 So let us start today's session with ellipse. Good evening, everyone. Good evening. Ellipse. So first of all, we'll understand the locus definition of an ellipse, right? So what is the locus definition of an ellipse? So how do you define ellipse as a locus? Okay, there are actually two definitions. Okay, one is something which I'll be talking about after some, you know, point in time in today's class. But let me begin with the very basic definition that is applicable for all types of phonics. So the locus definition of an ellipse is it's basically the path traced. It's the path traced path traced by a point moving in a plane moving in a plane in such a way, in such a way that the ratio of its distance is a distance from a fixed point, from a fixed point to that from a fixed line, from a fixed line is a constant. And this value of this constant is between zero to one. So this constant value will lie between zero to one. Okay, so let's read the definition once again. It's the path traced by a point moving in a plane in such a way that the ratio of its distance from a fixed point, sorry, from a fixed point, this fixed point is normally called as the focus to that form of fixed line. This fixed line is called the directrix, directrix is a constant and this constant is called the eccentricity, eccentricity. Okay, so let me just show you in a very brief diagram in a very small, simple diagram. Let's say this is the fixed line and this is the fixed point. Okay, let's say there's a point P which moves in such a way. Let's say this is your point P and this is a moving point. It's a moving point. Okay, and this is a fixed point. This is a fixed line, fixed line, fixed point, fixed point. And this point P is moving in such a way that this distance from S divided by this distance from the fixed line. So let's say I call it as the foot of the perpendicular as M. So SP by PM is a fixed value is a constant value. And this constant value is called the eccentricity. And the eccentricity value is between 0 to 1. Okay, so for a particular ellipse, this value will be fixed. Let's say it is 0.5. Right, so wherever it moves, the ratio of SP by PM will be 0.5. Okay, so it comes here also, the ratio will be 0.5. It comes here also, the ratio will be 0.5. It comes here also, ratio will be 0.5. It comes here also, the ratio will be 0.5. It goes far also, the ratio will be 0.5. Okay, so if you start tracing the path of such a point, you would realize that the structure that it will trace will be a oval shape structure like this. An oval shape structure like this. Just pardon me for some crooked, crooked diagram that comes up because of this. Yeah, so I did my best to, let me just draw it. Okay. So now this path as you can see is such that wherever the point P is, let's say it is here, then this distance, then this distance divided by the foot of the perpendicular or you can say the length of the perpendicular from P on to this line, that will always be the same value E, no matter wherever it goes. Okay, anywhere on this path, this ratio will always be E and that is fixed for that particular ellipse. Okay, now it was figured out that this graph is so symmetrical that the same E could also be applied if I had taken another point S over here, let's say S dash, I would call it for the time being and had I taken another line over here. Okay, so now there could be one more such point and one more such line. In short, there could be one more line and point pair such that again, this ratio, let's say I call this as M dash. So S dash P by P M dash will also be the very same E that you have written. Okay, that means there could be two pairs of focus and directresses. Okay, remember this directrix, let me name it D1 equal to 0 and S they are paired up, they are married. That means if you're referring to the distance of the point from this focus, you should refer to the distance of that point from D1. Similarly, S dash and let's say D dash or you can say D2, they are married. Married in the sense that if you're referring to the distance of P from S dash, you should refer to its distance from the line D2. Do not swap the positions. Do not swap the partners. For example, I cannot say SP by P M dash is equal to E. No, that would be wrong to say. SP by P M dash would not be, that would be some other value. Similarly, I cannot say S dash P by S dash P by P M1 or whatever you want to call as the foot of the perpendicular. Okay, that will not be. Are you getting my point? Sorry, somebody was trying to say something. Is it clear? Okay. So now let us talk about some basic terms associated with an ellipse. So let me draw one. I'll use my tool to draw. Okay. People with weird names are joining. Is it? Who is that? No, Eric Arbaiza. Who is this Eric Arbaiza? Hello me. Oh, God, God, God. Hello. Oh, my God. Who is this? Oh, my God. Removed. Zoom IDs are weird, actually. Chalo, I removed it. Who is this Zerif? There's Neil. Drag Neil. Yes, who's that? Okay. Any unwanted guests? Please leave the session. I don't know. I don't know who are these. There's somebody called Sam. Who is this Sam? Okay. Any unwanted, if any of you is keeping a name which is not familiar to me, be prepared to be removed from the session and then you'll not be able to join back. Okay. So I hope our students are not keeping such names and coming. Okay. Anyways, I think everybody is gone. You're sitting with this mask on all those things. Oh, my God. Okay. So first, we'll talk about the certain names that, you know, we'll be using when we are trying to talk about critical points and critical things. As you already know, they are two foci. Okay. I will refer to them as S1 and S2. Okay. And corresponding to S1, there is a directrix as we have already discussed. Oh, I'm so sorry. Yeah. So corresponding to S1, there is a directrix. Okay. Let me call it as D1 equal to zero. Okay. And corresponding to S2, corresponding to S2, there is another directrix. Okay. Let me call it as D2 equal to zero. D2 equal to zero. Okay. Now the line which joins the two foci is basically called as the major axis. Now, please be careful of the names being used. Okay. The line which connects the two foci, this line is called as the major axis. Okay. Later on, we'll see why the name major was given. Okay. The midpoint of the two foci is called the center of the conic. Okay. The midpoint of the two foci is called the center of the conic. Okay. Kinshukh can also draw the same ellipse with S1 to, from S1 to D2 using the same other ratio. No, no, no, no. As I told you, Kinshukh, I think you are not paying attention. S1 is basically linked to D1. So when you're talking about that E ratio, it is the ratio of the distance of the point P from S1 divided by the distance of the point P from D1, not from D2. Okay. So S1 and D1 are married. S2 and D2 are married. With some other ratio then, no, no, no. Remember the foci which you are dealing with, right? The directrix is the one which is closer to it. Okay. So D1 is the one which is associated with S1. You cannot associate the S1 directrix with D2. If you're associating S1 with D2, D2 should be on the S1 side or there could be a bigger ellipse coming up such that S2 is farther away from them. Are you getting my point? If you are trying to take that ratio of an ellipse with S1 and D2 as the foci and the corresponding directrix, S2 should not come in between them. S2 should be then further apart from S1. Okay. Yeah. All right. So what I was talking about, yes. This is the center of the ellipse. Okay. A line which is perpendicular to the major axis and passing through the center is called the minor axis. This is called the minor axis. Okay. This is called the minor axis. Okay. Now, these two points, they are called the vertices of the ellipse. They are called the vertices of an ellipse. So vertices of an ellipse are the points where the ellipse is being cut by the major axis. Okay. They are called the vertices of the ellipse. And other terms that we have already learned in parabola still hold to be true even for an ellipse. So if you join any two points on the ellipse, that will be called as a cord. This will be a cord. Okay. C-H-O-R-D cord. Okay. If you join two points on the ellipse which passes through, which passes through the focus or any one of the two foci, that will be called as the focal cord. That will be called as the focal cord. Okay. Any line which is perpendicular to the major axis, that will be called as a double ordinate. Okay. That will be called as a double ordinate. Double ordinate. Okay. And any double ordinate which happens to pass through the focus. Okay. Any double ordinate which happens to pass through the focus, that will be called as lattice rectum. And since there are two lattice rectum, we call them as, we call them as letra recta. Letra recta. It's a plural of lattice rectum. So we don't say lattice rectums. We say letra recta. Okay. And both will be of the same length. Don't worry about it. No worries, Ruchita. We just started with the session. I'm sure you would have done this in school also. Okay. So what all we learned? We learned about what is the center of an ellipse. We learned about, first of all, we learned about, there are two foci and there are two corresponding directices. Corresponding directices. The word corresponding is very important. The midpoint of the foci is basically your center. A line passing through the foci and the center is called the major axis. A line passing through the center and perpendicular to the major axis is called the minor axis. Line segment connecting any two points on the ellipse is the chord of an ellipse. If the line segment passes through the focus or any one of the two foci, it is called as the focal chord. And a line perpendicular to the major axis is called the double ordinate. If a double ordinate happens to pass through the foci, it is called the latter symptoms. If there are two latter symptoms, we call it as a lettera recta. Okay. I hope these definitions. Yeah, they're very much familiar to your bio words. Yeah, so any confusion regarding this terms being used? Okay, now let's introduce coordinate geometry to it. So let's start talking about equations. Let's start talking about coordinates because that's why we are here. Okay, we are here to learn the coordinate analytical geometrical aspects of this particular clinic. So let me take you to the next page. I hope the terms are clear to you. So I'll name the sheet as the terminology. Okay. Now time to introduce coordinate geometry. So our first attempt would be to derive the standard forms of the equation of an ellipse. Now there are two standard forms of the equation of an ellipse. I think you have already learned that in school. So we'll be talking about two standard cases. So standard forms of the equation of an ellipse that we'll learn how to derive. So before that, I would just do a quick construction. I just do a quick construction. Again, our ellipse. Okay, this is our why am I making it with pink always? What's that pink color and all you're using? No, I like pink. But I have to make it colorful. That's why. Yeah, yellow line, yellow line. And we'll do one thing. We'll keep up photo of this after we draw it. Okay, now, so let's say these are the two foresight, I'll name them as s one and s two. Okay, D one equal to zero D two equal to zero. Okay. And at the same time, I'll also draw, I'll also draw the minor axis. Okay. Now, in order to introduce coordinate geometry concepts to it, I'll first have to decide where is my origin and what are my reference axes. So I'm taking the major axis as the x axis, and I'm taking the minor axis as the y axis. So it's very obvious that your center will now become the origin. The center will now become the origin. Okay. Now, next thing what I'm going to do is I'm going to put a coordinate to one of the vertices. Let's say I call it as even vertex. I'm going to call it as a comma zero. So it's very obvious that we two will be minus a comma zero. Okay, we two will be minus a comma zero. Now a few things I would like to name as of now with some variables C comma zero minus C comma zero. Let me name this point as n one and call it as d comma zero. And let me name this point as n two and call it as minus d comma zero. Okay. Now there are too many unknowns a, c, d, and there's already an e in the definition of the of the ellipse. Okay, so I'll try to reduce them first of all. So basically, I'm going to use the definition of an ellipse over here. Ellipse definition is what it is the locus of a point which moves in a plane in such a way that the ratio of its distance from a fixed point, which is one of the four side actually to that form a fixed line, which is its corresponding directrix is a constant and that constant is e. Correct. This is the definition. Okay. So now, once you know this definition, what I'm going to use here is the point to be your v, v one point. Okay. So that moving point, let's say I take it to be at v one position right now. Right. So v one will also satisfy the same definition. Right. That any other point will satisfy on that ellipse. Even is no exception. It is also a part of the ellipse. Isn't it? So can I say v one s one by v one n one is equal to e. Yes or no. Let's call it as one. Okay. Now why I'm doing all these things. I'm going to use all these concepts to reduce the number of unknowns here. So C and D I want to remove and make only the use of A and E. Okay. I don't want, I don't want too many unknowns to be used as you will get confused. Okay. So let us try to find out our C and D in terms of A and E. That is my agenda right now. So for that, I'm using this first equation that is v s one or sorry, v one s one by v one n one that will be e only. Similarly, v one s two, v one s two, v one s two. That is the distance of this from s two divided by v one n two. That will also be e. Remember if you are referring to s two for the distance, you have to refer to the D two as your denominator of that ratio. Okay. You cannot swap the partners. You cannot say v one s one divided by v one n two is e. You cannot say that as e. Okay. That ratio would be something else. Okay. All right. Now these two equations I will use here. So let's say a v one s one. What is v one s one? What is the distance between v one s one? You will say a minus C. Correct. What is v one n one? V one n one, you will say d minus a. This is e. So can I say a minus C, a minus C is equal to d e minus a e. Okay. Similarly, v one s two, v one s two will be a plus c by v one n two. v one n two will be a plus d. Okay. That means a plus c will be d e plus a e. Correct. Now what I'm going to do is I'm going to first of all subtract these two expressions. Let me call this as three. Let me call this as four. Let's do four minus three. So four minus three, if you do, I get two C is equal to two a e. That means C will become equal to a e. Right. So I've been successful in writing C in terms of a and e. Similarly, I will also use the equation third and fourth to get my d in terms of a and e. So now what I'll do, I'll do four plus three. Four plus three will give you two a is equal to two d e. So d is equal to a by e. Okay. So what I'm going to do now, instead of saying instead of saying C comma zero, I will call it as a e comma zero. And this has minus a comma zero. So a e comma zero minus a comma zero. And instead of calling it as d comma zero, I will call it as a by e comma zero. Okay. Is this fine? In school normally, they just directly give you that this is a by this is a but they don't give a reason for it. Okay. So here I have tried to come from very basics. I've come started. I've started my journey from the very basic, you know, definition of an ellipse and given to you the coordinates of the two four side. And I will also give you now the equation of the two directresses. Now, remember here, as for the given diagram, your directresses are parallel to the y axis. And it is passing through a comma zero. So it is very obvious that the equation of this directress is x equal to a by e. And the equation of this directrix is x equal to minus a by e. Is this fine? Any problem with finding the focus and finding the directresses equations so far? So I have reduced everything in terms of a and e for my convenience. Okay. No c no d. Now, I just wanted to ask from you in a school, have you been calling the focus as c comma zero? Because in some schools, they call it as c comma zero. What is the trend? I mean, both are same things. You're calling it as a comma zero, right? Shadda for the standard cases. Okay, then then it's good. Then see many times what will happen? My terms and the school terms will vary. But don't worry about it. Ultimately, our end result would be the same. All right. Now, time to derive an equation. So for that, I will use the locus definition. I will say, let's say h comma k is a moving point says that the distance of h comma k from s one. Let me write. Yeah, s one divided by the distance of h comma k from the line x equal to a by e, that ratio is always e. Okay, so using that locus definition, I'm going to derive the equation. Okay, so s one p by p m one is equal to e. Now what is s one p s one p is nothing but under root of h minus a e the whole square k minus zero the whole square. Okay, so distance of h comma k from a comma zero is this is equal to e times PM. PM will be what PM will be h minus a by e modulus by under root of one square. Okay, so what are the distance of h comma k from the line x equal to a by e? That's h minus a e mod by under root of x square coefficient. That is one only. Okay, so I hope there's no problem in understanding this. Now, there's no point writing one square under root because it is one only. So I'll just remove that off. Okay, now he is a positive quantity. Remember, he is a positive quantity between zero to one. So this e I can introduce inside. Okay, so what I will do is it'll come inside here and remove this fine. In order to get rid of the modulus, I'll square both the sides. So let's square both the sides. On squaring both the sides, I'll get h minus a the whole square plus k square is each minus a the whole square. This will give you h square. This will give you a square e square minus two h a e plus k square. And also e square h square minus two h a e plus a square minus two h a e minus two h a e will go off for a toss. Okay, let's bring this term to the other side and let's bring this term to the other side. So this will give you h square one minus e square. Okay, plus k square. And this term on the other side will give you a square one minus e square. Right, let's divide throughout by a square one minus e square. So when I do that, I end up getting h square by a square k square by a square one minus e square equal to one equal to one. Okay. Yes or no? Now, for the purpose of getting the equation, we will generalize this. How do we generalize? We generalize by replacing our h with x and k with a y. So this will become x square by a square y square by a square one minus e square equal to one. Correct. Now this term here that you have under the denominator of y square. Okay, this term I will call it as a B square. Now the first question arises in the mind of why B square, why not B? How I'm so sure that it is going to be a positive term. Right, because if I'm calling something as square, I'm confident that it is a positive term. So how I'm so confident that it is a positive term, it is very obvious that it is a positive term because let me just do a small analysis over here. E is what? E is between zero to one, isn't it? So e square will also be between zero to one. Correct. So one minus e square will be positive term. Yes or no? And if you're multiplying a positive term with a square, that will also be a positive term. So there is no harm in calling this as a B square. Right. So ultimately the equations that we give the students as a standard form, as one of these standard forms is where x square by a square plus y square by b square is equal to one. This is our final equation. Okay. Now, one small thing I would like to add over here. Out of a square and B square, or for that matter, out of a and b, which one do you think is bigger? Okay. Let's find that out first. So again, I'll do a small analysis. Probably over here. Okay. So here we all know that e square, okay, e square is a term which is greater than zero. Correct. Yes or no? And e square is what? If you look at this term, our B square was a square one minus e square. So e square will be one minus B square by a square. Check it out. Okay. So if this sum is greater than zero implies that one minus B square by a square will be greater than zero. That means one is greater than B square by a square. That means a square is greater than B square. In other words, a is greater than B in this case. And because a is greater than B, let's go back to the diagram. Okay. This is your a. This distance is your a and you realize that actually this distance becomes your b. Why? It's very obvious. In this equation, if you put your x as zero, y becomes plus minus b. Right? So this point is actually zero comma b and this point is zero comma minus b. So this distance is b and this distance is a. And since this distance is a, this whole distance from vertex v1 to v2 is more than the distance from let's say b1 to b2. And that is why the name major axis was given to it because it is bigger of the two. Okay. Major minor concepts is very, you know, the terms are already known to us. They are dictionary related words. Okay. So because a is more than b, this length v1 v2 is more than b1 v2. And that is why that axis is called the major axis. And this axis is called the minor axis. By the way, a few things I would like to add over here. The distance between v1 and v2 is actually called the length of the major axis. This is surprising to many students. They say, sir, how can a infinite line have a length? No, this is how it is defined. Okay. So somebody says what is the length of the major axis? Later on, you will learn something called the length of the tangent. Okay. So tangent is a line. It's an infinitely extending geometrical figure, but still be defined something as a length of the tangent in the same way here in the diagram. The distance between the two vertices is defined as the length of the major axis. Okay. And the distance between the b1 v2 point is defined as the length of the minor axis. Okay. Length of the minor axis. Is that fine? Any questions? So before I seal off this equation, I would like to just put a quick note over here. a is greater than b over here. Okay. So in this equation, I have considered my a to b greater than b. And this is the diagram for it. And you already know the position of the foci. You already know the position of the directories. Few things we are we will be finding out over here. What is the length of the lateral vector? Length of the lateral vector is very important for us. So let us figure that out. So I'll move on to the other page. By the way, everybody please make a note of this. So let me summarize it in the next page. Let me summarize this in the next page. So for the standard case, x square by a square plus y square by b square is equal to one where you know a is more than b. Okay. This is very important. I'm writing it along with it so that you know that you are basically looking at an ellipse whose major axis is along the x axis minor axis is along the y axis. Okay. So for such axis, please for sorry for such ellipse, please remember the center is at zero zero. Okay. The foci are at a e comma zero and minus a e comma zero. The equation of the directories is x equal to a by e and x equal to minus a by e. Okay. A centricity is as we have already discussed under root of one minus b square by a square. Okay. One minus b square by a square. Now one thing I have to figure out what is the length of of latter sector or letter a recta. Let's figure it out first. So let me draw the diagram once again. Oh, I forgot to take a snapshot of it. See, I told you and I forgot myself because I don't want to draw it again and again. Okay. So let's say this is your major axis. This is your minor axis. Yeah. Okay. So let me just quickly take a pick of this a e comma zero. So let's say I draw a one of the letter a recta. Okay. Remember letter a recta is nothing but it's a double ordinate passing through the focus double ordinate passing through the focus. Okay. So we all know that this will be parallel to the y axis. Correct. So what will happen? Let's say L one and L two are the end points of the latter sector. So these end points will have the same x coordinates because they lie along a line which is parallel to the y axis. Okay. So what is the y coordinate at this point? Let us try to figure it out. For that I will use my equation of the ellipse, which I've already derived. In this equation, I'm going to substitute x as a e and y s k x as a e and y s k and I'm going to derive the k value from here. So this will give me e square plus k square by b square equal to one. So k square by b square is one minus e square. Now remember one minus b square was what? Again, I would like to take your attention towards this formula that we had or this assumption, I would not say formula. This was the assumption we had started calling this as b square. So can I say one minus e square is b square by a square. So this is b square by a square. So k square by b square is this. So k square is going to be b to the power four by a square. So k is plus minus b square by a. So this gives us, this gives us the coordinates of the end points of the lattice rectum. I would request, I'm sorry. Yeah, I would request everybody to remember these results because sometimes they may be required for you in a problem. Oh, why am I writing a square to say? Okay. So if you know your coordinates of the end points of a lateral rectum or one of the lattice rectum, so basically the distance between the end points will give you the length. So what is the distance L1 L2? So distance L1 L2, you will admit that that will be two b square by a units. So please keep this in mind. So just for your purpose of notes, I will put it over here also. So sorry. Any questions? All right. Now a few more things. One is what is the length of the major axis? Length of the major axis is two a units for this part of ellipse. What are the length of the minor axis? Length of minor axis for this ellipse will be two b units. Is this fine? Any questions? So everything is clear with respect to this standard case of ellipse. Could you show the derivation of this again? Derivation of the length of the lateral rectum? Shradha, that is what you wanted. Yeah, this is the derivation. I hope everybody has joined in now. So let me take the attendance. Done. Shradha, done and dusted. Could I move on to the next case? Shradha, you're able to hear me? Done. Okay. Thanks. Thank you. So we'll move on to another case, another standard form, another standard form. Now in this another standard form, it may just happen that your ellipse may look like this. Your ellipse may look like this. Okay. So in this case, you have your foci located here, let's say S1, S2. Okay. And you already know that the midpoint of the two foci is your center. Okay. So the major axis will be now oriented in this way. The major axis will now be oriented in this way. Minor axis will now be oriented in this way. Okay. And the two directories corresponding to S1 and S2 would be oriented here. So this will be your directrix corresponding to S1. This will be your directrix corresponding to S2. Okay. Now, so let's call this minor axis as x axis this time and this let's call this major axis as y axis this time. Let our center be still the origin. Let our center be still the origin. Okay. Now, all of you please listen to this. There is a small change from the method that is mentioned in books like Arihant, Adi Sharma and any other, you know, book that you are following for your JEE preparation and NCRT. NCRT somehow follows a different nomenclature. I don't know. NCRT is the odd man out when it comes to, you know, picking up the names, but don't worry about it. It doesn't change the concept. Okay. In NCRT, I'm sure they start calling this as 0 comma a right, but I'll still call it as 0 comma b actually. That's the small difference which I would like to bring in. In fact, because of this lot of things will change. So please somebody confirm me. I mean, I've actually vaguely remember that that is what NCRT does. It calls this point as, let me name it V1 V2. It calls V1 V2 as 0a and 0 minus a. Am I right? Somebody please confirm it. Kinshu. Yeah, Ruchita confirmed it. Okay. But I will be calling it as 0b only. Okay. And I will call this point as 0be and I'll call this point as 0 minus b. Okay. Right. In this case, the equation of your directories will become y equal to b by e and y equal to minus b by e. And this point, I will keep it as a comma zero and minus a comma zero. So let me call this as B1 V2. Right. Now the benefit of this particular nomenclature is that your equation doesn't change. That means you still follow the same equation of the ellipse that we followed for the previous standard case. Yes, one small change but a very vital change is b here is more than a. Okay. So no change in the equation per se, but be here will now become more than a. Okay. So kindly note down this figure very, very carefully. So where are the two for psi s1 s2? Where are the two vertices V1 V2? What happens to the equation of the directories? Right. And I'll also tell you what happens to the eccentricity. What is the length of the major axis? What happens to the length of the lattice victim? All of those will slightly undergo a change. Okay. So please note this down first. Yeah, for naffle, it is fine naffle. I'm only the teacher. So okay. I was very about R and R. Okay. Now here, a small change will happen. Here your a square will be b square 1 minus e square. Okay. So few changes. So what are the changes? I will write the name as changes. Okay. So first of all, in the case of a greater than b, we used to call it b square as a square 1 minus e square. Remember, in the previous case, your b square was a square 1 minus e square. But in this case, in this case, where b is greater than a, a square will become b square 1 minus e square. So as a result, 1 minus e square is a square by b square. So that means e square is 1 minus a square by b square. So here a small change in the formula will happen. But this is just a superficial change. The bottom line is basically the eccentricity is always, please note down, this is a universal formula for eccentricity. eccentricity is always 1 minus the semi minor axis length, semi minor axis length whole square divided by the semi major axis length whole square. Always. Okay. It doesn't matter what kind of an ellipse you have. Okay. Whether a is greater than b, whether b is greater than a, it doesn't matter. It is just going to follow this formula 1 minus semi minor axis length whole square by semi major axis length whole square. Now remember, in this case, your semi major axis is b. Okay. So I'll write it down. I'll actually write down b 1 and b 2 is equal to the length of of major axis. In this case, it is 2 b units. So semi means half semi means half. So semi major axis, somebody says, what is it? You'll say b. Okay. If they ask you what the length of the major axis 2 b. Okay. So b 1 b 2 is the length of the semi minor axis. Sorry, is the length of the minor axis. And that in this case will be 2 a. So semi minor is a and the full minor is 2 a. What the point? So this is a universal formula that is applicable for all the ellipses. Okay. Now what about the length of the latter sector? Will that change? Yes, that is going to also change. Your length of the latter sector, you can easily find out. Okay, by assuming that, let's say I take l 1 and l 2 be the extremities of this latter sector. Remember, y coordinate will be b for both the cases, something comma b x coordinate. Let's say I call it as h. Okay. So put it in this equation. So x square by a square plus y square by b square equal to 1. Put your x as h. Put your y as b e. So this will become e square. So your x square by a square becomes 1 minus e square. 1 minus e square is actually a square by b square. Okay, as you can see from here, 1 minus e square will be a square by b square. So your h square becomes a to the power 4 by b square and h becomes plus minus a square by b. So your x coordinates of the extremities of the latter sector becomes a square by b and minus a square by b. Is this fine? Okay. However, I would like to give you a universal formula for length of latter sector also. So in this case, it is to a square by b units. But there is a universal formula. Universal formula is no down. Length of the latter sector is given by two semi minor axis length whole square by semi major axis length. This is a universal formula. So these two things I would definitely want you to note down and keep it because it is a universal formula for eccentricity. And this is a universal formula for the length of the latter sector. Is this fine? So I hope you are able to compare the two standard cases from each other. Normally for ready reference, I will call this as vertical ellipse. And I will call the previous one as a horizontal ellipse. Okay, just to make sense to what ellipse I am referring to. So I will name one as the horizontal ellipse. And I will name this as a vertical ellipse. So in the next page, I will just do a quick comparison of the two standard cases. So please allow me to change the slide. Okay, so let me draw a quick partition over it. I hope you are able to see this partition. So x square by a square plus y square by b square equal to one where your a is more than b. And in the right side, I have kept x square by a square plus y square by b square equal to one where b is more than a where b is more than a. Okay, so let me just switch off my camera for the timing. I think it's coming into being. Now for this ellipse, the diagram is a horizontal one which I have already shown to you. Okay, so this is what I will call as the horizontal ellipse. And for this ellipse, the diagram is going to be a vertical one. Okay, so while you are noting down this, you should have an image of these in your mind. Okay, keep an image of these in your mind. Okay, next is the first point of comparison is the center. So I'll just write down here center. Okay, next I will compare them on their vertices, then I'll compare them on their foresight. Okay, even before foresight, I'll talk about eccentricity for the timing then foresight comes because without eccentricity, you will not be able to find foresight. So eccentricity first and then foresight. Okay, then I will compare them on their equation of the directories. Okay, then we'll compare them on their length of later sector, then length of major axis, major axis and length of minor axis. This line I'll make it slightly more bigger. So for this ellipse and for the other one, remember the center will be at origin. Okay, there's no difference. Both will have center at origin origin. Okay, for this ellipse, the vertices is at a comma zero and minus a comma zero. For this ellipse, it is zero comma B and zero comma minus B. For this ellipse, the eccentricity is under root of one minus B square by A square. For this ellipse, the eccentricity is one minus A square by B square. But however, as I told you, both will follow the same universal formula, which is under root of one minus the semi-minor axis whole square by semi-major axis one square. Osai here is a comma zero and minus a comma zero. Osai here will be zero comma B and zero comma minus B. Okay. Directices equations for this ellipse is x equal to A by E and x equal to minus A by E. For this ellipse, it is y is equal to B by E and y is equal to minus B by E. Okay. Length of the lattice victim here is two B square by A units. And for this, it is two A square by B units. However, the universal formula is still true, which is two times semi-minor axis square by semi-major axis. Length of the major axis is two A units. Length of minor axis is two B units. Here it is the opposite. This will be two B units and this will be two A units. So I hope this comparison chart is clear in your mind. And whenever we are solving a problem, please keep these in your minds before you are solving the problem corresponding to it. Any question, any concerns, anybody who has joined a little late because the count was one less, please let me know on the chat box. Could you scroll down? Yeah, sure, sure, sure. Okay. Time for some questions. For the following ellipses, find the length of the major axis, minor axis, coordinates of foci, vertices, vertices and eccentricity. Let's do all of them one by one. We'll start with the first one. Our third one we will not take right now. We'll take it in sometime after 10 or 15 minutes or so. First one, if you're responding to this, please write down all the answers in a single line and then press enter. Don't like put the major axis, press enter, minor axis, press enter. What will happen? If you give one answer, five other answers will come after that. So I'll not be able to judge you on your response. So make sure you write all of them in one single line and then press enter. Okay, Kinshukh. Okay, Archit. All right, let's discuss. So first of all, we will write it in one of the standard forms that we have discussed right now. Okay, so divide by 400 throughout. So this will become x square by 25 if I'm not mistaken and this will become y square by 16. Correct? Now, if you see, a is more than b over here, right? So it's a clear case where your a is more than b. The moment you get that information, automatically your image should recall, your mind should recall that image of a horizontal ellipse. Okay. So everything that you know about a horizontal ellipse, that should appear in your mind. Correct? So the moment you see a is more than b, it's a case of a horizontal ellipse. So a rough diagram is this one. So this is your ellipse that should come in your mind. Now, what all are required? One length of the major axis. So length of the major axis, I will write it down over here. Length of major axis, that will be 10 units. Okay. Because a is five over here, b is four over here. Correct? Length of minor axis will be eight units. Okay, better to write units after this. Okay. Next is coordinates of the foresight. Without a centricity, we cannot find the foresight, but we'll talk about vertices. Vertices is a comma zero minus a comma zero. So five comma zero and minus five comma zero. Okay. Next, we'll find out a centricity. A centricity is always one minus semi minor axis the whole square by semi major axis the whole square. Right? So this will give you three by five. Correct? Foresight is what? Foresight is a comma zero minus a comma zero. So a comma zero is three comma zero and minus three comma zero. A centricity we have already figured out three by five. Now, one thing I would like to talk about a centricity, there are some people who get confused between one minus a square by b square and one minus b square by a square situations. Let me tell you in an ellipse, it will never allow you to go wrong because let's say if you try to do one minus five square by four square, you will get a negative answer and negative under root will become non-rear. Right? So thankfully ellipse will never allow you to make a mistake in finding the centricity. Hyperbola will. So we'll talk about that in today's class only. Is it fine? Any questions here? Any questions? So let me check who all got it right. Excellent. Anusha. Excellent. Ruchita. Good Rohan. Good Shambo. Very good Ritu. I think everybody got this right. No mistakes. Good. So you know your basics very well, that means. Okay, let's do the other one. Second one. Three x square plus two y square is equal to six. The response I've already started to pour in. Very good. Okay, let's discuss it. I think most of you have got it. So in this case, let us first write down. Let us first write this down like this. Okay, so let's convert it to a standard form. And this will give you x square by two plus y square by three is equal to one. Now, as you can see here, a square is two. That means A is root two and B is root three. So this is a case where B is more than a, right? So B is more than a is a case of a vertical ellipse, right? So when you are trying to solve this question, you should get an automatic figure in your mind of this ellipse. Okay, so it's a case of a vertical ellipse, right? So for vertical ellipse, let's try to answer these questions. What is the length of major axis? I'll just write major axis for the sake of gravity. So major axis will be two B over here. Two B is two root three. Minor axis will be two A, which is two root two. Okay, ascenticity will be under root of one minus a square by B square, a square by B square will be this, which is one by root three. Okay, now your vertices will be 0 comma B E, 0 comma B E and 0 comma minus B E. What was the other thing that was required? I think we have, yeah, we have found everything. Yeah, sorry. So, oh, sorry, this is your foresight, not the vertices, my bad, foresight, foresight. Your vertices will be, your vertices will be 0 comma B, 0 comma minus B. So those who have got these as the answer, they are absolutely correct. Okay. Aniruddha, there is a big judgment error over here, my dear. This is not a case of a horizontal ellipse. So your minus one comma zero, one comma zero and root two comma zero minus root two comma zero are all wrong. Okay, Aniruddha, Aniruddha H A, I hope you are able to know your mistakes. Ruchita, absolutely correct. A root five. Actually, one thing, Shambhu, you're saying ascenticity is root five by three. Do you realize root five by three will be more than one? How can ellipse have an ascenticity more than one? These are small things that will tell you that you're going right or wrong. Okay. Most of you have made mistakes. Most of you have made mistakes. Aniruddha, that's correct. Yeah, Akshadraga, correct. Is this fine? So I hope you are able to figure out where you are going wrong. Now, next, we are going to talk about generalized or shifted form of an ellipse. So what is a shifted form or what is a generalized form? Even before we do that, I would like to do with you one small question. Sorry, after that, I'll do this concept. So I would like you to first, you know, understand the locus definition of an ellipse and try to answer this question. So here's a question for all of you. The question is, I'll just give it from my side. Find the equation of an ellipse, whose focus is at one comma one corresponding directrix has an equation of 3x minus 4y plus 2 equal to 0 and eccentricity is half and eccentricity is half. Is the question clear to all of you? See, this is a focus, one of the four, one of the four side and there is a directrix whose equation is this. I hope I'm able to draw a correct positioning of this. So it's something like this. No, origin and one comma one are on the same side. Yeah, origin and one comma one, they are on the same side. Yeah, something like this. Okay. So your ellipse is somewhat oriented in this fashion. Okay. So what is the equation of this ellipse? Now, if you want to solve this question, I can give you one minute, but if you are not willing to solve it, do let me know. Anybody who's trying this, do you want time to solve this? Done. Okay. Kinshuk, can you put your answer on the chat box? Okay, Vavav. Okay. So your answers and Vavav's answer are different from each other. Okay. Anyways, we'll check it out. Anybody else who's trying? 100% wrong. Okay. So here all, please remember the basic definition. The basic definition is if you take any point h comma k on this ellipse, it's distance from the focus, which is one comma one divided by distance from the directrix. Okay. So let me call this as PSNM. So SP by PM should be equal to assenticity. Assenticity is given to you as half. Okay. So SP is equal to E times PM. Okay. So let me write SP. SP is this is equal to E times PM. PM will be mod of 3h minus 4k plus 2 by 5. Right. Now, you may try to simplify this. Take this 10 on the other side and square both the sides. So I'll be just jumping few steps because I don't want to write too much. Okay. 3h minus 4k plus 2 the whole square. Okay. So, right. So 10, 100 h square and from here, I will get 9 h square. Right. So that will give me 91 h square. k square will be 16 coming from here and is already 100. So it will be 84 k square. The figures are slightly ugly. And from here, I will get a 24 minus 24 hk. So they'll come on the other side and make it 24 hk. 24 hk. And I will get minus 200 h. And from here, you will get 12 h. Right. So that will give you 200 minus 12, which is 188 h. Okay. Similarly, minus, oh, I'm so sorry, minus 212 h. Am I right? Yeah. It was minus 200 and from here, I will get minus 12. So minus and minus will make it more minus. Right. Minus 200 k. And from here, I will get 16, 16k. Correct. So 16 and minus 200 will give you 184 k. Right. And constants will be 4 from here and I think 200 from here, which is 196 equal to 0. Any questions? Now, generalize this. When you generalize this, you have to put your h as x and k as y. Right. Now, normally, I like doing such problems after I have done the standard case of a ellipse or standard case of a conic. Just to realize that, just to realize that your ellipses may not always look that simple as what we had done in the previous two slides. Remember, they are conic at the end of the day. Okay. So they can take a structure like this. Please be aware of this. So any conic, including your ellipse, can always take a very, you can say, complicated structure like this. Getting my point. So why I'm telling you this is because I see most of the students doing a mistake. In the exam, they always start with an ellipse equation to be x square by a square plus y square by b square equal to 1. Okay. No, that is a very standard case. All the ellipses in the world do not look like the standard cases. Standard cases are the simplest of all the cases. Right. Let's not get this wrong. Let's not have a wrong notion that whenever an ellipse question comes, I have to start with x square by a square plus y square by b square equal to 1. I just have to find A and B. No, it is not about finding A and B. It's about knowing the definition properly. So what you did was a standard case, standard case where you had a very special positioning of your major axis and you know, minor axis and your directices. This is a case of an oblique ellipse, right? The biggest identification mark of an oblique conic is the presence of an xy term. This is very important for you to know. Okay. For a shifted case or for a standard case, the xy term is normally missing. What normally always missing. So before I was going to the generalized case, I thought I should do one question based on this so that you are basically grounded with respect to the fact that ellipses do not always look like x square by a square plus y square by b square equal to 1. They can take this form also. Make sense? Any question? Now, a question comes from the students many a time. Sir, will it ever happen that I will be given this kind of an ellipse equation? And they would ask me to find out the focus equation of the directices, et cetera, et cetera, like the way we did in the previous. In J main and in school, no. But in J advance, you may be. Okay. So that option, I'm not ruling out. Is this fine? So don't worry. We'll go again through this chapter once again. All right. Now coming to the generalized form. Now coming to the generalized form which I wrote and I raised because I wanted to do that problem, generalized form or the shifted form of an ellipse. So what is a generalized form or a shifted form? Please note down. It is those forms or these are those forms where the center is not at origin. The center is not at origin. Your major or you can say major minor axis, minor axis are aligned parallel to the coordinate axis. Okay. So this is what we'll call as a shifted form of an ellipse. See again, the previous example was the case of a rotational ellipse only. But as I told you, that will not be tested. Even this form will not be tested. I think in Rajaji Nagar, if I ask the students, have you done these kinds of cases where there's a shifting or happening of the conic? I'm sure you would have not done. See, I knew it. In school, the teacher will keep it only restricted to the standard form, which becomes very, very easy. Okay. But I'll just go a step more and I'll cover this case as well because we have time. We don't need to rush too much. Okay. So we'll have time. So we'll take these cases. So as a matter of diagram, let's say I take a small case of a standard ellipse. Okay. Let's say I have an ellipse where A is greater than B. So we all know that this ellipse looks like this. Now for such an ellipse, we know that the center is at origin. Okay, this is your center and your major axis is along the x-axis and your minor axis is along the y-axis. What if I decide to shift this ellipse? What if I decide to shift this ellipse to this position? Okay. So as you can see, I have not made any dimensional changes to the ellipse, but now the center has come to alpha, comma, beta. Okay. And your new major axis and your old major axis, they are still parallel to each other. That means your major axis is still aligned parallel to the coordinate axis. Okay. Similarly, your new minor axis is still parallel to the y-axis. Okay. So in such cases where there is only shifting of the origin happening without any kind of a rotation. So there's no rotation involved. I'll repeat once again. In these kind of conic, there is no rotation involved. I'll write it down over here. No rotation involved. No rotation involved. That means you are just shifting it. Right? Shifting it by maintaining the major axis and minor axis still parallel to the coordinate axis or still parallel to the x and the y-axis. There is no rotation. There's no oblique nature provided to the conic. Okay. So now you tell me how would such an equation or how would the equation, so this equation is this. So what would be the equation of this guy? What will you say? So you'll say it's as simple, right? In this case, your equation will become x-alpha the whole square by a square plus y-beta the whole square by b square equal to 1. Okay. Remember when you are shifting the curve alpha right and beta up, x should be replaced with x-alpha, y should be replaced with y-beta. Absolutely. Okay. It is also akin to saying that you are shifting your origin to minus alpha minus beta. So either of the two ways, if you can, anyways, you can choose to follow the end result would be the same. Okay. Now a question comes in the exam that if I give you this ellipse, will you be able to locate its center? Will you be able to locate its foci? Will you be able to locate its vertices? Will you be able to locate the equation of the direct view directices? Will you be able to find out the equation of the major axis and the minor axis? Can we all do these stuff through a question? Let's take a question and through that question, we'll try to find out the answers to all those critical coordinates and all those critical equations. Okay. So please allow me to take you through those concepts by the help of an illustrative problem. So can we go to a question now directly? Okay. So I would like to ask from you a simple question. Find number one center, number two vertices, number three, eccentricity, number four, foci, number five, equation of directices, number six, equation of major axis, equation of major axis, not length. Don't get confused, equation of minor axis. And let me also ask you length of major axis, length of minor axis, length of lattice rectum. In short, basically we're doing everything possible right for that particular ellipse. And what is that ellipse that is there on your screen? I'm going to write it down. X plus one the whole square by nine plus y minus three the whole square by 16 equal to one. Would you like to try this out first or should I solve this as an illustrative problem? What do you want to do? You will try. Okay. Very good. Go ahead. Go ahead. Try it out. Try it out, please. Yes. Length wise, nothing changes. Dimension wise, nothing changed. Okay. Dimension wise, lattice rectum length, major axis length, minor axis length that will not change when you're shifting it. The shifting only changes coordinates. Shifting only changes equations. Shifting never changes dimensions. It's like you moving, right? If you move, your coordinates will change. And what you will be known by different names in different locations, right? So that is your equation and that is your coordinates, but your height, your weight, etc. That will remain the same. Isn't it? Now don't start applying physics. If you go to the equator, it will be less, if you go to the pole, it will be more, all those things. Let's keep physics aside. Okay. Let's try to solve this question now. I'm sure most of you are trying it. Okay. So let's do this question. Now one way to solve these questions is by making a diagram. Okay. So if you're able to make a diagram, okay, if you're able to realize how the shifting is happening, everything is basically relatively placed to the center of the corner. Okay. And one way is actually by a method which we call as the role change method. Okay. So I'll discuss what is this role change method where you will not require to make any kind of a diagram. Right. So what is this role change method? So what do we do is we first compare this to we first compare this to a standard form of a ellipse. Okay. So what I'm doing is I'm using capital X, capital Y, just to denote your small x plus one and small y minus three. Right. So basically I'm trying to see that this is a case or I'm trying to say that this is a case where your standard ellipse of this nature. Okay. Let me put your a square also as nine and v square as 16. So let's say this is three square. This is four square. So basically this ellipse has been shifted to attain the given ellipse in our question, isn't it? So here the role of capital X is actually being played by small x plus one. Role of capital Y is being played by small y minus three. Okay. All of you please pay attention to this. Even if you're trying to solve it on your own, I would request you to take a break from there as of now. A is three, B is four. Right. So automatically you come to know that this is a case where a vertical ellipse because B is more than A, right? B is more than A. So there was a vertical ellipse which was shifted in this case, isn't it? Okay. So first of all, recall everything that you know about the vertical ellipse. What all you will recall? Where is its center? Where is the foci? Where is the vertices? What is the eccentricity? Et cetera, et cetera. So first we'll talk about center. Okay. So for this ellipse, for this ellipse, try to recall where was the center? You will say, sir, 0, 0. Yes. So what do you will do instead of 0, 0, you write x is 0, y is 0. Okay. Don't write 0, 0 just like that. Just write one more alphabet. I'll ask you to take that pain of writing one more alphabet x equal to zero and y equal to zero. And now do a role change. Change your capital X with x plus one, change your capital Y with y minus three. So automatically your x becomes minus one, y becomes three. That means your center of the required ellipse that I have is minus one comma three. I'm sure everybody would have got this. Yes or no? So I'm basically telling you a method by which you will never require a diagram to get your answer. Okay. This method is called the role change method. Next, vertices. Let's talk about vertices. So this was your first part of the question. Second part of vertices. Where are the vertices for this? The ones which have certain, where are the vertices for this ellipse? You say 0 comma b, 0 comma minus b. Correct. Now, instead of saying 0 comma b, say 0 x is 0, y is b. Okay. And 0 comma minus b. Okay. A small step you have to take here whenever you're solving this question. Instead of writing 0 comma b, 0 comma minus b, you have to write x equal to 0, y equal to b. And again, x equal to 0, y equal to minus b. Now do a role change. Now do a role change. So x will become minus one, y will become seven. Here again, x will become minus one, y will become minus one. Clear? So have you got this as your vertices? If you have got this well done, you are on the right track. Okay. Next is your eccentricity. Let's find out the eccentricity. That is the third thing. So I'll just, oh my God. So I'm almost bottom of the sheet. Yeah. Accentricity is under root of, in this case, one minus a square by b square because it is a vertical, it's a case of a shifting of a vertical ellipse. Even the eccentricity is not going to change because of shifting. Okay. Because eccentricity is basically the ratio of the distances. So both are going to change accordingly. Okay. So it's going to be one minus a square by b square. What is a square? a square here is going to be three square and b square is four square. So which is nine by 16. So this will give you a root seven by four. So that will be an eccentricity. Any questions? Next. Four psi. Four psi for that ellipse is zero comma b e as I requested you, write it like this and zero comma minus b. Okay. Now do a rule change. So capital X is small x plus one. Capital Y is small y minus three. Be will be root seven. Okay. Similarly here also x plus one is zero y minus three is equal to minus root seven. So this will be minus one comma three plus root seven and this will be minus one comma three minus root seven. Okay. One very common mistake which I've seen in students, they blindly change the sign of the y coordinate over here. So once they have got one, they will say the other one is just minus of three plus root seven. Please do not do that. Don't apply this strategy on the shifted one. On the standard case, you can just change it to be plus minus. But when it is a shifted one or when it is a generalized form, you cannot apply that method. Okay. So don't be in that rush to know complete your answer. You will end up getting a negative one. Okay. I know there are some people joining John by and all. Oh my God. Who is this John by John by you are removed. Who else? Random guy. You are also removed. Anybody else? Jeff. Jeff Hardy. Oh, sorry. Where is Jeff? Jeff, Jeff, Jeff, Jeff, Jeff. Oh, Jeff ran away. Yeah. I know some, some random people are joining. Okay. So I was like, all of a sudden three people joined in. So, okay. Yes. So this is the story behind your ascenticity. Oh no, sorry. Forsyth, Forsyth, Forsyth is done. Equation of the directories. So fifth one, equation of the directories. For our case, the standard case, the directories equation was y is equal to b by e and y equal to minus b by e. Right. So your y is nothing but your y is nothing but small y minus three. B is four, I believe four divided by root seven by four. So it will become this. And again, y minus three is four. In fact, minus four divided by root seven by four. So one, one will become three plus 16 by root seven. And other will become three minus 16 by root seven. So this will be your diagnosis. Okay. Any question, any question, any concerns, please do highlight because we'll be taking a few questions in some time based on this. Now, equation of the major axis. What is the equation of the major axis for this case? Your major axis was along the y axis, right? So equation of the major axis, let me write it down, six one, equation of the major axis was x equal to zero. So in this case, it will be x plus one equal to zero. Okay, so in our case, actually, I should write that in yellow because I have been doing that in the previous instances. So x plus one equal to zero. Okay, this will be the length, sorry, equation of the major axis. Equation of the minor axis would be your x axis, x axis is known as y equal to zero. Okay, so this will be y minus three equal to zero. Correct. So this is also done and tested. Length of the major axis, length of the major axis will be, length of the major axis will be two B, two B, two B is nothing but eight units. Okay, length of the minor axis will be two A. So I think that was six units. Okay, next, length of the latter sector in this case will be two A square by B. Correct. So that is nothing but nine by two units. How many of you got everything correct? How many of you got everything correct? Excellent. Excellent Sharda. Very good. Okay. All right. So now we'll take one more case and this time you will do it completely. Let's do this one. x plus two the whole square by let's say four plus y minus one the whole square by one equal to one. I'll not ask you to find a lot of things over here just few of them. Find number one center, number two vertices, number three, eccentricity, number four, four side and number five, equation of the characteristics. That's it because once the process is clear, there's no point doing all those stuffs. If at all you realize that there are some random people entering, do let me know. None. All right. So let's discuss it. So again, we'll do a role change. First of all, we all realize that it's a case where you're, it's a case where your A is more than B. Correct. So it's a shifted version on applied to a horizontal ellipse. Isn't it? So your role of x is being played by x plus two. Your role of y is being played by y minus one. Okay. Now let's try to find out the center for this kind of an ellipse is at zero zero. And remember what I have said to you instead of writing zero zero, you better write it as capital X equal to zero and capital Y equal to zero. Okay. So that will become x plus two equal to zero and y minus one equal to zero. That means minus two comma one will be your center. I'm sure everybody would have got this right. Okay. Next, what this is for this kind is a comma zero and minus a comma zero. Right. A comma zero and minus a comma zero. So first write it like this, then do a role change. So x plus two is equal to a is two in this case. Okay. And y minus one is equal to zero. And x plus two is equal to minus two and y minus one equal to zero. So that'll give you that'll give you the coordinates as zero comma one. And this will give you minus four comma one. I hope you people have got this. Again, some some people are joining. Oh my goodness. Yeah, I removed them. One second. Let me lock this system. Locking will happen. What will happen? If somebody drops off, you'll not be able to join again. All right. So this is your vertices. Ascenticity is nothing but under root of one minus d square by a square. So b square by a square is one fourth, which is nothing but root three by two. Okay. Somebody has joined. The foci here will be the foci here is x equal to a let me write it in white capital X equal to a and y equal to zero and capital X equal to minus a and y equal to zero. Okay. So that'll give you x plus two is equal to a will be root three and y minus one is equal to zero. And this will give you x plus two is equal to minus root three and y minus one equal to zero. So the coordinates that will come out is root three minus two comma one. And this will come out to be root three minus root three minus two comma one. Okay. Equation of the directories is equation of the directories here is x equal to plus minus a by e. So that'll give you x plus two is equal to plus minus a by e will be four by root three. So x equal to minus two plus minus four by root three will be your answer. Is this fine? Any questions? All right. Now your examiner must be a very kind hearted person to give you a question which is actually well written like this. Many case they may not give you like that. Totally 20 uninvited guests. Yeah. Now I've locked my meeting room. So now nobody will be able to join. So in many cases what will happen? They will not give you the equation in this simple looking format. They will probably expand it or they may probably open up the brackets, complete the square and all those things. I mean open up the squares and all those things and give it to you. In those cases it becomes slightly more lengthy to solve these questions because we have to convert it to such a form. So what I'm going to do is I'm going to take a simple question where I'm going to give you a form where your ellipse is not going to be in a very simple looking format. For example, let's take this question. Okay. Find the assenticity vertices, foresight, directories, length of data, set them equation of the latter set them for this ellipse. Now, when I give this question, the first reaction that the student gives me is that said, didn't you say that we do not have oblique kind of an ellipse? Please let me tell you this is not an oblique ellipse. Only oblique ellipses are those ellipse where your axis is slightly rotated. No, that is not this kind of an ellipse. Remember the one way to identify an oblique ellipse is the presence of x y term. There is no x y term over there. There's x square, y square, x, y and constant. That means it is the case of a shifted ellipse. No rotation is happening in this case. Okay. So how will you solve this question? I would request first of all everybody to give it a try. Then we'll discuss it. Okay. So first thing is we have to complete the square my dear. So there's a process also to complete this square. Normally what I do, I take the all x related terms together. I take all the y related terms together. Okay. And I put my constants slightly away. Okay. So when you see this term, the first feeling that you get is I could have written it like this. Okay. In this case, if you take a four common, I could write it like this. Okay. By the way, when you do that, you get an extra term of 16. So minus 16, you can write it over it. Correct. Anything that I'm missing, please do highlight. So this is nothing but x plus two the whole square plus four times y plus two the whole square equal to four, equal to four divided by four. It becomes x plus one the whole square by four, y plus two the whole square by one equal to one. Now you know that it's a case of a shifted ellipse. Right. So just do a role change. It's basically a case where you have shifted this ellipse. Okay. You may write it as two square also. This you may write it as two square equal to one. Correct. So now what all is required? First, assenticity is required. Assenticity is one minus b square by a square. That's actually root three by two, same as the previous question. Next is vertices. In this case, the vertices will be x equal to a or you can say plus minus a comma zero. So you can write it like this x equal to plus minus a y is equal to zero. So your x is x plus one. Let me change plus minus two. Okay. So one vertex will be if I'm not mistaken, two comma minus two. And the other vertex will be minus three comma minus two. I hope you have got this. Anybody who's getting a different answer? Next, what else is required? Forci. Forci is capital x equal to plus minus a and capital y is equal to zero. So capital is x is x plus this plus minus a plus minus a will be plus minus root three and y plus two equal to zero. So your coordinates will be a root three minus one comma minus two and minus root three minus one comma minus two. If we name it vertices. Forci. What else we need? Equation of the direct cases. Equation of the direct cases is capital x capital x is equal to excuse me, sir. Yes, sir. This is Shraddha here. Shraddha. Sir, for the vertices, shouldn't it be one comma minus two for the first one? One comma minus two. Oh, yes, yes, yes, thank you. Thank you for correcting that. Thank you, Shraddha. Oh, I didn't read your messages. Sorry. Yeah. Now, direct cases is x equal to plus minus a by a. Okay, so x plus one is equal to plus minus a by two divided by root three by two is four by root three. So from here, you can get two equations. Okay, please find them out. I'm not going to do that. Next is length of the ladder system. Length of the ladder system. In this case, it is two b square by a. So two into one square by two. That is actually one unit only. Next is equation of the ladder system. Okay, now for the first time, you're doing equation of the ladder system. Equation of the ladder system is very, very simple. It is x equal to plus minus a e. That is the equation of the ladder system. Okay. So your x is nothing but x plus one plus minus a e. So plus minus a is two e is root three by two. So x plus one is equal to plus minus root three. Just solve for just find out your two equations from here. Those will be your equation of the letter. Right. So my main point here was just to tell you that you have to complete a square like this. This is the most critical part of our problem solving. Okay, so more than this, nothing will be asked in your school exam, but still I'll tell you a few more extra things. Before that, we'll take a few questions also. Anything that you would like to ask, any confirmation that you would like to see from here, please do so. I'm going to the next page. All happy with this? Okay. All right. So another concept that I was trying to discuss with you when I was talking about the locus definition, I said that there are two locus definitions of an ellipse. One I already discussed with you in the beginning of the chapter. Now that I'm going to discuss with you, but before that I would like to talk about focal distances. Focal distances. In many books, you will see the word focal radii being used. Okay. Focal radii, focal distance both mean the same thing. In this concept, we are trying to find out that if you have been provided with, let's say, a standard form of, let's say ellipse. Okay. Let's say I'm taking a very standard form, x square by a square plus y square by b square, where a is greater than b. Okay. And I have a point x1, y1 on this ellipse. Okay. My question to you is, my question to you is, what is the distance of the point p from s1? And what is the distance of point p from s2? Your answer should be only in terms of, it only be in terms of x1, a and e. No, y1 should appear. No, no, no. My question is only s1, p and s2, p. Pakul, I'm not asking you to add that. If you are trying to use the distance formula of x1, y1 from s1, please don't waste your time doing it, because there are other alternatives to achieve the result. The other alternative is we all know that this distance, this distance is e times the distance, isn't it? Let's say I call it as m1. So it is e times pm1, right? So finding the distance of p from m1 is quite easy, because you know this equation is x by x equal to a by e. So pm1 distance is a by e minus x1. So it is actually a minus e x1. Okay. So the distance of s1 from p is a minus e x1. Similarly, the distance of p from, I'm sorry, the distance of p from s2 will be e times pm2. Okay. So this will be e times pm2. So e, what are the distance pm2? pm2 is a by e plus x1. So that is nothing but a plus e x1. Okay. Now, how am I going to use these two information? If you see the sum of the two focal radii, that will always be, that will always be 2a. That means it is a constant. It doesn't depend upon where is the point taken. It depends upon the ellipse dimension. It doesn't depend upon where you have taken a point, x1, y1. So this is another definition of an ellipse, which says that ellipse is defined as the locus of a point. If you want to note down, you can note down, I'm writing it. Ellipse is defined as locus of a point moving in a plane, moving in a plane such that the sum of its distances, the sum of its distances from two fixed point, from two fixed point is a constant. So many times we say that if you want to draw an ellipse, if you want to draw an ellipse, you just take two fixed points. Let's say you drill two nails in the wall of your bedroom. Take a string. Keep a pen in such a way that the string is taught. That means the string is like this. So put a pen here. Put a pen over here. And start rotating this pen in such a way that the two ends of the strings are always taught. That means tight. When you start rotating this, you will actually generate a curve which will be that of an ellipse. So if you start rotating it, keeping the string length you will see that the beautiful structure will come on the wall of your bedroom and that will be that of an ellipse. Please don't try this at home. Do it on a paper or something. Another important concept that is related to this is if you keep a source of light at one of the four sides, it is going to hit the surface of the ellipse and let's say the surface is silver, then after hitting it, it will pass through the other four sides. We know the reversibility of the direction of light. So if I light a bulb at the four sides, after hitting the surface of the ellipse, it will pass through the other four sides. Getting my point? Any questions here? Lot of questions have been framed on this concept also in JEE exam. So meanwhile, we'll just take a question on the same. I hope you have copied this down. No, it has to be the four side only. If I keep a light source over here, after hitting, let's say if I keep it over here, after hitting the surface, it may not pass through the four side. It may pass through some other random point. But if I keep a light source at the four side, any one of the four sides, after hitting the parabolic reflector, it will always pass through the other person. Varik Pakul, is it clear? All right. So let's take a question. Say emanating from 0,6 is incident on the ellipse at a point P with ordinate 5. Sorry, it is written S over here, but it is actually 5. After reflection, the ray cuts the y-axis at B. Find the length of PB. Let me put the pole. Very good. We are all two minutes about to get over. I have got just one response. Good. Now I'm sorry, I'm getting a few more responses. We'll stop in exactly three minutes. I mean, not from three minutes from now, but after 30 seconds from now. Okay, 5, 4, 3, 1. You don't need all those stuff, Pakul. Okay. C is the most chosen option. C for Chennai. Okay. Now let's try to solve this question. So first of all, if you see this ellipse, what type of an ellipse is this? First of all, it's a standard case. Now, in standard case also there are two cases. Is it horizontal one or is it vertical one? Let's try to figure it out. So this is going to be 64 if I'm not mistaken and this is going to be 100. Oh, so here we know that A is 8, B is 10. So B is more than A. So it's a case of a vertical one, isn't it? It's a case of a standing ellipse. Standing ellipse, vertical ellipse, whatever you want to call it. Okay. So it's a case like this. Now for this ellipse, where is the foci? Foci we know is 0 comma B, 0 comma minus B. Correct. So B is already known. What is E here? Let's check. E is under root of 1 minus A square by B square. Okay. A square by B square is going to be 36 by 10, which is 6 by 10. So B E will be, B E will be 6. So foci will be at 0 comma 6 and 0 comma minus 6. Okay. Now coincidentally, the ray is emanating from the focus. So 0 comma B and it hits a point where the ordinate is 5. So something comma 5. Now that's very easy to find out the x coordinate. If you know the y coordinate, finding the x coordinate is very easy. So x square by 64 and y is 5. So 25 by 100 is equal to 1. That's actually 1 fourth if I'm not mistaken. So x square by 64 is 3 fourth. So x square is going to be 48. So this guy is going to be 4 root 3. Yes or no? Okay. It could also be minus 4 root 3. Doesn't make a difference. Okay. Now after hitting this point, it has to pass through the other foci. It has to pass through the other foci, which they are calling as B and they're calling this point as a P. So they're asking you what is the distance between P and B? They're just asking you the distance formula. So 4 root 3 minus 0 the whole square. And let's say 5 plus 6, which is 11 the whole square. So I think this is going to give you 48 plus 121 under root, which is 169 under root, which is 13. So option number D is correct. Option number D is correct. Ayurama Krishna, what is this? He has voted for D. Nobody has voted for D. Look at the result. First time in the history of symptom academy. Nobody picked the right option. Have I done correctly or have I done a mistake? That makes me doubt my own working. Is it clear? Any questions? Strange. Rarest of rarest. Rarest of rarest of rarest occasion. Okay. So before we wind up this chapter, a few more things I would like to discuss. One being the parametric form. At least I would like to discuss with you the parametric form of the minus one square. Where sir? Oh where Gayatri? No, it's 11 square. No Gayatri. See, it will pass to other point. Other point will be 0 comma minus 6. So what is the distance between 5 and minus 6? What is the difference? 11? Okay. Any questions? I'm just checking if some unwanted people are joining. Yeah. Parametric form. Parametric form of the standard ellipse. Why does it touch the y-axis at 0 comma minus 6? What did I tell you, Prakul? One ray of light coming from one focus after hitting will pass through the other focus. Other focus is 0 comma minus 6. So one of the recommended parametric form for such ellipse will be x equal to a cos theta and y equal to b sin theta. Okay. Where theta is a parameter. Where theta is a parameter. Okay. Now, many people ask me, can you show this on the, can you show this on the diagram? Where is this theta located? Now, let me ask this as a question to you. If I have a point over here, let's say I call it as a point P, which is a cos theta comma b sin theta. Okay. Let me call this as O. Let me call this as A. Okay. Which angle do you think is going to be theta? Which angle is going to be theta in this case? If let's say I have a point, which is A cos theta comma b sin theta, which point, which angle do you think is basically theta? I would request you to give me a response on the chat box. Anybody? Okay. Now 99% of the people, I don't know about you, but they will say this angle is theta. Okay. There you go. Prakul has said P O A is theta. But let me tell you if you are claiming that as theta, that means you are claiming that this length is A cos theta. Correct? That means you're claiming that this is A. If this is A cos theta means you're claiming this as A, because this is 90 degree. Correct? And you are claiming that for the same A, this is b sin theta, but ideally it should have been a sin theta. Correct? So the response that angle P O A is theta is absolutely wrong. Okay. Now let me show you what is this angle theta. You'll be slightly surprised and taken back by that. So let me erase all these things. Okay. So first make a circle, make a circle, make a circle with the major axis as a diameter. So let me make a circle with major axis as the diameter. Okay. On this circle, let me pull this P up to meet over here. So let's say it meets at Q. Okay. If you bring this Q down and connect it to O, then actually this angle is theta. Got it? Let me repeat the whole process. What I did first was I made a circle with the major axis of that ellipse as the diameter. Now from that point which you're calling as A cos theta, V sin theta, take a vertical up. That means go straight away up till you hit the blue circle. Okay. From where it hits the blue circle, call it as Q. Connect OQ. The angle Q O A will be theta. Now what is the reason for it? I'll justify it. It's very obvious that since you've drawn this blue circle by keeping O A as the radius and O A is supposed to be A. That means OQ will be A. And since you're calling this as theta, you know that this length is going to be A cos theta, which is very well justified that this is A cos theta. And in the equation of the ellipse, if you put X as A cos theta, you will see that Y square by B square becomes sin square theta. That means Y will become plus minus B sin theta. In other words, B sin theta will be one of the points. Of course, there's another point which will be right down over here. Okay, which I'm not talking about. Let's say P dash. So this justifies the fact that the choice of this angle theta is absolutely correct. Okay. So please remember this. That's the biggest mistake which people do. When I give them that there is a point on the ellipse A cos theta comma B sin theta, 99% people wrongly take this angle to be theta, that is P O AS theta, which is not the case, Q O AS theta. This circle is basically called as the auxiliary circle. This circle is basically called as the auxiliary circle. Now we'll talk more about auxiliary circle when we are doing this chapter in more detail. But as of now, just one statement I would like to give about auxiliary circle. auxiliary circle are basically locus off. If you want to write down, you can write down auxiliary circle are basically locus off the foot of the perpendicular drop from the focus onto the tangents drawn to the ellipse. Okay. For example, let us say I sketch a tangent at some point. Let me sketch a tangent. Let's say I sketch a tangent at some point. Okay, let's say at this point I sketch a tangent. Okay. If I drop a perpendicular from one of the four side, let's say this four side onto this particular tangent, it will meet on the auxiliary circle. Okay. Yes, I'll repeat once again. I'll write it down also. auxiliary circle is nothing but it is the locus off the foot of the perpendicular from one of the four side on the tangents drawn to the ellipse. I will demonstrate this on GeoGebra. I will demonstrate this on GeoGebra. All of you please pay attention. Let me open my GeoGebra. Yes. So I hope you can see a screen of mine and I will quickly write down the equation of ellipse over here. Okay. As you can see, this is the ellipse equation. I'll find out the focus also. Okay. So as you can see, A and B are the two four side. Let me choose any point on this ellipse. Let's say I choose a point C. On C, I will draw a tangent. As you can see, I've drawn a tangent. Now I'm going to drop a perpendicular from B onto this tangent. Okay. And the foot of the perpendicular is this guy D. Correct. Now all of you see the motion of D. I'm going to trace the path of D. As I move my C, look at the path of D. There you go. What do you see? What do you see? Do you see a circle? That black circle, do you see that? That is what you call as the auxiliary circle. Okay. So I'll just take a snapshot of this and put it in your notes. Okay. Is it fine? So this is all we have for ellipse. Okay. And I have actually taught you more than required. Okay. Even these things will not be asked. You'll be mostly limited to your standard cases. Let's take a break now. On the other side of the break, we'll quickly talk about hyperbola.