is an auxiliary circle? A very circle is nothing but it's the locus of the foot of the perpendicular draw from the focus onto any tangent. Okay, we will study about it in more detail. This is a very important concept actually for from your completed level exam point of view. So this circle is called the auxiliary circle. Now, all of you please pay attention. From this point P, okay, make a vertical line like this. Okay, wherever this vertical line cuts the auxiliary circle, okay, connect it to the center, connect to the center of the syllabus. This angle is what is theta. Is it clear? And not this angle which many people wrongly believe it. This is not theta. Please note that this is wrong. Okay, so if you realize for this point, your theta will anyways be zero. Okay, and for this point, your theta will anyways be pi by two. So many people get confused because of these two values which I have written. Okay, that makes them believe that PCA is theta which is not the case. Okay, so theta is this angle and I can justify it also. See, as you have drawn a circle whose radius is A, this length is A. Okay, so this length is A, this length is going to be A cos theta. And that's the reason for your x coordinate over here, A cos theta. Okay, now for the y coordinate of this point, for the y coordinate of this point, you just need to use your equation of the ellipse, x square by a square, y square by b square equal to one, put x as A cos theta. So this will become cos square theta. So y square by b square will become sine square theta. So y value will be plus minus b sine theta, but I'm just taking the one which is given on the figure over here. Okay, so this is your eccentric angle theta. So eccentric angle is always the angle made from the auxiliary circle, okay, point connecting the auxiliary circle. Now this point q is exactly on the top of b, so exactly over b. So from p, if you extend the line to cut the auxiliary circle, from that point, the angle here is calculated as theta and not from p. Is it fine? Any questions with respect to what does this parameter actually represent in the diagram? So A cos theta and B sine theta, theta is the eccentric angle given by this angle. Clear any questions? Any questions, any concerns? Could you scroll up? Yeah, this is the part of it. Okay, great. Let me check if I have questions from parametric form. Unfortunately, questions are not there, but okay, let me just, you know, small questions. Suggest a parametric form for form for x minus 1 by whole square by 4, y plus 2 the whole square by 9 equal to 1. Okay, yes. Okay, very good. So here, see, it's just like saying x square by a square, y square by b square equal to 1, correct. So for such cases, our parametric form, suggested parametric form used to be x equal to A cos theta, y equal to b sine theta, correct. Now, here you just have to write capital X as small x minus 1, A in this case is 2, okay, and capital Y is y plus 2, b in this case is 3, okay. So one of the suggested parametric forms that you can write is x equal to 1 plus 2 cos theta and y equal to minus 2 plus 3 sine theta, okay. So this is one of the answers. I think some of you have also written that. So we'll talk more about it. See, parametric form is not a part of your school syllabus. That is why we are not, you know, going into the details of it. But definitely, life is going to be very difficult without parametric form in coordinate geometry. So we'll have to use parametric form in many of the situations. So we'll take that up once we are doing the chapter in more detail. Next concept that we are going to talk about is the focal radii or also called as the focal distances. So let us take a standard form here. So this is our standard form, okay, x square by a square, y square by b square equal to 1, okay. Now let me ask a very simple question to you. Let's say there's a point x1 comma y1, okay. What is the distances of this point from the two four sides? That means what is s1p and what is s2p? So please give me a response in terms of whatever has been provided to you in the question. So equation you already know, a, b is known, x1, y1 is the point. So give me the length s1p and s2p in terms of x1, y1, a and b. Of course, not everything will be used. You can also use e also. e can also be used. Okay, so let me just take this up, sort of into the difficult one. See, let's say this is your directrix corresponding to s1. So we already know that every point on the ellipse will satisfy the fact that s1p by pm1, that will be e. So s1p would be e pm1, okay. And we already know that this equation is x equal to a by e. So this gap, which is pm1 gap is actually a by e minus x1. So I can say this is e times a by e minus x1. In short, it becomes a minus e x1. Similarly, if I make a, let's say, directrix over here and I want to know what is s2p. So s2p is again e times pm2. Let's say I call this as m2. So s2p is e times pm2. Now, please note, please note this equation is x equal to minus a by e. So this gap will be x1 and a by e. So add x1 to a by e. So e times this, which happens to give you a plus e x1, okay. Now, what do I plan to do here by using these focal radii? If you see, if you add the two focal radii, you will end up getting a constant value actually, which is your 2a, okay. Now, this brings me to the second locus definition of an ellipse. Remember when we started our chapter, I told you there are two locus definitions. One of them, which I already told you, that is the ratio of its distance from fixed point to a fixed line is always a constant whose value is between 0 to 1. That is one locus definition. Now, the second locus definition, please write it down, everybody. Locus definition 2. So ellipse is also defined as the path traced by a moving point, such that the sum of its distances from two fixed points from two fixed points is a constant, okay. So this is another locus definition, yeah. So here, we have already seen that these two fixed points are the two foresight and this constant is the length of the major axis. Please note it is applicable to any type of ellipse, even though I have used my standard ellipse to derive it, many people asked me this question. So having to derive this result only for a standard ellipse, maybe it will not work for a non-standard ellipse? No, this is going to be working for any type of an ellipse, whether it is standard or non-standard, okay. So make a note of this. So many a times, we give this suggestion to students to try this activity experiment at home.  Preferably you can put a permanent marker or something like that. And just pull this taut like this. So this is your marker and start rotating the marker like this, okay. When you rotate it, keeping the string taut, okay. So the string should be kept out. Taut means the both ends should be tight, okay. Then you'll end up getting an ellipse. By the way, I was just joking. Please don't use a marker at all on the walls of your home. Don't spoil your home walls, okay. You can try doing it on a cardboard sheet or something. So just take two points, put two pins, take a small thread. Small means it should be bigger than the distance between the two. Tight to the two points. Take a pencil or a sketch pen or whatever. Keep the ends of this string taut and start rotating it like this. You'll end up always getting an ellipse. Is this fine? Any questions? Any questions, any concerns? Should we take a small question based on the same? I think this is not a right question. Sorry, this is not the question to be solved. Sorry about that. X comma Y be any point on the ellipse, okay. And this is your standard form, 16 X square plus 25 Y square is equal to 400. F1 and F2 are two points given to you, okay. They're asking you, what is PF1 plus PF2? By the way, the point itself is not known to you. But still they're expecting you to find PF1 plus PF2. So that should ring a bell in your mind. Let's do this question and give me a response on the chat box. Okay, Ashita. Okay, Satya. Satya has a different answer altogether. Okay, Satya. Okay, Vishal. Okay, let's discuss this out. Now, if you look at this ellipse given to you, okay, let's write it as a standard form. So for that, you need to divide by 400 on both the sides. Okay, so when you divide by 400, you get X square by 25. Y square by 16 is equal to 1. So clearly A is 5, B is 4. Okay, A is more than B. That means your eccentricity would be under root of 1 minus B square by A square, which happens to be 3 by 5. Okay. Now, let us try finding the four side. Four side in such cases is at A comma 0 and minus A comma 0. Right, so as per our given data, it actually comes out to be 3 comma 0 and minus 3 comma 0. Now, it's just not a coincidence. We had already speculated that these two should be the four side. Okay, not because they use F1 and F2, but since the point was not known to me and they're still expecting me to find PF1 plus PF2, that means it must have been inspired by the theory which we have discussed. So once you are sure that it is the two four side, then SF1 plus SF2 will be 2 times A and 2 times A will be 10 units. So no matter whatever is the point, it doesn't actually matter. Okay, that X comma Y could be any point on that ellipse. This distance is going to be fixed 10 units straight away. Is it fine? Any questions? Let's take another question. Find the lengths and equations of the focal radii drawn from this point on the ellipse. Now, let's not talk about equations as of now. Let's talk about only lengths. Okay, this part we are not going to do as of now. Okay. I mean, it doesn't take much time to do it. It's just two point formula, but let's not waste time doing it because that is easy part. Let's find out the lengths of the focal radii drawn from this point on this ellipse. Okay, Harshita. Okay, Satyam. Anybody else? Okay, let's go back to the board where we had discussed about the lengths. In this case, it was a horizontal ellipse and our lengths came out to be A minus EX1 and A plus EX1. But do you think this expression remains the same even for a vertical ellipse, which I can see over here? So this is actually a case of a vertical ellipse. So if you divide by 1600, if I'm not mistaken, this is 64. So this is a case where A is 8 and B is 10. So B is more than A. Okay, so it's a case of a vertical ellipse. Let me draw one. Let's say I take any point X1, Y1 and let's say these are the two four psi S1, S2. So what is this distance S1, P? And let's say this is our directress, one of the directresses. So S1, P is equal to E times PM1. Now, this directress's equation is Y equal to B by E. So it's E times now B by E minus Y1. So that's going to be B minus EY1. So that will be S1, P. And similarly, S2, P. S2, P will be B plus EY1. Okay, so what is the distance of or what is the focal radii? So the answer is going to be 10. Okay, first of all, E, we have to find E out. Okay, so E value in this case is under root of 1 minus A square by B square. Okay, or you can write it as under root of 1 minus 4 square by 5 square, which happens to be 3 by 5. So E is 3 by 5. So it's going to be 10 minus 3 by 5 into Y1. Y1 value here is 5. Okay, so that answer will be 7. And this answer will be 13. So 7 and 13 are your answers for this question. Well done. I think the first one to get this right was Harshita. Okay, any questions, any concerns? So please do not blindly remember your formula. That's what many people do. Okay, so that will prove very costly to you. So the last thing that we are going to discuss here is the intersection of a line with an ellipse. So that will be our last discussion agenda. By the way, that point is also beyond your school, but we'll still do it because we anyways have to cover it later on. Okay, so we are simply allotting one class each as of now, but we have many more classes required to finish off our chronic section. So the next part that we are going to touch upon, should I go to the next slide? Are you all done with this? Anybody wants to copy from here? So the next thing that we are going to talk about is intersection of Y equal to MX plus C with standard cases or standard forms of the ellipse. Okay, just like we had done it in case of a circle. Okay, so let us say this is our standard form of an ellipse. Okay, and there is a line Y equal to MX plus C. So Y equal to MX plus C will interact with this ellipse in three ways. The first way is where it will be a secant cutting at two distinct points. Second case, it will be a tangent or touching at only one point. And the third case is your NSNT case, neither secant nor tangent case. Fine. Now, let us try to derive the condition for the first situation to take place. When you think such a line will be a secant to such an ellipse. So what should be the condition on MCAB? Such that the line becomes a secant line. Now, how will you solve this question? Any idea? How would you solve this question? Anybody any recommendation from your side? Okay, so this is a strategy which you can adopt to solve this question. In circle, you were saved actually because in a circle, there was a concept of radius. Okay, so in case of an ellipse, what are you going to use? Any idea? Distance of line from both sides. So how does that help you? Distance of line from both sides. What will I get from this? Give me a complete statement. I will use this and this and I will get my answer like that. Right now, you are giving elements separately, separately. Distance is less than major axis. Okay, so if the ellipse is like, if the line is like this, distance is less than major axis. It is not a secant. So how does that logic work? Okay, so the idea here is the distance geometry is not going to work at all. What is going to work is your discriminant of a quadratic equation. Okay, now let me show you how. See, if you simultaneously solve this equation with the other one, that means if you start putting your y as mx plus c in the second equation. Okay, let me simplify this. You will see here, this is a quadratic in x. All of you agree that this is a quadratic in x? Okay, now this quadratic must have two real and distinct roots for secant to happen. Are you getting my point? Now, see how am I associating the discriminant of a quadratic equation with the situation number one. So situation number one means there should be, these two points will have some x coordinates, which will be different from each other. In other words, the two x values which correspond to the x coordinates of a and b, they are actually roots of this quadratic. And those two points should have a distinct x coordinates and they must be real. That means they must exist. In other words, this quadratic equation must give me real and distinct real and distinct roots for the situation number one to take place. Means the discriminant here should be greater than zero. Is it understood by all of you? Any doubt why I am claiming that the discriminant should be greater than zero in this case? So again, I'll repeat for the secant two, for this line to be a secant, it must cut the ellipse at two distinct points. So the two distinct points must have two distinct x coordinates, which happens to be the root of this quadratic. So this quadratic must have two distinct real roots. And for that to happen, the discriminant of this quadratic must be greater than zero. Now, let's write it in a proper way. Right now, it's written in a very, you can say mix and match way. So let's collect x square coefficients together. So if I'm not mistaken, x square coefficient will be a square m square plus b square. x coefficients will be two m c a square. And constant terms will be a square c square minus a square b square equal to zero. Am I right? Correct me if I have missed out on anything. Okay, now, discriminant is what b square b square minus four a c minus four a c. This must be greater than zero. Correct. So let's simplify this. So four m square c square a to the power four must be greater than four a square a square m square plus b square c square minus b square. Okay, four, four gone.  Now let's expand the right side. Let's expand the right side. This will give you a square m square c square minus a square m square b square plus b square c square minus b to the power four. So this two will also get cancelled off. Correct. Now, drop your b square terms throughout. So from here I can write c square is less than a square m square plus b square. So please note this down. This is the condition for the line to be secant for this to be a secant. Okay, please make a note of this. Is it fine? Any questions? Now, this condition that I have derived is only applicable. See, name of the topic I have categorically mentioned so that nobody gets confused. This equation is going, this condition is only going to be used when you have a standard form of an ellipse. If you don't have a standard form of an ellipse, you are going to get even much complicated condition. Okay, are you getting my point? So the condition should not be universally used. This condition should not be universally used. I'll repeat, this condition must not be universally used. If your ellipse is not in the standard form, this condition is not going to be applicable. So now without much waste of time, can you tell me what should be the condition for the second case to happen? So for the second case, yes, the discriminant must be great equal to zero. By the way, you can just save your time because the inequality will just become a equality here. So what you do is, instead of a less than, put a equal to sign. So please make a note of this. This condition is called the condition of tangency. Condition of tangency. Now, this condition, if you try to recall the circles condition, the circles condition was something like this. Okay. So if you expand this, it is actually a square m square plus a square. The only difference now here is instead of an a square, now you have a b square. So if a becomes equal to, or sorry, if your b becomes equal to a, that means the ellipse will become equal to a circle or become, becomes a circle. So the condition of tangency will become the same as what we had for a circle. So that is a comparative study that you can do to remember this result. Okay. So this was for a circle. In a circle, you just change the last a square with a b square, you end up getting the condition of tangency for an ellipse. Then note it down. Okay. So last, the third case, which is your ns int case, your discriminant should be less than zero for the third case. So your c square, your c square will become greater than a square m square plus b square. Okay. So less than we have already used it. As you can see, you have already used less than. Okay. Now we are, we are already used equal to. So only thing we left is greater than. So you don't have to rework the whole thing again and again. So this is your condition for ns int condition for neither secant nor tangent. Any questions? All right. So let's take a few problems based on this concept. Prove that the straight line lx plus my plus n equal to zero touches this ellipse. If this condition a square l square plus b square m square equal to n square is satisfied. Just say it done. If you're done. None, Satyajit. Very good. This is a problem directly based on the condition of tangency which we have already derived in the previous page. Satyam also done. Okay. Great. Okay. See here, we have already seen that if this is the line and this is your ellipse, then the condition of tangency, I like it in short form condition of tangency is given by c square is equal to a square m square plus b square. But now the problem is my line is not in y equal to mx plus c form. So I have to convert it to that form first. So what I'll do is I will write my is equal to minus lx minus n divided by m throat. Okay. Now if you see here, this is playing the role of your slope. Okay. And this is the playing the role of your y intercept. So you can write it like this capital M plus C. Okay. So now your condition of tangency will look like this c square. So it will become capital C square capital C square. This is your Cot. Let me write Cot. Capital C square is equal to a square m square plus b square. Correct. So this is nothing but n square by m square. And this is l square by m square plus b square. Okay. Just multiply with m square throughout. This is what we had to prove, right? a square l square b square m square equal to n square. Yeah, then proved and proved. Is it fine? Any questions? Any questions? Any concerns? Let's take this question. For what value of lambda does this line y equal to x plus lambda touches the ellipse nine x square plus 16 y square is equal to 144. Again a super easy straightforward question. So I'm not going to do any difficult question till your board exams. Okay. Sorry. This is your semester exams. After that, we'll take phonics to much more details. Okay, Satyam. Okay, Satyajit. Easy question. Aiyu. Vishal's answer is completely different. Okay, Vishal. I don't know which is the right or wrong answer. Okay, enough. Okay, Arsita. Okay. So let's discuss it. So if you see this is your x square by 16 y square by nine equal to one. Correct. And as per the given line, if you see this, you're sorry. If you see your m is one, okay, lambda is c is lambda. Just compare. And a is four and b is three. So as per our condition of tangency for it to touch, we should have this equation, this we should have this relation satisfied. c square is equal to a square m square plus b square. But before you're using it, please ask yourself, is the ellipse given to us a standard form of an ellipse? Yes, it is a standard form. This is a standard form. So this condition is definitely going to work. So c square is equal to a square m square plus b square. So lambda square is going to be 25. So lambda is plus minus 5. So what does plus minus 5 signify? It signifies that there could be two possibilities of tangent with the same slope. One would be, let's say here and one other would be on the other side. Okay. And both will be having a slope of one, but one will be having a y intercept of five, other will be having a minus 5. So that is how two answers are going to come up. Does it find any questions? Okay, so we'll take some generic question here. I think this question was a good question. I had skipped it. No, I could see it while I was scrolling down. Find the equation of an ellipse whose axes are parallel to the coordinate axis and having its center at two comma minus three. One focus is at three comma minus three and one vertex is at four comma minus three. So center, one focus, one vertex and the fact that the axes of this ellipse are parallel to the coordinate axis is given to me. Okay. Actually, that was also needless because that you can easily infer from the fact that, okay, I'll not give you the hint. Okay, let's do this problem and give me a response on the chat box. Very good, Satyam. Anybody else? Yeah, let's guess the scenario. I think the vertex is at four comma minus three. This is your vertex and this is your focus and this is your center. So vertex focus center. So basically the scenario looks like the ellipse is going to be like this. Now, of course, the center, the focus and the vertex are in such a way that they are like this. This vertex cannot be here because if this vertex is here, center doesn't seem to lie between vertex and the focus. Okay, so it has to be this position. Now, I know that the equation of such a ellipse would be x minus two whole square by something square and y plus three whole square by something square. Right? So without knowing my A and B, my equation will not be completed. So how do I find my A and B? Very simple. The distance Cv will be A. So whatever is this distance, Cv, that is going to be A, which is actually two units if I'm not mistaken. Okay. And the distance between the focus and the center, that is actually one unit, is actually Ae. Okay, so E is half and you can now find B square by using the fact that B square is A square one minus E square, which means which means B square is A square, which is four, one minus half square, which is going to be three in our case. Get me family. So A is two and B is B square is three. That means the entire equation will now become, I'll rewrite it over here, x minus two whole square by A square, which is four, y plus three whole square by B square, which is three equal to one. Of course, you can expand and simplify to whatever extent you can. Right? So we have completed ellipse from our school point of view, but to be very frank, we have actually gone beyond school also. Okay. So I think whatever comes in school must be definitely within your reach and we'll continue with the next part or you can say the remaining part, which is almost going to be 70, 80% remaining in our after board after your school semester exams. Okay. By the way, any notification on when is going to be your school exam?