 of a focal cord connecting two eccentric points. Okay, so let us say I have a hyperbola. I'm not very sure whether I have the picture of the hyperbola with me or it's gone. Oh yeah, it's there. Okay, so let's say this is a hyperbola and we want to draw a cord which connects two points whose eccentric angles are phi 1 and phi 2. So let's say this is point p and this is point q. So point p has the eccentric angle of phi 1 and point q has an eccentric angle of phi 2. Okay, now when I say eccentric angle of phi 1 and eccentric angle of phi 2, it is obvious to all of you that I am basically I'm talking about a point whose coordinate is a seek phi 1 b tan phi 1 and the other point q will have coordinates of a seek phi 2 comma b tan phi 2. Okay, and this is our one of these standard cases x square by a square minus y square by b square equal to 1. Okay, so the equation of pq and I'm sure most of you will be able to derive it so I don't want to waste time deriving it. It is very similar to what we had done for the case of an ellipse. Okay, so you know two points and you can easily write down the equation of the line connecting the two points and that equation is given by, please note down, the equation is given by x by a cos phi 1 minus phi 2 by 2 minus y by b sin phi 1 plus phi 2 by 2 is equal to is equal to cos of phi 1 plus phi 2 by 2. Please note the differences here, very, very important. There is a phi 1 minus phi 2 here and there is a phi 1 plus phi 2 sitting here also and here also. Okay, later on when we are doing the equation of a tangent in parametric form, we will show that if phi 1 and phi 2 become equal, it will actually become the equation of a tangent at the eccentric point phi 1 which is basically the parametric form of the equation of the tangent. Okay, just note this down and please, please prove this as homework. Doesn't take much time, it's just a two point form of the equation of a line. You already know these two points, you just have to write down the equation of a line, find the slope, use one point and slope and get the equation. Okay, now what is important are two important results, a results associated with it, important results I should write, important results associated with it. Very similar to what we had done in case of an ellipse, in case of a hyperbola also, tan of phi 1 by 2, tan of phi 2 by 2, this will be equal to 1 minus e by 1 plus e. If the chord passes through, let me write it down, if the chord passes through, passes through, the focus A e comma 0. So if this is a focal chord, so this is a chord in general, okay, this is a chord in general, and if that chord becomes a focal chord, but the focus here should be A e comma 0, then the eccentric angles of the two points, which are basically the points at the extremities of the focal chord will satisfy this relation, tan phi 1 by 2 into tan phi 2 by 2 equal to 1 minus e by 1 plus e. Now I would request people to tell me what was the similar relation for ellipse, does anybody remember that? Does anybody remember that? We are meeting on Sunday morning, taking an x-ray effort, do you remember the result that we had derived? Right Aditya. So in case of an ellipse, remember there was an e minus 1 by e plus 1, okay, and of course if you practice you'll end up remembering it, I'm not asking anybody to mug up any formula. If you have, would have practiced a few questions, you would have definitely come across that. So for an ellipse, remember there was an e minus 1 on top and e plus 1, of course 1 plus e and e plus 1 are same thing in the denominator, okay, and if your chord passes through minus e comma 0, if your chord passes through minus e comma 0, this will become ultra reciprocal, 1 plus e by 1 minus e, okay, this is when your chord passes through chord passes through minus e comma 0. Is this fine? Again these two results are very easy to figure out, all you need to do is put the point A as x and y as 0 and get this relation after a bit of simplification, you all know your trigonometry very well by now, okay. So these two are very easy to figure out from this equation itself. Now I'm not wasting time and energy figuring it out because we have already done a similar exercise in ellipse, okay. So let's directly now go on to the equations of tangents. Last class I think I had stopped at the concept of intersection of a line with a hyperbola, right. So now we'll be going ahead with the different forms of equation of tangents, but before that please note this down. Any paper tomorrow? HSR, Kormangala, Rajaji Nagar, YPR, physics in DPS, okay. Objective, Aditya. Full objective, okay, do well. Okay, Thursday you have physics for Kormangala, okay. All right. So now we'll talk about equations of tangents in various forms for a hyperbola. So I'll be referring to the standard case of a hyperbola first of all. So let me follow the dynamic. So let's say this is our standard case of a hyperbola x square by a square minus y square by b square equal to 1. First we'll talk about how do you get the equation of a tangent when you have been asked to make a tangent at a given point x1, y1 on that given hyperbola. See this process is the age old process which we have been doing since our circle days. You need to just generalize. What do you need to generalize? I already told you to get the equation of, let me write it down like this. So if you have any standard, sorry, any conic and you want to find out the equation of a tangent tangent at x1, y1, okay. You find the t and equate it to 0. And what is this t? t is obtained by t is obtained by doing the following changes. Replace x square with xx1, replace y square with yy1. I've been telling this since beginning of conic. I think the more number of times I write the better it will be in your mind that this is important. Okay, so these are the changes that we normally perform while getting the equation of a tangent. So x square is replaced with xx1, y square with yy1, xy with xy1 plus x1y by 2, x with x plus x1 by 2, y with y plus y1 by 2, and constant remains constant. So here also if you want to do that replacement in this equation of the hyperbola, it will become xx1 by a square minus yy1 minus yy1 by b square equal to 1, okay. And you are already aware that we call this term, we call this term as s, okay. We call this term as s1. That means in s, if you substitute the point, you get your s1 and we call this expression as a t. I mean, I'm talking with respect to this hyperbola. Depending upon the hyperbola, this expression will keep changing, okay. So t equal to 0 will become your equation of the tangent in this case. So this form of the equation is called the point form of the equation of a tangent, okay. No issues. Now adding to this, we can also talk about the parametric form. So if the very same point, if the very same point is having a eccentric angle of let's say theta or a phi, let's say phi, okay. All you need to do is replace your x1 with ac phi and y1 with b tan phi. So in this equation, if you replace your x1 with ac phi and y1 with b tan phi, you'll end up getting the equation as x ac phi by a square minus y b tan phi by b square equal to 1. And these all things are pretty much known to you already. So all you need to do is keep this equation in your mind, parametric form of the equation of a tangent, very, very useful. This is called the parametric form. So point form, parametric form and both are related is just that you're replacing your point with the parameter. No, this is not the first slide. There was a slide before it also, right? You joined late. This was a slide. We talked about the equation of a focal chord and we talked about the relationship between the eccentric angles of the extremities of the focal chord. Sorry, we talked about the equation of a chord and then we talked about the conditions of the eccentric angle for focal chords. If you want, you can take a snapshot of this part only. This is the one which is important. And of course, I'll be sharing the notes as well. Okay, done. Now, going back to the next slide. So this parametric form, if you, let's recall what we had done for the equation of a chord, just now I've shown you the equation of a chord. So what was the equation of a chord? Equation of a chord was x by a cos phi 1 minus phi 2 by 2 minus y by b sine phi 1 plus phi 2 by 2 equal to cos phi 1 plus phi 2 by 2. Okay. Now, in this chord, if you change your phi 1 and phi 2 to be the same quantity, let's say phi, then you would realize that on doing this, you will end up getting this equation from the chord. That means that chord will start becoming a tangent, right? That we all know. So if the two points, the two secant points are coming very close to each other or overlapping with each other or coincident with each other, the chord will become a tangent. So this will happen here. So x by, now see here, this will be cos 0. Cos 0 will be 1, which I will substitute later on. This will become sine phi and this will become cos phi. So divide by cos phi throughout, you will get 1 divided by cos phi, which is c phi. As you can see, this is our first term over here minus y by b sine phi divided by cos phi will become tan phi, which is your second term over here equal to 1. So divide by cos phi. Is this fine? Any questions? All these equations that I am writing, that is only for the standard form. If you want to write it for the general case of hyperbola, you need to start with this fact that here you have to make these substitutions and depending upon if you know their parameter, if you know the parametric version of the point, you can substitute it in the equation, this which I will write it on top. So t expression is given as AXX1, BYY1, HXY1, X1Y plus GX plus X1. So here you have to do all your substitutions. Whatever you want to get. So this is your t expression. So t equal to 0 in that, you substitute the point or you substitute the parametric version of the point and get your desired equation of the tangent. Is it fine? Any questions? Any concerns here? Quad equation. Okay. This is the quad equation. Only for a standard form. Guys and girls, please understand this. This is a quad equation only for that standard form and that to one of the standard form x square by a square minus y square by b square equal to 1. Please do not use this for any generic form of a hyperbola. See, that is why derivation is important. When I derive it, you understand that, okay, this result itself was derived for a standard form, right? Shradha, let me know once you're done. Yes, Aditya, figure that out. Okay. So for the other standard form, please, are you all aware that for the other standard form, x square by a square minus y square by b square equal to minus 1, the parametric form is a tan phi and b c phi, correct? Now, choose two points of the same format a tan phi 1 b c phi 1 a tan phi 2 b c phi 2 and just get the equation of the tangent once again. Okay. So figure out what changes comes out. Get the equation of a chord, get the equation of the relationship between the eccentric angles passing through the focal chord 0 comma b and 0 comma minus b. So that would be a good exercise. Okay. That would be a good exercise to just to get an idea about that. Okay. Anyways, now we'll go to the third form, which is the slope form. In slope form, we basically will be provided that there is a standard case of a hyperbola. Okay. So the same diagram, I'll just pull it out. And I want to sketch a tangent to it. Whose slope is m? Okay. So the slope of this line is known to us. Then what is the equation? Simple. We already know that if y equal to m x plus c has to touch, if this is tangent to the standard case x square by a square minus y square by b square equal to 1, what is the condition of tendency? What was the condition of tendency? It must satisfy c square is equal to a square m square minus b square. Again, please do not use or do not misuse this relation for any other case other than this standard form. Okay. That means c is plus minus under root of a square m square minus b square. So the equation of the tangent would become y equal to m x plus minus under root of a square m square minus b square. And I already told you for such a tangent to exist, a square m square minus b square should be greater than equal to zero. In other words, m should belong to the interval minus infinity to minus b by a union b by a to infinity. Okay. This is already discussed in the previous class. So please make a note of this. Is it fine? Any questions? Any questions? Any concerns? Okay. So in the interest of time, we will start talking about equation of normals. Equation of normals. Okay. So let's first talk about point form of the equation of a normal to a standard form of a hyperbola. So let us say I have a point x1, y1. I have a point x1, y1. And at this point, I sketch a normal. Okay. What is normal? Normal is nothing but a line perpendicular to the tangent at the very same point. Okay. Now I would request everybody to prove this that the equation of the normal at x1, y1, equation of the normal at x1, y1 to the standard form of a hyperbola. I want you to do it on your own because then only you'll remember the result because this result is very important. a square a, a square x by x1 plus b square y by y1 is equal to a square plus b square. Now can anybody tell me how is it related to the equation of a normal at x1, y1 to a ellipse x square by a square plus y square by b square equal to 1? What is the only difference? What is the only difference? Recall for an ellipse it was for the standard ellipse it was a square x by x1 minus b square y by y1 equal to a square minus b square. So all you need to do is in order to attain this is replace b square with a minus b square everywhere. So if you replace this b square with a minus b square and this b square with a minus b square, this is what you end up attaining. So this is how you can basically remember the equation of the normal. So this is something that you need to derive right now prove right now very simple. I'm sure all of you will be able to do it. Let me know once you're done with the proving of the equation of a normal. Yeah, so basically the slope of the normal is negative dx by dy at x1, y1. And that would become negative a square y1 by b square x1. So the equation of the normal will be y minus y1 is equal to negative a square y1 by b square x1, x minus x1. So just do a minus simplification here. Just divide by y1 down and multiply with b square. So b square y minus y1 by y1 is equal to minus a square x minus x1 by x1. Okay, simple mathematics. So this will become a square x by x1 plus b square y by y1. And on the other side, you'll end up getting b square y1 by y1, which is b square. And there's already a plus a square sitting over it. Okay. Please remember this result, you will not get time to derive all these things in the examination room. So tangent equation, normal equation to all the conics, at least in their standard forms, must be in your mind. And that will come with practice, not with sitting and reading with a formula book. Nowadays, many people are following that Black Arihant formula book. That formula book is a waste till you are aware of the process till you have derived it on your own. Okay. So this is related to when the point is x1 y1. If you have been provided with the very same point in the parametric form, that means this is ac phi, ac phi comma b tan phi. The same equation. Okay, so this is called the point form. I'll let you down point form. The same equation will become a square x by ac phi plus b square y by b tan phi is equal to a square plus b square, which can be further simplified as x by a cos phi y by sorry, further simplified as ax cos phi by cot phi is equal to a square plus b square. So this equation is what we call as the parametric form of the equation of a normal. This is called the parametric form of the equation of the normal ax cos phi by cot phi is equal to a square plus b square. Now let's talk about the slope form just like we did it for the tangent. All right. So for the point form, sorry, for the slope form, let me first ask you this question. If y equal to mx plus c is a normal, the standard form of a hyperbola, this then prove that then prove that c square is equal to m square a square plus b square whole square by a square minus m square b square. Now, do you remember I had done something called condition of normality to an ellipse. This is actually the same for a hyperbola. And how is this result different from the result that we had seen for an ellipse? Anybody remembers? Anybody remembers the, the similar relation of condition of normality of y equal to mx plus c to be a normal to the standard form of an ellipse? Does anybody remember that? If yes, can you tell me what are the, yeah, exactly. Just that your b square is a minus b square in that case. So that will actually help you if at all you want to remember it, but remembering this is not important. So if at all you want to relate the results, you can always relate it by changing your b square with a minus b square. Okay. Now, please prove this. First of all, please prove this very simple proof and do let me know once you're done with the proof. Okay. So basically we'll use the parametric form that we had just now derived. So if you remember, we had got the parametric form as by cot phi plus ax cos phi is equal to a square plus b square. Now, as per our given scenario, these two equations are same. Okay. So if these two equations are same, we can say b cot phi by one is equal to a cos phi by minus m is equal to a square plus b square by c. Okay. Now, this is more than enough for you to go ahead and just eliminate your phi. Okay. Just eliminate your phi. You are done with the problem. Okay. So how would I eliminate my phi? So from here, I can say my cot phi is obtained as a square b square by bc. Okay. My cos phi is my cos phi. My cos phi is obtained as minus m by a a square plus b square by c. Okay. Now, how do you eliminate it? Very simple. If this is your cot phi, tan phi would be bc by a square plus b square and c phi will be minus ac by m a square plus b square. Correct. And we already know that c square phi minus tan square phi is equal to one. Correct. So this is going to give me a square c square m square a square plus b square whole square minus b square c square by a square plus b square whole square equal to one. Correct. So this is simple. You just take the LCM off. This is going to become c square a square minus b square m square equal to one. So from here, you end up getting c square as m square a square plus b square whole square whole divided by a square minus b square m square. Or you can say m square minus m square b square. That's make a difference. What is important more than the result is how it is obtained because these are all your, you can say, you can say a working ground towards practicing locus base question also. So when you know what to do, when you know, okay, this is what I have to achieve to get this, basically that is important. The process, the way you make a roadmap of how to solve something, this is more of a practice to that rather than the result. Many people think that, why do I need to know all these things? I just remember the result. That is not important because the question given to you will not be a straightforward one. Maybe it will require a bit of thinking from your side where you need to know that, okay, if I want to get this locus done or this kind of a question solved, I have to write, compare, eliminate my parameters. That process itself is more important for solving questions. So let me check if I have some questions which I have. Okay, maybe we'll take this question. A simpler one. Prove that this line is a normal to this hyperbola if this condition is satisfied. Start from scratch. I mean, think as if this is a new question. You don't have to remember the result at all. I mean, if you happen to remember it well and good. But for solving this, you don't have to remember that condition of normality. Start from scratch the same way as you prove that formula. Just say done on the chat box once you're done. Okay, and the equation that we have over here is this. So as per the equation, these two should be the same because both represent the normal. Now, in order to get this condition, what is the thing that you'll be running in your mind? Okay, somehow I have to eliminate that eccentric angle phi or that parameter phi. Okay, same process. A cos phi by l is equal to b cot phi by m is equal to a square plus b square by n. Okay. So from here, I can see c phi. I'll directly write it down. C phi is going to be a n by l a square plus b square. And tan phi would be tan phi would be b n by m a square plus b square. Now use your elimination. Use your Pythagorean identity that c square minus tan square is going to be a one. And there you go. I think the problem is done here itself. Just to rearrange the terms, I think a square plus b square, the whole square divided by n square could be taken to the right side. Is it okay? Any questions? Any concerns? Any questions? Any concerns with respect to this? Next. A normal to the hyperbola x square by a square minus y square by b square gone meets the axes in m and n and the lines m p and n p are drawn perpendicular to the axes meeting at p. Prove that the locus of p is another hyperbola whose equation is given by this. Another easy question. Please solve it. Question is clear to everybody, right? Okay. So this is the normal which meets the axes in m and n. So these are your m and n. And from here, you draw n p and m p perpendicular to the axes meeting at a p. We have to find the locus of p. So if you keep changing this point, let us say the point of tangency. Okay. Let's say I call this point of normalcy as r. If you change your point of normalcy, how would this point p move? That is what you need to figure out the equation of. Okay. Aditya is already done. Shritya is already Shritya. So we'll start with, okay. What about others? Should we discuss it? Locus-based question. You should never skip. Okay. Locus basically sentence your idea of coordinate geometry. Others, Ritu, Adarsh, Amruta, Anand, Atharv, Bhoomika, Gurman, Viva Manigar, Rashmika. Okay, still solving. Okay. The best you can say, bet when you are talking about such problems is choose the parametric form. Because in parametric form, you get angles to eliminate. And for angles, there are so many identities available in the market. Not in the market in your trigonometry chapter. Okay. So let's say I call this point to be a-seq-phi, a-seq-phi, comma, b-tan-phi. Okay. So what is the equation of the normal at r? Normal at r, you can use your parametric form, ax, ax cos phi, by cot phi is equal to a squared plus b squared. Okay. Now, where does this line meet the x axis? So for that, put y as 0. So m coordinate will be, if I'm not mistaken, a squared plus b squared by a cos phi, comma, 0. Okay. Where does it meet the y axis? Or you can say the conjugate axis of this hyperbola, where you put x as 0. This will become a squared plus b squared by b cot phi. Okay. Now, looking at these two coordinates, what can you comment about coordinate of p? You say, sir, p will have the same x coordinate as m. Okay. And same y coordinate as n. Okay. So this is your, sorry. I'll just write it like this, h comma k. Then just writing h comma k. Okay. So this is your h. This is your k. Now, in order to get a direct relationship between h and k, you have to eliminate phi. That is very simple. From here, I can say, from the comparison of h, I can say, seek phi, seek phi is a squared plus b squared by h. No, sorry, other way round. Seek phi is a h by, my bad, a h by a squared plus b squared. Correct. And from here, I can say tan phi is bk by a squared plus b squared. Okay. Now use your derived Pythagorean identity, seek and square phi tan squared. Same thing that we had used to get the condition of normalcy. That is why I said the idea is important. You never know where you will be requiring it. What kind of a locus problem. You would need that approach. So that is important. So this is your equation of the, I can say, condition between h and k. Okay. And all I need to do is do just slight simplification and generalization. So generalize your h with an x, k with a y. So this will become a squared, x squared minus b squared, y squared is equal to, and this is what you wanted to prove. Any questions here? But then it's done very good. Okay. Now, all the things that we had done with respect to a chord of contact, pair of tangents, equation of a chord whose midpoint is known, they all remain the same even for hyperbola. So that generic, you can say, formula that we had seen still holds true for a hyperbola as well. So that saves a lot of time because we spend a considerable amount on these concepts in circle chapter. So that pays off over here. So if I talk about pair a chord of contact, okay, so let's say from a point x1, y1, okay, you are sketching two tangents. Okay. Okay. This is your chord of contact, COC. So chord of contact equation remains the same as what we had done for the other conics, t equal to zero. Okay. Pair of tangent equation remains the same. What is that? t square is equal to SS1. Okay. But remember, this will give you a combined equation of the pair of tangents, not separately. For separately, I had already told you what to do. Does anybody remember what to do for getting the equation? Let's say, I know this point x1, y1. This point x1, y1 is given to me. Okay. So what do you do to get the equation of the pair of tangent separately? Simple. For separate equation, for separate equations of tangents, assume a line having a slope of m passing through x1, y1. Okay. So this becomes a line mx. Correct me if I'm wrong. Okay. So this will become your C kind of a thing. Use the condition of tangency, because this is a standard case of a hyperbola. C square is equal to A square, m square minus B square. Correct. So from here, you end up getting, you end up getting a quadratic in m. Okay. So that will give you two slopes, m1 and m2. Let's say these are your two routes. I'm just giving you a roadmap for how to proceed. I'm not solving it. I'm just giving you a roadmap that how will you approach such kind of question. So once you get the equation of a tangent, you use the condition of tangency. And from the condition of tangency, you get a quadratic in m. From there, you get two routes. Just put it back over here. You end up getting the two tangents. Okay. I'm not repeating. I'm not giving any kind of a question based on this, because this is already tried and tested in our previous phonics. Okay. Any questions? Any concerns? So when will it happen to give only one real m? You have to check the condition from the fact that this quadratic discriminant is equal to 0. On the graph, if you're talking about, see, if you have, if you want only one pair of tangents, okay, then where should be the point be? Of course, it cannot be inside. If it is inside, it will not give you any pair of tangents. Okay. Kinshukh, we can also factorize the above expression. Ah, factorization is not a, you can say, time efficient work. It takes time. Okay. Of course, when we do pair of straight lines, I will tell you how to split the pair into separate lines from which they are formed. That is there always. But that comes with the cost. We will do that exercise in the pair of straight lines concept. But yes, that can also be done. Now, Aditya has a question. When will the pair of straight line become a single line? When will that pair of straight line become a single line? When the point, when the point itself is on the hyperbola. Okay. So what will happen? Let's say I'm not taking a case where you have a, you have a situation like this. There's a tangent here. Okay. There's a tangent here. And there is a tangent coming from this side. Okay. It's not, it's not very easy to basically sketch tangent always. Okay. Now here, if this point slowly goes on the hyperbola, what will happen? If this point goes on the hyperbola, that pair of tangent will just reduce to the equation of a tangent at that point. Are you getting my point? Now, why I'm saying that is because we have already seen that if I take a point on the hyperbola through that point, I cannot draw a tangent which touches both the arms of the hyperbola. Right? So what did I discuss when I was discussing with you the line intersection with a hyperbola or the condition of tangency? If I have a point on the hyperbola, I can only draw one tangent at that point only. They cannot be a pair of tangents. So just a second. Yeah, sorry. So when the point becomes on the hyperbola, you can only get one pair of tangents. So in that case, your S1 should become a zero and this will boil down to t square equal to zero giving you only one equation. And that too is nothing but the tangent at x1, y1. Okay. Now let's talk about the equation of a chord bisected at a given point. Bisected at a given point. Let me pull the figure once again. So let us say we have a chord and we have been provided with a midpoint. Let's say midpoint is x1, y1. Again, the same story repeats the equation of a chord whose midpoint is x1, y1 for a standard case or for any case. Okay. This is a generic formula. This is a generic formula and that formula is t is equal to S1 already aware to you. Okay. We are repeating it time and again so that you are well versed these with these mnemonic formulas t equal to zero, t square is equal to S1, t equal to S1, et cetera, et cetera, et cetera. Okay. Now using this concept, we'll be talking about the equation of the equation of not this one. We'll use this concept pair of tangent or whatever is this concept, the ones which we did in the previous board. Okay. So here we'll talk about through the use of pair of tangent, the equation of director circle. Okay. Maybe I'll take it enough. Next slide. What is a director circle? The locus definition of a director circle remains the same. Director circle is locus of points from which tangents drawn to that conic, in this case, the hyperbola are perpendicular, are perpendicular. Okay. So I would request you all, I would request you all to give me the locus of such points from where the tangents drawn are perpendicular. Just as a question, treat this as a question. So you're drawing perpendicular. So you're drawing tangents which are perpendicular to each other. Let's say it may be like this also. This is also perpendicular. You need not be always like this. So what is the locus of this point p, h comma k, from which tangents drawn to the hyperbola, let's say pt1, pt2, they are at right angles to each other. This is our standard case x square by a square minus y square by b square equal to 1. Please find this and let me know the result on the chat box. The method which I told you for getting the separate equation of the tangents, use that method only. That is the hint. Done anybody? Circle equation, it'll come out. Can I have something looking like a circle, okay. Let's say the tangent equations, I mean, let's say a tangent passing through a tangent through h comma k be this. That means y is equal to mx plus k minus mh. And if this line is a tangent to the given hyperbola, what is the condition of tangency? c square is equal to a square m square minus b square. c here is this. Now, as I told you, this gives us a quadratic in m. That is for an ellipse. Now it is correct. Okay, now if you see this equation is, this equation is a quadratic in m and I'll just write it down. Okay. In a proper format. Now this root is m1, m2. Let's say the roots of this equation are m1 and m2. So these are your, let's say two roots of this equation. Now, as per the question, as per the question, since the two tangents are perpendicular, m1, m2 is equal to minus 1. Because your tangents are perpendicular to each other. So m1, m2 in a quadratic, the product information is given by c by, so c is this. A is this. Okay, this is minus 1. In short, k square plus b square is equal to minus x square plus a square. In short, x square plus k square is equal to a square minus b square. Okay, now generalize it. So when you generalize it, you end up getting x square plus y square is equal to a square minus b square. Now, this is a very interesting piece of information which says that if b is greater than a, then there is no director circle. So is director circle always possible? No. So if you realize that in the hyperbola, the length of the transverse axis is more than the length of the conjugate axis, there is no such point. There is no such point from where you can draw perpendicular tangents to the hyperbola. That means there is no director circle existing. Okay, so please understand director circle need not always exist. Similarly, the director circle for the director circle for the conjugate case. Okay, this is given by similar equation x square plus y square, but just the difference is it is b square minus a square. And if a is more than b, then there is no director circle. Then there is no director circle. Is it fine? Any questions? Any concerns? Do let me know. Please copy this down if you want to. Mirror image, yes, correct. Mirror image about, now you have to tell me mirror image about what? Is it mirror image about y equal to x line? If it is that, only x and y positions will get stopped. It will not give you the same equation. No, why will y equal to x line? If it is y equal to x line, x and y positions will get stopped. So your question should look like y square by a square minus x square by b square equal to 1. But even a and b are stopped. Be careful. In fact, you can take it as an exercise to tell me about which line the stopping must be done. Okay, let's take another question. I think I had few questions on t equal to s1 also. Yeah, maybe we can take this one. Tangents quarter the middle points. Yeah, we can take this one. From the points on the circle x square plus y square is equal to a square. Tangents are drawn to the hyperbola x square minus y square is equal to a square. Prove that the locus of the midpoints of the chord of contact, the locus of the midpoints of the chord of contact is a curve given by this equation. Good question. So there is a circle. Okay, and I've taken a point on the circle. Tangents are drawn to the hyperbola. This is the chord of contact. You have to find the locus of the midpoint of this. This point can move on the circle x square plus y square is equal to a square. Simple question. Do let me know with a done on the chat box once you're done. Let's see how many of you are comfortable dealing with locus based questions. Can tangent at one arm of the hyperbola pass through its other arm? No, no, no, no, no, no. I showed you that. If it is cutting any one of the arms, it no longer remains a tangent. In fact, what you said is a tangent at a point cannot pass through the other arm. Okay, that is for sure. Maybe I'll just show you a demonstration on GeoGibro also. In fact, I've already shown this to you, but still I will just repeat the whole process. That will never happen. Never happen that a tangent is cutting another x square by let's say nine. It will never happen. Let me just move around this point at very far and also it will just become asymptotic. That means you will touch the other arm at infinity, but never cut it. Yes, is this done? Done. Aditya is done. Good, Aditya. So guys, November 7th is KVPY also for you. November 15th will be your board exams will start. I hope you are all geared up for this challenge. Okay. Please don't fall sick. And any guesses on when the notification for the first phase of JEE main will, I mean, what will be the dates for the first phase of JEE main? Any guesses? I mean, even I'm guessing. What do you think would be the logical date? Should it be as early as January? Or will it be after your second semester? Okay, Aditya says it should be in the first week of Feb, Feb mid. Okay. Jan would, Jan would not make sense. You know why you'll be having back to back exams. Just the board is over. And then again, Jan, you'll be writing here this thing. And that too, because of this one month of gap in between CBSC and ISC schools have not been able to complete the full syllabus. Okay. Till last year what used to happen? I tell you the timeline because I keep on doing every year. The syllabus for the entire class 12 gets over by November, latest by November, third week or November end. Right. Now what has happened? You people will prepare for the board exams at least 15, 20 days before your actual exam. So your teacher has to stop, you know, and I mean, your teacher has to stop at semester one topics only. Okay. Then when will she get time to complete semester two topics only after December mid when your first term is over? And how come somebody starting December mid? And let's say, even if there are three chapters to be covered, it will take at least one month or two months. Jay main cannot happen in January because of that syllabus of school only will not be over. How can, how can Jay expect a national level exam on those topics to happen? I mean, of course everybody will do badly other than the ones who have already completed it. So chances of it happening in January is very, very bleak. Who knows? Is there any notification saying there's four attempts? And maybe if the Jay and the other boards will ask you to get vaccinated also before your second semester, if at all you're turning 18, four attempts, you know, what is the problem with four attempts? They were not able to complete the four attempts as per the speculated time. So it backfired for them. For the Nata, not Nata, NT as you'd say. It backfired. They were not able to complete it. Feb seems to be okay. I mean, fair deal. But I believe it is going to be April only. The first phase will happen in April. That's what I feel. I may be wrong. We'll have to wait for the notification. Yeah, March. March year of the second semester will happen. No, just after that. That's what I am speculating. So 40% chance for me is in Feb. 60% is in April. April towards the end and last week of it. I mean, it's my guesswork, logical guesswork. No, no, no, don't think like that. More time is not good. More time means your competitors are also getting more time. So if you, if you will get time, others will also get time and people elsewhere, they will utilize that time to get a better rank. So the best deal for us is it happens as early as possible because we have covered all the topics to great depth and all. Many people are not able to cover to such depths. So the earlier, the more better it is for us because we will have an upper hand on the other competitors. Okay, anyways, this question is done. Okay, so let's take this point, the point which is moving. Let me call this point as a point R. So let point R be a sin theta or a cos theta comma a sin theta. Okay, basically I have taken a parametric form off of this particular curve. Okay, a cos theta comma a sin theta. Now, if you want to draw a chord of contact to this point, so let us say this is my x1, y1. So what are the chord of contact equation, chord of contact equation is t equal to 0. This t will be obtained from the hyperbola equation. So x, x1 minus y, y1 minus a square is equal to 0, where x1 is a cos theta, y1 is a sin theta. Okay, and of course a square, which is as good as saying a cos theta minus a sin theta equal to a. Now, let us say I assume that the equation of the chord bisected at h comma k is t equal to s1, which is nothing but xh minus yk is equal to a square. Oh, sorry, is equal to h square minus k square. Okay, t equal to s1. Please remember, s1 is when you substitute h comma k in the equation of the hyperbola. I should have widely written minus a square minus a square at both the positions, but you know they are going to get cancelled off, so let us not waste time. Now, as per your given scenario, these two equations are same. That means the chord of contact and the chord whose midpoint is h comma k they represent the same equations in this case. So the same methodology we will be adopting, we will be comparing these two equations, we will be comparing the coefficients of these two equations. So cos theta by h is equal to sin theta by k is equal to a by h square minus k square. Now, what do you have to do here? You have to eliminate theta because this point, this point is moving. So this theta should not be appearing in your answer because theta is a parameter, it is changing. Now, it is a child's play to basically eliminate theta. You can use your Pythagorean identity. Okay, so cos square theta plus sin square theta is equal to 1. You can use it. Okay, so you will get a square at square by s square minus k square whole square plus a square k square whole square equal to 1. So I think it gives us a square, I mean I am just writing down back in terms of x after generalizing. I do not want to waste time doing it. So is this the equation which we want to get a square? Yeah, they have taken it common. Yeah, correct. Okay, so this is what we wanted to prove. Is it fine? Any questions? Any questions? Any concerns with respect to this? Now, next we are going towards the concept of diameter and conjugate diameters. We can just put x square plus y square is equal to a square. Here if you want, here you want to put. Why? This is that h comma k point. How do you know that this h comma k will also lie on the same circle? This is the locus of this point. H comma k is not this point. H comma k is that bisector of the chord. That cannot be written as a square. Yeah. Anyway, so the last, I mean not the last concept for the topic but the last concept for the day. We'll spend some time talking about the diameter and conjugate diameters. Okay, diameters and conjugate diameter. Let's see how much we are able to cover up with this concept. If you're not able to complete it because there's some properties also to be taken up, which can, we can take it in the next class along with rectangular hyperbola and asymptote concept. So in this topic, what is left? The concept of rectangular hyperbola and asymptotes. And I think next class is more than enough to complete it. Maybe next class, we can also start with the concept of pair of straight lines. Okay, we'll see how much we are able to complete. So what is the diameter? First of all, the definition of a diameter is already evident in our last chapter of ellipse. So it's the locus of the middle points of a system of parallel quads. Okay, so it's the locus of the middle points of a system of parallel quads. Okay, now using this definition, I would request you to give me the equation of a diameter which bisects a system of parallel quads having a slope of m. So let me make a system of parallel quads. Okay, I will not be able to make a lot of them, maybe a few of them. Yeah, so let's find out the equation of, the equation of, maybe if I make it like this, it'll be more. So equation of the locus of the midpoints of this set of parallel quads, maybe. So this is your diameter, which is passing through the midpoints of these parallel quads. So these are all quads which are parallel and having a slope of m. So please give me the equation of the diameter. Very easy. Hint is, use the t equal to s1 concept, the equation of a chord bisected at a given point. And let me know with the done on the chat box. I mean, I have purposely made it pass through the center, but I'll leave up to you to figure out from the equation that whether it is actually passing through the center or not. Just don't go by my diagram. Should not take more than 30 seconds. I don't know why you're taking so much time. Are you? Nobody? Okay, let's say this point is h comma k. So what are the equation of this? You can say the chord, okay, whose midpoint is h comma k. So it's a t equal to s1, correct? So t equal to s1 is xh minus yk is equal to h square minus k square, correct? Now this, this is basically the equation of any one of the quads, okay, equation of one of the family members, one of the parallel quads. And you already know that this, this should have a slope of m, right? So what is the slope? What is the slope from this equation? I'm so sorry, I missed out a b square, a square b square, sorry about that. a square b square. I was still in the previous problem, actually, there was only a square, that's why I missed out. Yeah, so this is the equation, this is the equation of one of the parallel chords. So what is the slope here? Slope here is minus a by b. So that is nothing but correct me if I'm wrong, h a square, sorry h b square by k a square, correct? k a square. And this must be your m, this itself is the equation, right? Generalize it, generalize it by putting h as x and k as y. So this itself becomes your equation of your, okay, so you can write it as y is equal to b square x by a square. Again, one very big similarity with that of the ellipse. In ellipse, do you remember what was the equation of the diameter? Recall in ellipse, there was a minus b square, right, right? So see, again, this is what I told you long back ago that there are a lot of connection between ellipse and hyperbola, where if you just replace your b square with a minus b square, you get a lot of your work done, okay? And this clearly passes through the center, okay, passes through origin, that is the center of the hyperbola. And it is true in general also, okay? Any diameter will always pass through the center of the hyperbola. Is it fine? Any question with respect to this? Now tell me what was stopping you from solving it? I got no response. What happened? Okay, so talking about conjugate diameters, what is this? Maybe I will not be able to, okay, conjugate diameters. So what is the concept of conjugate diameters? We talk about it. May not be able to do a lot of things here. So two diameters are said to be conjugates of each other. Two diameters are said to be conjugates of each other. Definition is the same as what we have seen in an ellipse. Conjugates of each other. When each by six, all chords parallel to the other. Okay, so I'll just take a case here. Your space is not there. Give me a second. I'll just copy paste this to the next slide. Okay, so let us say I have this standard hyperbola with me, okay? And let me just erase this. Directesis is coming in here. Okay, so I have this standard hyperbola with me, okay? And let me just erase this. Directesis is coming in between. So let us say this is one of our diameters. This is one of our diameters. Now, if I make all chords parallel to this, if I make all possible chords parallel to this, okay? There would be a chord, there would be a line which will be passing through the midpoint of all the chords which are parallel to this diameter. Okay, and let me make it in pink color. Okay, that would be, let's say a line like this. Okay, so this line, pink line is bisecting all the chords. It's bisecting all the chords. This is not looking even parallel. Okay, so think all of these are parallel. Okay, so if you connect the midpoints of all these chords, you'll get another diameter. So this pink line and this green line, they are conjugates of each other, pair of conjugate diameter, pair of conjugate diameters. Okay, now just a question for all of you, a very small question, I would like you to attempt. Let's say this line is y equal to m1x and this line is y equal to m2x. Okay, prove that if m1 and m2 are the slopes of the pair of conjugate diameters, the product of m1 m2 will always be b square by a square. And again, very similar to what we had done in case of our ellipse. In ellipse m1 m2 was negative b square by a square. Here it is b square by a square. That b square is replaced with a negative b square to get this condition. Okay, so let's prove this first of all very simple and let me know with done on the chat box that you are done with this proof. Dancetej is done. Sorry, Kinshukh is done. Just needed the previous formula exactly, Kinshukh. So what we can say is that if let's say there are all chords parallel to the pink one. Okay, let's say y equal to m1x by 6 all parallel chords with slope m2. Correct. So ideally the equation should have been y is equal to b square x by a square m2. Right. Correct. So this behaves as the slope of this line, which as per your question is m1. So m1 is equal to b square a square m2. So that means m1 m2 is equal to b square by a square. Okay. Is it fine? Now I have a question for all of you. Just think about it. Maybe we'll discuss this in more detail in our next class. Do you think the pink diameter, which is the conjugate of the green diameter over here, will it ever meet the hyperbola? See, I'll just give you, let's say pnp dash. This lies on the hyperbola. Correct. So will this, let's say I take some point here, q and q dash. Do you think q and q dash can ever be on the same hyperbola? Kinshukh is saying, okay, I will not give out his answer. What do you think? No, no, no. Is it intuitively or is it any mathematical observation that you did, which makes you believe that? That's intuitive, I believe, Kinshukh. It gives a higher slope and the word higher, etc, they are all m1. Okay. Okay. Now, the answer to this is very simple. I'll do it mathematically for you to understand it. Let's got the maths behind it. Okay, great. So in the interest of time, let me take it forward. See, let's say, let's say y equal to m1x and y equal to m2x, they are the equation of the pair of conjugates, pair of conjugate diameters. Okay. Diameters. Okay. So let's say this point, this line meets, this line meets the hyperbola. Okay, this meets the hyperbola. That means this diameter is basically your green line, the one which I've shown in the figure, green line y equal to m1x. Okay. So let us solve it simultaneously with the equation of the hyperbola. So y square, I will write it as m1x square by a, sorry, b square equal to 1. So this is nothing but if I just solve for x square, I'm just skipping some steps, it will be something like this, b square minus a square m square. Correct. Are you getting my point? Okay. And let's say at the same time, the other guy is also meeting the hyperbola. I mean, I'm just assuming that the other guy is also meeting the hyperbola. As of now, I'm not sure whether it'll actually meet or not. So similarly, similarly, let y equal to m2x meet the hyperbola, meet the hyperbola, hyperbola. If at all it meets, if at all it meets. Okay. I mean, this is just subject to my further mathematics. So we'll see what are the mathematics behind it. In the same way, x square should have been a square b square by b square minus a square m2 square. No difference, just m1 and m2 are same. Okay. Now please understand here, this is a positive term. This is a positive term. So ideally, this should also be positive. Should be positive. Correct. Am I right? y equal to m1x is the green line, Shradha. And y equal to m2x is the pink line. This is your pink line. And this is the green line. Okay. This is your green line. I mean, green and pink as per the diagram, can you show that this diagram which happens? Ayurama, I'm so sorry. It's orange. Sir, you are colorblind, sir. No, I have no shame in admitting that yes, I'm colorblind. Thank you, Kinshuk. Yeah, you're right. It was actually orange line. Okay. Now see, this is positive. It means b square minus a square m1 square should be positive. Correct. Now let's try to do some maths behind it. Correct. Which means m1 square should be less than b square a square. Okay. That means m1 should be less than b by m1 should be less than b by a. Correct. Now please understand here that m1 and m2 are related by this relation. m1 m2 is b square by a square. Correct. So can I say m1 is b square by a square m2? So this m1 I will replace with this term. Correct. Yes or no? So here you end up getting m2 is greater than b by, correct me if I'm wrong. Yes. That means that means m2 square is greater than b square by a square. And in short, you're getting something like this b square minus a square m2 square is negative. Now see here, if this is negative, let's come back. Let's come back to the right side x1 x square expression. Let's come back to this expression. So here what you'll see that this guy is a negative. Sorry, negative. This guy is a positive and this guy ideally should be positive if they're meeting at the real points. But this is not possible. How can it happen that a positive quantity is equal to a negative quantity? That means this assumption that they will meet itself is false. That is the ends of the conjugate of y equal to m1 x will not lie on the original hyperbola. In fact, we'll discuss it in the next class. It will actually lie on the conjugate of that hyperbola. So if one diameter is connecting two points on this hyperbola x square by a square minus y square by b square equal to one, the conjugate of it cannot touch this hyperbola. It will in fact touch the conjugate of this hyperbola, which I will discuss in the next class. Don't worry. This is on our cards in the next class. So with this, we end up today's session. Can both of them lie on the same hyperbola? Are you talking about the vertices of both the pair of conjugate diameters? The answer to that is no. One at a time, either pp dash lies or qq dash lies. Both all the four vertices, in fact, the pair of vertices cannot lie on the same hyperbola. Is that what you wanted to ask? No, no, no. The diameter, one of the diameters will definitely meet the hyperbola at two positions, which we call as the vertices of the diameter. Kinship, what's your question? Let me just end the recording.