 So, first of all, I would like to start with the concept of asymptotes, asymptotes of a hyperbola. So what is asymptotes? As we have already seen asymptotes when we were talking about graph and its transformation in class 11th beginning itself. So asymptotes are basically tangents to a curve at infinity, but these lines are at a finite distance from the origin. So obviously any line which is at an infinite distance from the origin may be touching a curve at infinity, but that is not the definition of an asymptote. So asymptote is basically a line located at a finite distance from the origin, but touches a curve and touches a curve at infinity is called an asymptote to the curve and asymptote to the curve. Hi Anusha, good morning. How are you feeling today? Much better. Good. Good to know that. Yes, very important to take care of your health at this point of time because now you are heading towards exam phase. All the knowledge, everything that you have acquired, all the practice that you have done will go in vain if you are not healthy. So sleeping, eating and slightly exercising is very, very important more than anything else. If these three things are shaky, everything will fall. So please take care of these basic routines. Okay, now how do we derive the equations of asymptotes of a standard hyperbola? So let me begin with the standard hyperbola. Let me begin with one of the cases. In fact, for any of the two cases, the equation is going to be the same or the equations of the hyperbola is going to be the same. So let us say we have a standard case, equations of asymptotes. So let me take one of the standard case. I am sure I would have a diagram for the same. Let me just pull it out on this one. This one. Yeah. Okay, so basically, let's say this is our one of the standard cases. Even though later, we'll figure out that for both the standard cases, the hyperbola, the asymptotes are the same. That means even if you have a conjugate of this hyperbola, the asymptote equations are not going to change. Okay. So what's an asymptote? It's a line which touches the curve at infinity. So if you see this red line, this red line is basically touching both the arms. In fact, it's a tangent, which is common to both the arms. And the touching point of both the arms will be at infinity. Okay. And there is one more asymptote to this hyperbola. So there are a pair of asymptotes. Okay. So we will be deriving the equation of these asymptotes. Let me call this as asymptote, asymptote one, asymptote one, and asymptote two. Asymptote two. So we'll be deriving that. I didn't get that. Could you try again? What is? Sorry about that. Yeah. So we'll be deriving the equations of these asymptotes. So what I'm going to do, I'm going to say let y equal to mx plus c be an asymptote. Okay. Be an asymptote to the hyperbola. So let us try to simultaneously solve the equation of y equal to mx plus c with the equation of the hyperbola. So when I do that, I end up getting x square by a square minus mx plus c the whole square by b square equal to one. Okay. Let us try to collect it as a single expression in x. Basically, we will get a quadratic expression in x. So you'll have b square x square minus a square mx plus c the whole square minus a square b square equal to zero. Okay. So same activity as what we did for finding the condition of tendency. So here, if I collect all the x square terms, it is going to give me b square minus a square m square x square. You will end up getting minus two a square mcx. And this is going to give me minus a square b square plus c square taken common a square. Now this quadratic, this quadratic in x must have infinite roots has two roots located at infinity. Okay. So root should be at infinity. So what would you, you know, what could you comment about this equation? If both the roots of this quadratic equation are at infinity. What would you say? So first thing you would say, sir, the coefficient of x square itself should be zero. So if both the roots has our infinity, number one, I could say the coefficient of x square, this must be zero. In other words, m is nothing but plus minus b by correct. Everybody agrees. So if in a quadratic equation, the coefficient of x square is zero, it becomes actually a linear equation. Isn't it? And linear equation is basically trying to say that. So I'm just making a small diagram out of it. So if let's say this bend, this bend of the quadratic, I'm just taking a gender quadratic. If this bend of the quadratic is basically nullified because it has been made a linear. So basically it becomes something like this, then the other root will be at infinity. But the problem is if I make the coefficient of x square as zero only, then my other root could be at a finite distance. Right. So if I make this as a zero, then I would be ending up getting minus, let me write it in white. So if I make the coefficient of x square as zero, I'll get minus two a square mcx is equal to a square b square plus c square. In other words, my other root would be this one. Right. But I don't want the other root also to be at a finite distance. That means this term should also be zero. So second, you can say a conclusion that we can draw is two a square mc must also be zero. And because a square and m are not zero, basically c has to be zero. Yes or no. So these are the two takeaways that we can have from the fact that if this particular quadratic has both the roots at infinity, not only the coefficient of x square must be zero, but also the coefficient of x must also be zero. So in light of this, I got the information about the slope and the y intercept of the asymptotes. So I could easily frame the equation of the asymptote now. Correct. So when I do that, the equation of the asymptotes would become equation of the asymptotes would become y is equal to mx plus c. Please note that I have taken m as b by a first and then minus b by a plus c. Okay. That is, you can say your two asymptotes are I'm writing it in a slightly fancy way x by a minus y by b equal to zero and x by a plus y by b equal to zero. Okay. So please make a note of this. These are the two asymptotes of these are the two asymptotes of our standard hyperbola x square by a square minus y square by b square equal to one. Please make a note of this. Okay. Now I would request you, I would request you to figure out what would be the equation if my standard hyperbola, let me change my pen color here. Yeah. If my standard hyperbola is the conjugate of the given one, let's say if I take the standard hyperbola to be x square by a square minus y square by b square equal to minus one, what would be the equations of the asymptotes? Will it change? Will it be any different? Or will it be the same? See, if my both the roots are at infinity, I cannot have x value as a finite value because if I make the coefficient of x square only as zero, I get a linear equation. But that linear equation can have one finite root. But that root also I want to be infinity. So I have to keep the denominator here as zero so that that root also becomes at infinity. And that's why your C has to be zero. Does that answer your question, Chitish? That's the whole point. I'm trying to simultaneously solve the equation of a line and a hyperbola just to see where they interact. And I keep the equations in such a way or I keep my parameters of this quadratic in such a way that they has to meet at infinity. So I have kept my coefficients of this quadratic equation in such a way that this line and the hyperbola interact with each other only at infinity. Okay, so some of you maybe you have not solved it and you're directly saying, sir, it'll be the same. Okay. If you want, we can spend some time finding it out. So in the same way, let the equation of the asymptote be, let the equations of the asymptotes be y equal to mx plus c. So same thing I will do here, x square by a square minus mx plus c, the whole square by b square equal to minus one. So this will be b square x square minus a square mx plus c, the whole square plus a square b square equal to zero. Okay. Is it fine? And if I just again take the coefficient of x square and make a quadratic like this minus to a square mcx and then you have plus a square b square minus c square. Please note that here, the conditions are not going to vary because still you are going to say b square minus a square m square is zero. That means your m value will be still plus minus b by a. And you're also going to say that two a square mc should be zero. That means c value should be zero. In short, the equations are not going to change. The same equations, the same equations of the asymptotes are going to be obtained. The same equations are going to be obtained. Is it fine? Any questions? Okay. Now, one important thing that I would like to discuss over here are certain remarks or certain important points with respect to the hyperbola. Important notes. The first thing that you would observe here is that if you take the equations of the asymptotes, which is this and this and create a pair of asymptotes, create a pair of asymptotes. Now today, I'll be talking about pair of straight lines also. So how do you create pair of straight line equation, very, very simple. In creating a pair of straight line equation, we write the two lines with in the general form or you can say write the two lines in such a way that there is zero on the right side and we multiply both the equations. So if you multiply both the equations like this, you will end up getting something of this nature. Okay. Now, here is an important observation. The observation says that the equation of the hyperbola or its conjugate or its conjugate differ from the equation of, when I say equation, basically I mean to say this part, not the equal to zero part, okay, the equation of the pair of asymptotes because they both share the same pair of asymptotes, pair of asymptotes by a constant, by a constant and that constant need not be one. In this case, you see that there was supposed to be a one for the case of a hyperbola or minus one for a case of its conjugate. Here, there is a zero. That doesn't mean they differ only by one. It is for the standard case that they're differing by one. But in general, if you talk about any hyperbola, any general hyperbola and its conjugate, they will differ from the equation of the asymptotes only by a constant. Okay. See, what I want to say is that, see, normally the equation of the hyperbola that we have is this, correct? Okay, this part, let's call it as H, okay? The asymptote, the conjugate has the equation, this, correct? This part, let me call it as C, okay? And asymptote equation is this, pair of asymptote equation is this, okay? Let me call this as A, okay? So what I want to say is that, H and A differ by a constant. Differ by a constant, okay? And C and A, they also differ by a constant. Now, the best analogy here is one more analogy, which is one more takeaway, which is if you add the H and C, if you add H and C, it will be twice of A. Now, many people ask me, sir, whatever you are saying, is it true for any type of hyperbola? Yes, it is true for any hyperbola, okay? So this relation, this relation, this relation. So these are the important takeaways. So the H part of the hyperbola and the A part of the asymptote, as you can see on the board, I have shown you what is H and A, okay? The same thing that we used to take as an S, okay? S and all, when we were dealing with the conics. So that thing I'm calling as an H as of now, okay? So that H and this A will differ by a constant. C and A also will be differing by a constant. And not only that, a further important takeaway is if you add H and C, you'll get twice of the A. This is a very important relation, which helps us to figure out the equation of the asymptotes, conjugates, etc. Same constant. Yes, that's why that constant will go off for a toss. H plus C, let's say H was more or less by a constant lambda. Then C will be more by that constant lambda. So that when you do H plus C, your constants go off and it becomes to it. Exactly. In this case, that is one, actually, in the case of standard hyperbola, that is one, but that one is only for the standard case. It may not be true for other cases. Okay. This is one important observation. There are a couple of more, which will be, I'll be writing down over here. So first note this down. When I say equation, please read it as an expression, which is to the left of the equal to sign. Okay. Equation doesn't mean you have to put all equal to zero and all expression, which is on the left side of the equal to, I'm talking about that. So that expression differs from the equation of the pair of asymptotes. And again, this equation is also the whatever is on the left side. This, this is what I'm talking about. It differs from this by a constant. Okay. And always remember a hyperbola and its conjugate have the same asymptotes, which we have already seen. Okay. Now a few takeaways other than this, let me go to the next slide. The angle between the asymptotes, the angle between, between the asymptotes will always be two tan inverse B by A. Okay. At least this is very obvious in the case of the standard hyperbola. So in the standard hyperbola, let me just pull in the diagram again. Yeah. So in case of a standard hyperbola, at least this part will be correct. And this part will be true, obviously. So this slope itself was B by A, right? And this slope itself was minus B by A. Okay. So as you can see this angle and this angle will be same. So let's say this is theta, this is theta. So theta itself is tan inverse B by A. And so the angle between the asymptotes, which is two theta is two tan inverse B by A. Okay. I'm just talking about this part of the angle. This part of the angle will be two pi minus that, sorry, pi minus two theta. Okay. Now I would like you to figure out as your, you know, just a deeper analysis. Does it hold true for any hyperbola? Does this thing hold true for any hyperbola? Just try to figure it out. Okay. Next. Third observation or third important note that all of you should be having. If, I mean, it's a corollary of the same thing. If the angle between, if the angle between the asymptotes, the asymptotes, let me write it asymptote is two theta. Then the eccentricity of that asymptote is given as seek of theta. In fact, I should write a mod of this because the angle may be mentioned as an obtuse angle also. So please figure this out that if the angle between the asymptote is two theta, then the eccentricity of that asymptote is given by seek theta. Very easy. I would request everybody to derive this or prove this particular fact. The answer is actually there on the screen itself. It doesn't take much of a time. Request you to write it done on the chat box if you're done with it. Done. Super easy. This result itself will help you. Ritu is also done. A lot of, eccentricity of hyperbola. Ascindicity of pair of asymptotes is infinite. Oh, I'm so sorry. If I said so, sorry. Okay. So from here, from this expression, you know that theta is equal to tan inverse b by correct. So tan theta is b by what is the eccentricity formula? Ascindicity formula is one plus b square by square, which is under root one plus tan square theta. And that's nothing but under root of secant square theta, which is mod of seek theta. Why is the infinite for a pair of asymptotes? See, in pair of asymptotes, what is the definition of e? What is the definition of e? It is the ratio of the distance from a fixed point to that former line, right? So in the case of, in the case of the hyperbola, what happens? The fixed point goes to infinity. Okay. Those two fixed point, the two centers, they go at infinity. So the ratio of that infinite divided by any finite distance is considered to be infinite value. Okay. So as the hyperbola starts, you know, coming closer and closer to each other, what will happen? That ratio will start becoming more and more and more. So for a pair of straight lines, the two four sides, they go to infinity. Okay. All right. So having taken up the basic concepts of asymptotes, we are ready to take up a few questions based on the same. So let's try to take some questions. Let me begin with this question. Find the asymptotes of this hyperbola. Find the asymptotes of this hyperbola. You could also use the basic idea here, but I would like you to take a smarter move to solve this question. The basic idea that I'm talking about right now is the one which I took to get the equations of the hyperbola that is getting a quadratic in X and making the coefficient of X square and X as 0. That is definitely a way out to solve this question, but I would request people to try another way. How do you get the equation of the asymptotes from the properties that we have seen in today's, you know, a little while ago? Okay. Try that out. If you're done, you can share with me the equations. You can give me the combined equation or you can also spread the equations and tell me also. But as per the question, they want the two separate equations. See the question is asymptotes, not pair of asymptotes. So I would request you, if possible, give me the equation of the two asymptotes separately. No idea. Okay. Should I make, I'll make your life easy, very, very easy. See, what did I tell you a little while ago? The hyperbola and its asymptotes, the expression to the left side of zero, they differ only by constant, right? So you can safely assume that the equation of the asymptotes could be this plus some constant. You can put as a lambda. Oh, excellent, Raghav. Excellent. Raghav has already solved it. Very good. Okay. Okay. That's, that's even that is better. Okay. Now see, if you're claiming that this equation is the equation of pair of asymptotes, that means it is basically the equation of pair of straight lines, correct? So a pair of asymptotes is pair of straight lines. So if our equation is of a pair of straight lines, in fact, today I'm going to talk about this also. The derivation of this part, most of you would be eagerly awaiting that what is the derivation of this part set always uses this fact that ABC plus two FGH minus AF square minus BG square minus CS square equal to zero right. So if this equation represents, represents, represents, pair of straight lines, then this expression must be satisfied. Right. By the way, since you have only learned determinants, the better way to keep it in mind is this determinant should be zero, isn't it? Okay. Now, if you are claiming that this equation is also a equation of a pair of straight lines means it should meet this criteria, correct? Now, let's try to analyze what is my A here, B here, H here, F here, G here, C here, etc. So for this equation, it is very obvious that A is zero because there is no X square term. Okay. B is also zero. There is no Y square term. H is half because two H XY, it's only XY, so H is half. Okay. G is minus one because minus two is there, minus two is two G, so G is minus one. So F is negative three by two and our C is lambda. Okay. So if I put it over here, this will be a straight away zero. Two FGH, two FGH, if I'm not mistaken, that will give me two. Okay. AF square, BG square will be zero. CS square will be minus lambda into half square. Correct. Anything that you would like to ask here, please do so. Anything that is bothering you, what is going on, what is happening, you want a reconformation, please do ask. Okay. All set. Great. So this is nothing but three by two is equal to lambda by four. So lambda value is going to be a six. Okay. In other words, as you can see here, the difference between the expressions of the asymptote and the hyperbola is six. Okay. So the equation of your pair of asymptotes, equation of the pair of asymptotes will become this. Okay. And now can we actually get the separate equation from here? It's very easy. You can easily factorize it also. Take y comma and x minus three, take minus two comma and x minus three. So as you can see the two lines which constitute this pair of asymptotes are respectively x minus three equal to zero and y minus two equal to zero. So these are your pair of, sorry, these, these are your separate asymptotes equation of asymptotes, asymptotes separately. Is this fine? Any questions, any concerns here? Yes. A very good question. Now, how would you get the equation of the conjugate from here? Use the fact that H plus C is equal to two A. Okay. So H equation is already known. C i is not known to me. Two A is also known to us. Okay. So let's, let's do it. I mean, that's a good question to solve actually. So H is x y. I think it was minus three y or something. Minus three. Yeah. Minus three y. Minus three y. Minus two x. C equation, I don't know. I'll keep C expression. I don't know. I will write it as a C only and twice of x y minus three y minus two x plus six. So as you can see here, one of the x y will get cancelled. One of the three y's will get cancelled. One of the two x's will get cancelled and you will end up getting C as this. That means the equation of the conjugate of this hyperbola, conjugate of the given hyperbola will be x y minus three y minus two x plus 12 equal to zero. This is just an expression. We just put a zero. You'll get the equation. Is it fine? Shatish, that answers your question. Can we get it from the original hyperbola? Directly, that would be difficult. Directly, that would be difficult. So Shatish, I think asymptote is a good route by which you can actually get the conjugate of that. Okay. Now, I'm not saying it is impossible. You can always find out, but you have to take in a lot of things into your account. You have to find out the conjugate axis, transverse axis. You switch their positions. Ascending will be changing as per the given relationship. And from there, you can get the equation. That is good enough. I mean, if you can spend that much of time, you can get the equation of the conjugate, but this is a good medium. This is a good method by which you can actually reach the conjugate equation. Okay. Now, these students, not these students, these small children. Now, one important property is also there with respect to asymptote, but I will not give you as a property. Maybe I'll give you this as a question. Okay, so let's take that as well. So let me change the slide. Any question with respect to getting the equation of the asymptotes? Any questions? Any concerns? Anybody? Okay. So let's talk about one more property. In fact, I'll give you that as a question. So let me, yeah. So last class, if you recall, we had done the concept of conjugate diameters. Correct? So with respect to the conjugate diameters, there's a very important property, but I would like you to take that as a question. Yeah, the property is this, but take this as a question. The question is the parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes, and that parallelogram has a constant area that parallelogram has a constant area. So let me do one thing. Let me make a diagram, first of all, so that visually you can imagine what the question is demanding from us. So let us say this is our hyperbola and let us say this is one of our extremities of the conge, extremities of a diameter. Okay. So blue line that you see on your figure right now, maybe I'll make it once again. Sorry for that. Yeah, this blue line is a diameter. It's not coming symmetrically, but please consider it to be a touching the hyperbola. Yeah. So let's say this is the one of the diameter. Now the conjugate of it, let me make it in pink color. Okay. The conjugate of it, I think in the last class, I had discussed with you that the conjugate of this hyperbola will be lying, the conjugate of this diameter would be lying on its conjugate. Right. That means this extremity will lie on the conjugate of this particular hyperbola. So let me make the conjugate diagram. Okay. So the dotted yellow one that you see, that is the conjugate of the white hyperbola. Okay. So please note that the extremities or you can say the vertices of a pair of conjugate hyperbola, one of them if it lies on the original hyperbola, then the other extremities of the conjugate will lie on its conjugate. That means this hyperbola is conjugate. Are you getting my point? So let's say this hyperbola is x square by a square minus y square by b square equal to one. Then it's conjugate equation. You already know x square by a square minus y square by b square is equal to minus one. And both of these are pair of conjugate diameters. So let me call them as p, p dash. Let me call this as d, d dash. Okay. So p, p dash lie on the original hyperbola. D, d dash will lie on its conjugate. Okay. So the conjugate diameter lies on its conjugate. Okay. This was clear to us in the last session itself. Right. Now, let us say this point p, this point p has a coordinate. Oh, let me, let me just make the requirement of the question. The question requirement is if I sketch tangents at these points. So I'll just make some tangents at these points. Maybe my diagram will not be very, very good, but I'll try my best to do justice. Okay. So they will meet at, in fact, for the second case also, the similar thing will happen. So let's say this is my, again, as I told you, my diagram may not be very, very crisp here. Okay. As you can see, it creates a parallelogram over it. This parallelogram in gray color. Okay. Now this property says that these two points here. Let me call it as Q and Q dash. And these two points here, let me call it as R and R dash. Okay. They will actually lie on the asymptotes for this pair. So these two hyperbola, they share the same pair of asymptotes. Right. So let's say the asymptotes are this and this. I'm purposely making it pass through R and R dash. Okay. So these will be your asymptotes, these will be your asymptotes of this particular hyperbola pair. Remember a hyperbola and its conjugate hyperbola share the same pair of asymptotes. So if you make this parallelogram, these two will lie on, let's say Y equal to BX and these two R and R dash will lie on Y is equal to minus B by X. Okay. That is something that we need to prove also. And this line that you see, which is making the parallelogram, they are tangents at P, P dash, D, D dash. And this area of the parallelogram is always a constant area that we need to figure out how much is that area. Okay. So let's say this point P is A seek phi, B tan phi. Okay. Then P dash will be the negative of this. So P dash will be negative A seek phi, negative B tan phi. Okay. I would like you to tell me, I would like you to tell me what would be the coordinates of D and D dash? First of all, let us do that. D and D dash coordinates. I want you to write on your chat box, at least D you write D dash is negative of that that everybody knows. K, aditya, okay, fine. I mean, did you guess that or did you figure it out? It's actually simple. I mean, everybody can find it out. See, remember the relationship between the slopes. So let's say this line has a slope of M1 and this line has a slope of M2. So what is M1, M2? In case of hyperbola, B square by A square, correct? That is something that you need to know. Now, from the very fact that M1, M1 will be B tan phi by A seek phi. Okay. And M1, M2 should be equal to, M1, M2 should be equal to B square by A square. So what is M2 then? What is M2 then? M2 will be B by A seek phi by tan phi. Okay. So that, that conjugate, let's say equation of D D dash, that is the conjugate of our PP dash will be Y equal to M2 X. Okay. This you need to solve. This you need to solve along with the equation of its conjugate. So this and the equation of the conjugate X square by A square minus Y square by B square, which I will directly use here itself minus Y square by B square is equal to minus one. Okay. We need to get the X value from here. Okay. So let's get the X value from there. Not a rocket science. A square you can send to the other side. First of all, you can put like this. Okay. Here also you can take X square tan square minus secant square, which is known to be a minus one. So minus minus will get adjusted. So this minus sign will get cancelled with the same. So from here you can say X square is equal to A square tan square phi. So your guess work was actually correct, Aditya. That point X coordinate is eight and five. That was a smart guess. So this is eight and five comma B seek five. Okay. Similarly, the other point D dash that would be a negative eight and five negative B seek five. Now, this is as a matter of fact, I would request you to remember this result also because you know, that is going to be helpful for solving other minor questions. Are anything you take no minus will give you this part. Positive will give you this part. Doesn't matter. The extremities is what we need. Okay. Now the problem statement is something else. This is just an intermittent step. The problem statement is we need to find out first of all, the fact that the meeting point of the tangents drawn at P and D at least is on Y equal to B by AX. That's the first part of our question. Later on, we'll come to the area of the parallelogram and all that stuff. So I would request you to first write down the equations of equations of tangent at P uses are simple. We already know that we have to replace our X square with X X one. So X X one will be your a seek five. Replace your Y square with Y Y one, which is B and five and equate it to one. In other words, you end up getting a seek five by a, sorry, X seek five by minus Y tan five by B equal to one. This is the equation of tangent at P. Similarly, what is the equation of tangent at D tangent at D? Simple. Simple. All you need to do is just change the position of seek and tan. I think it will be minus plus minus here because there was a minus one here. No, that's fine. Is it fine? Any questions? Now, these two equations, we have to solve simultaneously. This two equations solve simultaneously. So I need to know what is the point of intersection and does it lie on Y equal to B by X? Please do that. Please do that. I would request everybody to work that out. An intelligent way of doing that is use Y is equal to B by AX in both of them in this equation also and in this equation also and see whether it does it lead to a similar expression. Okay. So let's do that instead of solving it and checking whether it's satisfying the equation of this line, it's better to put Y equal to B by AX in both the equations and check whether are we ending up getting the same expressions. So when I do that, I end up getting, let me just let me just substitute. Yeah. Minus B by A, no, sorry. Y by B, Y by B is X by A, equal to one. Correct. So this gives me this expression and here also I'll get minus X by A tan phi, Y by B will be Y by B will be X by a C phi. Same thing I'm getting. As you can, as you can check the expressions are same. That basically justifies the fact that the meeting point of this, that means the tangents at P and D, okay. P and D, sorry, P and Q, P and D, which is supposed to be your extreme points, or you can say the vertices of the diameter and its conjugate, they meet on, they meet on the asymptotes, they meet on the asymptote for the hyperbolic. Okay. So this is what we wanted to prove as a part of the property. Is this fine? Any question related to whatever we have done in this? Any question? Any concerns? Now, next part of the question is we have to also show that the hyperbola, the parallelogram, which is formed by these tangents at the extremities of the diameter and its conjugate diameter will be of a constant area. And my interest is in knowing what is this constant area? Now, you could actually guess this result from your ellipse part also. In the ellipse also, we did a similar exercise. There we had obtained something, one result. Try seeing whether you are getting the same result in this case also or not. So again, you can use the fact that the area of this parallelogram will be four times this area. As you can see, I'm shading one small parallelogram inside. So the full parallelogram area will be four times, let me name it as a C point, C P Q D. So area of our required parallelogram will be four times C P Q D. Use the fact to get the area of the parallelogram. So everybody work that out and give me a response on the chat box. Is the question clear? What I'm trying to ask you? Maybe I'll take the diagram on the next page also. Let me take it to the next page. So area of area of. Let me name the figure as Q R Q dash R dash area of the parallelogram area of the parallelogram Q. Q R Q dash R dash will be four times will be four times. Area of the parallelogram C P Q D. And what is the area of the parallelogram C P Q D? Can I say it is the base into height? Now for the base, can I choose, can I choose the length C P? Okay, because C P length, okay, I already know. And height is nothing but it is the perpendicular that you will be dropping from C on Q R line into perpendicular from C on Q R line. Okay. So first let us find C P here. C P is simple. C P would be the distance of origin from the point P, which is going to be a square C square phi and B square tan square phi and the perpendicular. Let me name it as C M. The perpendicular C M will be the distance of origin from the tangent at D. By the way, tangent at D, I've already figured out what was the tangent. Let me write it here also tangent at D. So what was the equation of tangent at D? X by a. So X by a minus X by a tan phi Y by B C phi equal to one. Correct. So this was the tangent at D. So what is the distance of origin from this line? Distance of origin from this line is mod one by under root of square of the coefficients of X and Y. Okay. So let's multiply the two results. But before that, a bit of simplification here would be appreciable. Okay. AB and inside the denominator under root, it will become a square C square phi plus B square tan square phi. Okay. So when you multiply C P into C M. So C P into C M will end up getting these two terms cancelled and it'll give you AB. Therefore, our required area of the bigger parallelogram that we needed was four times AB and this result is same as what we had. Same as what we had in our ellipse case. Is it fine? Any question? Keep the result in mind. You never know. You might require it in your, you know, J exams. Okay. So I think we have done more than enough with respect to our asymptotes. Now we'll be moving towards a rectangular hyperbola that should not take much of our time. Maybe another 45 minutes or so. And then we can take a break and start with a pair of straight lines topic. Is it fine? Any questions? Any concerns with respect to the area of this parallelogram? So area of this parallelogram is equal to the product of the lens of the transverse and conjugate axes. Now related to the concept of asymptotes is the concept of rectangular hyperbola. In the school, we normally talk about this concept, but not in great detail because when I ask students, what is a rectangular hyperbola and why it is called a rectangular hyperbola? All I need to hear is, sir, I don't know why it is called rectangular hyperbola, but I just know that A and B values are equal in this case. Other than that, people don't know anything about rectangular hyperbola. Why the name itself rectangular was given to it that itself students are not aware. Okay. So let me just talk about rectangular hyperbola. A rectangular hyperbola is actually called a rectangular hyperbola because it's a hyperbola whose asymptotes are at right angles to each other. Right. So a hyperbola whose asymptotes whose asymptotes are at right angles. Okay. Right angle that gives rise to the word rectangular. Okay. So the right angle to each other is called a rectangular hyperbola. So let us talk about a general hyperbola and we already know the equation of the asymptotes. So let me just pull in my diagram. So we had already seen that for the standard case, your hyperbola had to pair of asymptotes whose equation was y equal to b by x and y equal to minus b by x. If they are perpendicular to each other. Okay. So let's say I make these two perpendicular to each other. Then what will happen? The product of their slopes. The product of their slope should become a minus one. Right. Looking at this itself, you can say b square is equal to a square and hence B is equal to it. The fact that the asymptotes are perpendicular to each other makes B and a value is equal. That means the length of the transverse and the conjugate axis would be equal in such cases. In short, the equation of the hyperbola itself can be fine-tuned to. So this will be, can be fine-tuned to this. Okay. Either right in terms of a or right in terms of b doesn't matter. Okay. So that means your equation becomes something like this. Okay. For a rectangular hyperbola, e value will always be a root of two. That also is known to everybody. So for a rectangular hyperbola, e value will always be a root of two. This is also worth noting. Okay. Now, since the asymptotes of a rectangular hyperbola make 90 degrees with each other. Okay. So if you want to create a rectangular hyperbola, whose asymptotes themselves behave as the coordinate axes, then this equation that you see will slightly undergo a small change. We will derive that result right now. So what I'm going to do next is I'm going to create a hyperbola whose asymptotes themselves behave as the, as the coordinate axes. Okay. Then what would be the change in the equation of a rectangular hyperbola? That is what we are going to see next. But before that, if you would like to ask anything with respect to this, please do let me know. I think things are obvious here. Okay. So now if I want to create a hyperbola. Let's say I make the asymptotes themselves behave as the as the coordinate axes. Let's say I make the asymptotes. So this is the asymptotes as of now. Okay. We already know asymptotes are at 90 degrees for a rectangular hyperbola. So now if I create a hyperbola whose asymptotes are these two lines, the hyperbola would now look like this. I hope you can imagine. So this, this diagram is somehow rotated. Okay. This diagram that you see on your screen right now, this diagram is somehow rotated. So that now your asymptotes are basically behaving as your x-axis and the y-axis. So under such kind of transformation, you would realize that the diagram would look like this. Okay. Now, of course, when there is a kind of a rotation of axes, the equations undergo a change. So I would like to know how would this equation, how would this equation change? What would be the new equation? If your asymptotes are considered as your new y-axis and the x-axis, let's go back to the diagram here once again. Look at this diagram and tell me, look at this diagram and tell me if you want to make your new coordinate axes coincident with your asymptotes, how should you rotate it? This is 45 degrees. By the way, everybody could easily figure it out. If you want to make your coordinate axes coincident with the asymptotes, how should you rotate it? Can you give the full answer? Angle is not sufficient. Shrita is just saying 45 anti. No, sir. It is 45 degree clockwise. Guys, your old coordinate axes will rotate in this way to match with the new coordinate axes. Are you getting my point? What I'm trying to say? No, asymptotes don't rotate. Coordinate axes rotate. Your coordinate axes would rotate to match with the asymptotes. So how would you rotate it? It is 45 degree clockwise, not anti-clockwise. Everybody who said anti-clockwise, you realize what went wrong. Did you realize your mistake? Any questions? If you rotate 45 degrees anti-clockwise, you'll end up getting this shape. But I don't want this shape. I want this shape. The one which is on my next page. How are you getting my point? So here, we have to use the concept of rotation of axes by 5 by 4 clockwise. Now, just to recall of you of your class 11th rotation of coordinate axes concept. When you rotate your coordinate axes concept, how is your new x and old x related to each other? Just to make you realize this. We had done this in our matrices chapter also. We had done it. So you cannot say, sir, I don't remember it. So this is IEU. My pen stops writing at some places. I don't know why. So this formula is to be applied when your coordinate axes are rotated. So phi here is minus 5 by 4. So in light of that, my old x will be new x cos minus 5 by 4 minus new x. New x is capital. I'm just representing it with capital. Ultimately, I'll keep everything in small. This is just for your initial distinction between your old and new and using capitals. But ultimately, when you write things, finally, you should always write it in small. This is just for our initial distinction. This is going to give me x plus y by root 2. Correct me if I'm wrong. This is going to give me minus x plus y by root 2. So what I'm going to do in our old equation, which was x square minus y square is equal to a square. I'm going to make these two substitutions. So when you do that, you end up getting capital x plus y by root 2 the whole square minus negative x plus y by root 2 the whole square is equal to a square. Let us do some bit of simplification here. Capital x plus y the whole square minus. This is you can treat it as something like this is equal to a square on this side will give you if I'm not mistaken for x y is equal to 2 a square. That means capital x y is equal to a square by 2. And as I told you, it is not a practice to write things in capital. So please write things in small. Okay. So this would be your new equation, the new transformation of this equation. So this equation will transform to transform to this equation if your asymptotes are now treated as your coordinate axis. But normally what do we do? We write this a square by 2 term as another square just to keep it light and simple. So this becomes your equation of a rectangular hyperbola whose asymptotes are your coordinate axes with asymptotes are the coordinate axes. Does it find any questions, any concerns? Okay. So we'll talk a bit about this hyperbola x y is equal to c square. Let me ask you some critical questions based on your basic understanding of hyperbola. First note this down anything that you would like to ask here also please feel free to ask me. That is what the same things that you are asking me, I will ask you now. Okay. So my first question to all of you would be for this hyperbola number one, tell me where are the vertices? Where are the coordinates of the vertices? Give your answer in terms of c. Give your answer in terms of c. No vertices. Why? Even if you have a hyperbola like this, why doesn't have a vertices? These are the two vertices. C comma c minus c comma minus c. Okay. So these two will be the vertices c comma c minus c comma minus c. Yeah. Now the equation of the transverse axis in this case will become y equal to x and the equation of the conjugate axis here will become y equal to minus x. Okay. So this will be your transverse axis. This will be your conjugate axis. Okay. Good. Tell me. Transverse axis and conjugate axis have already written. So let me write it as a note also so that when you're transferring to it, you can realize this transverse axis. Now let me ask you, where are the four side? Where are the four side? Tell me. Tell me. Tell me. Correct. Correct. Correct kinship. Absolutely. Let's wait for this also to respond. See, if you already learned that this, this sense is Ae, no? Ae. Correct. Isn't it? So a role is being paid by C by root two, isn't it? Because a square by two, I'm sorry. A square by two role is being played by C square. Correct. So A is root two C. So in this diagram, let me, let me just make a small change in the diagram here. So in this case, in this case, this length, this length that you have, that is actually trying to play the role of, that is trying to play the role of. Your A, isn't it? That is root two C. Correct. The root two C. So A is actually a root two C. So this, this sense, why have I written this so high? This will be C comma C. Correct. So this, this sense will be A into E. A into E will be two C. Okay. And if this is two C, this whole thing is two C. This has to be root two C. And this also has to be root two C. That means I'm just drawing it separately over here. If this distance has to be two C. And you know, this is 45 degrees. This has to be root two C. This has to be root two C, which indirectly gives you the coordinate of this fellow, which is your focus. Okay. And here also that this is for this one. And for this one, it is minus root two C minus root two C. Okay. So let me write it over here for clarity. Your foci would be at root two C comma root two C minus root two C comma minus root two C or C root two, you can write it just to avoid the confusion, whether the root is over C or not. Okay. What would be the equation of the directories here? What would be the equation of the directories here? What would be your equation of the directories? Recall, what is the original equation of the directories? If my hyperbola was like this. If my hyperbola was like this. What was the equation of the directories? Let's say this directories I'm talking about. X is equal to A by E. Okay. Now, because your coordinate axis got rotated by 45 degree clockwise. Do you recall what we had done? We had replaced our X with X cos phi minus Y sine phi putting my phi value as a negative five by four. Correct. Precisely the same thing we have to do here as well. Isn't it? So when I do that, what do I get? I get X by root two plus Y by root two in place of X over here. So it'll be X plus Y by root two is equal to A by E. Now C here, A as I already told you is root two C. E is also root two. Correct. So root two, root two here will get cancelled off and you'll end up getting capital X plus capital Y. But again, it's not a good practice to write things in capital. This will become your one of the directories. Y divided by root two, can you similarly figure out for the other directories as well? What would be the other directories? Minus root two C. Correct. So X, let me write it in Y. X plus Y is equal to plus minus root two C. Not required. Yeah. Is it fine? I'm not overhead. Next part is what is the length of, what is going to be the length of the ladder system? Length of ladder system. I'm not teaching you anything new here. I'm just making you realize how to change your, how to change your critical equations, critical points, et cetera, in case of rotation. That is all I am teaching you here, which you are already aware of. So this is nothing new. Yes. In terms of C, I would like to know from you. What is the length of the ladder system for a rectangular hyperbola? Very good. Aditya. Length of ladder system is two B square by A. So P two B square by A. And you already know that B and A are equal. So it's as good as two A and it's as good as any questions. Next, again, I'm not done. What would be the equation of the auxiliary circle? I hope you already know what is an auxiliary circle. An auxiliary circle is basically a circle, which is obtained by the locus of the foot of the perpendicular drawn to any tangent from the foresight. An auxiliary circle normally has, normally has the radius as the length of the semi, semi transverse axis. I think I've already given the responses. So I will not waste your time. So this is your equation of the auxiliary circle, which is nothing but. An auxiliary circle is a circle whose. Center is. Same as. That of the hyperbola and radius is length of semi transverse axis. Okay. Next, what is the equation of the director circle? This is interesting. What is the equation of the director circle? Fast fast fast fast fast. X square? Hey, what? No shit. X square plus y square is equal to 0. Yes, it's a point circle. See the origin is the director circle. Because an we are equal. No, in this case. So X square minus y square x square plus y square is equal to a square minus b square and in the are equal to 0. is your origin. So origin is your director circle for x y is equal to c square hyperbola and asymptotes you already know so I will not be wasting time asymptotes here would be your coordinate axes. Another interesting aspect is the parametric equation. The parametric equation that we normally use for x y is equal to c square is x is equal to ct y is equal to c by t that means if I have to choose a point so if a point needs to be chosen is to be chosen on on x y is equal to c square we choose it as ct comma c by t okay so that the parameter t here gets cancelled off when you multiply it okay so this is how we choose a point on a rectangular hyperbola x y is equal to c square hyperbola is it fine any questions any concerns anybody has do let me know okay now I could also give you a list of equation of the tangent equation of the normal but I would you know request you to derive that out maybe in the next slide we'll take up 10th point what are the equation of a tangent equation of a tangent x1 y1 point first we'll take x1 y1 point and then we will change it to ct comma c by t also what would be the equation of a tangent at this point to the x y equal to c square hyperbola please do it give me the response can you guys a question so this rotated form is the general form of a rectangular hyperbola I would not name it as a general form it is no I would not say a general form I've never seen such kind of a word used we just say it's a case of a hyperbola where the asymptotes are behaving as the coordinate axis that's what we say can show okay please give me the equations here yes yes why not why not yes we can say that now okay so here the concept doesn't change my dear please remember the concept doesn't change the menu say x y equal to c square and you want to write the equation of a tangent we do the same remember I had given you in the conic that we replace our x y with x1 y plus x y1 by 2 the same thing I'll be doing here also so make this replacement over here you'll end up getting you'll end up getting x1 y plus x y1 is equal to 2c square okay so please note down and also note down now that this term will start behaving as your t expression that we were using for our general conic so this expression will start behaving as your t and s will be your x y minus c square so please note this is your s and this is your t now you already know it I mean I'm just making you realize it again and again okay can you take a small mistake 2c square will come on the right side okay so now let's do it for the parametric form parametric form you'll say that there is nothing now you just put your x1 as ct and y as c by t okay cancel off things a lot of things will get cancelled here so x by t plus y by y t is equal to 2c will happen okay so this will become your parametric form so this is point form this is parametric form many people say sir do we need to remember these results also see again I I leave it up to you okay all that is permitted by your capacity all right so let's now move on to equation of normal done copied anything that you would like to ask see by mistake so let's go to the normal equation I think it's the 11 point equation of normal first of all we'll find it at a generic point x1 y1 and then I'll take a parametric point so first let us find out at x1 y1 in fact I would request you to prove this it will be x square minus y square sorry it will be xx1 minus yy1 is equal to x1 square minus y1 square please prove this this is the equation of the normal will not take much of your time done yeah same thing at the very good anybody needs help in this does anybody require our help who's following this see you know a point x1 y1 you already know the slope slope of the normal would be what you can get it from here itself slope of the normal would be minus y1 by x1 okay so simple y minus y1 is equal to minus y sorry the negative reciprocals of this negative reciprocals so that will be x1 y1 so you just have to put x1 y1 you can put a y1 on this side so this is as good as saying xx1 minus yy1 and take the take the xx1 square to the other side will become x1 square minus y1 square okay now the same thing we will find out for a parametric position ct comma c by t simple I mean nothing different over here you just have to put your x1 as ct y1 as your c by t one of the c's will go off so it'll become xt minus y by t is equal to c t square minus 1 by t square on simplification this actually yields xt cube okay so you are multiplying forward t square so xc cube minus yt minus ct to the power 4 plus c equal to 0 so this is the equation in the parametric form this is the equation in the point form okay so this is all you need to know related to your rectangular hyperbola maybe one question I will take before we wrap up this hyperbola topic anything that you would like to note down here see we derived everything so there's no need to you know bother about forgetting any formula I know this topic Koenig itself is full of so many formula and still more formula are going to come in your pair of straight lines topic but if you have your basic understanding about all these concepts okay yes after the break nine I'll give you a break in some time so let's take a small question here if the normal at a point t1 to the rectangular hyperbola let me write it down if the normal to a point t1 see this is just like representing the eccentric angle okay so I have represented the parameter of that point and I'm calling it as a t1 today's class will be till 11 o'clock Shadda I hope I had dropped a message no read the message says 8m to 11m because October 17th there will be no class that Sunday there will be no class that's why I thought I would compensate for it by taking it now because next class will be a test only for j advance people so anyways they will not be able to sit for three hours yeah so the question is if the normal to the point t1 to the hyperbola x y is equal to c square meets it again at t2 meets it again again at the parameter or meets it again at a point whose parameter is t2 then find let me give as a proof this question then prove that even cube t2 will always be a minus one even cube t2 will be a minus one done guys I thought it will be a super easy question than shittish is on excellent kinshukh is on good okay so let us pull the equation of the normal in parametric form that we had just now written this form okay so let the equation of the normal at t1 that would be so the equation was x t cube so I will write x t1 cube minus y t which is my minus y t1 minus c t to the power 4 so I will write minus c t1 to the power 4 because I am drawing the normal at t1 okay now this particular normal would be satisfied by the other point which is c t2 comma c by t2 so you just put this in this equation see what happens so when I do that I end up getting c t2 t1 cube minus c by t2 into t1 minus c t1 to the power 4 plus why did I write c1 here unless this means plus c equal to zero plus c equal to zero drop the c's because we don't need them or let me drop it in the next step we'll still get confused yeah so you have t2 into t1 cube minus t1 by t2 minus t1 to the power 4 plus 1 equal to zero multiply throughout with the t2 you'll end up getting t2 square t1 cube minus t1 minus t1 to the power 4 plus 1 equal to zero okay so here oh sorry so here what we can do is we can take these two terms together and this is t2 and these two terms together other way around we can do we can take these two terms together and these two terms together so here if I take t1 cube common I'll get t2 square minus t2 square minus t1 am I right you can also take oh once again this is your t2 here also and here if I take one common you get t2 minus t1 so from here I can factorize it as t1 cube t2 plus 1 times t2 minus t1 equal to zero and I know that t1 and t2 are not the same points t1 and t2 are not the same points because the question says that the normal meets the curve again at some other point so t1 t2 are not the same points okay that means this cannot be zero which means the only possibility is this is zero and that means you have done with the proof t1 cube t2 will be a minus one no need to remember these derived results okay these are they will be asked as questions directly okay don't put a lot of memory pressure on yourself okay I'll remember this also remember that also no it'll not happen like that okay we have educated resources we'll have to you know use them optimally the last and not the least part of this topic is the very simple reflection property of the hyperbola but I will not be going into much detail of it I'll be leaving up to you to figure it out reflection property of hyperbola remember there was a reflection property of an ellipse what was the property in the case of an ellipse that if a rear flight emanates from one of the focus then after hitting the elliptical mirror surface it will pass through the other focus isn't it in case of hyperbola also there is a similar property okay and what is that property and why is that property true that is something that you would be proving let me just erase so this property says that if there is a rear flight okay if there is a rear flight which comes from let's say here okay it's the hyperbolic reflector so this is basically I can say silvered on the outer side okay then after hitting this after hitting this it will pass through a focus provided the extension of this passes through this focus okay so please note this down if a rear flight comes like this says that if allowed to move further it will pass through one of the focus one of the foci of this hyperbola then after hitting the hyperbolic reflector you can say it will pass through the other focus are you getting my point in other words we need to prove this and I'm leaving this up to you for your homework assignment in order to prove this basically we have to show that I'm going to draw tangent also at the same point so we have to prove two things here one this angle and this angle will be equal okay and second thing you have to prove that's a theta theta and the other thing that we need to prove is that this angle and this angle will be equal of course if you prove one the other is automatically proven okay so prove the reflection property of the hyperbola no the fact that this is happening that is why this you know the reflection property is obeyed right the fact that these angles are equal and hence the reflection property is obtained that means if you're a if a rear flight is basically passing through it then only it will pass through the other focus it is it is reflecting they are equal it's not the thing it is equal that's why they're reflecting other way around okay so this is all we need to learn about hyperbola and other this thing non degenerate conics okay now we'll talk about after the break the pair of straight lines that is another part of a conic which we call as a degenerate conic so we'll take that concept after a small break