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.  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?  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.  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.  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.  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.  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 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.  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.


Are you for waiting until the last second to merge in traffic? Usually? Yeah. But there is the situation where it's actually the right thing to do. It's called a zipper merge, and it's a road design that encourages drivers to merge as their lane ends. Many states now recommend using zipper merges as a more efficient way to guide traffic during road work. It decreases traffic delays by up to 40% and reduces accidents. But shifting lanes ahead of the merge point to be polite puts all the cars in one lane, which means traffic slows down to a crawl. And even worse, those that are doing it right get me mugged by the other drivers who won't let them in. There are plenty of circumstances where you're in the wrong for merging at the last second, but not at zipper merges.


It was a really good zombie movie, believe it or not, because they have all these zombies and then they had these other ones that were like these vicious skeletons and the zombies were just kind of out of it. The zombies reminded me of many ways of human culture because they were like the walking dead and they didn't have a lot of emotions and they didn't really articulate them. The thing is you're grown and grown, you know, and then it started out that way and then this girl gets a real human, gets taken by all the zombies and she's over there and she gets befriended by a zombie who is drawn to her humanness, to her likeness and it's a love story. It's a zombie love story. This is what I mean by the Holy Spirit's use. If you can use a zombie movie for waking up, you can use just about anything and the Holy Spirit can use anything in this world. So he starts to fall in love and she's freaked out by that.  He could hardly speak words. I don't know how attractive that would be.  And it's really not good with articulation. But amazingly, he helps her and she feels the zombies desire to be helpful and actually she draws him into communication. You know, he's able to come to life a bit with his communication.  You know, we know we can have swear words, we know how to scream at somebody, how to yell at somebody and the Holy Spirit's like, that's just mmm, mmm, to the Holy Spirit. It's not exactly helpful. We have to learn how to let the Spirit use the words, you know, and then we have to have a demonstration in our lives to back up the words, otherwise people won't believe anyway if we're not in integrity with our words. The humans who are afraid of the zombies, because they thought the zombies wanted to kill them, they had to go through a change of mind, too, and learn that they were coming alive and that they weren't to be feared anymore. So it was a whole movie about dropping the fear and learning to trust and starting to see the sameness. And eventually all of the skeletons, the kind of vicious skeletons were eliminated or dissolved away and everybody came back to being human. So that's a metaphor of coming back from being walking dead to coming alive. And those are the kind of metaphors that are used in the Course.  When your mind is happy, then there'll be more smiling going on, less frowning. He said, your forehead is serene. What's great? That's a symbol of a serene forehead instead of a frowning wrinkly forehead. It's a serene forehead. That's the way Jesus is talking. There's another time in the Course where he says, a smile has come on your sleeping face.  But a smile has come. He's describing a sleeping face with a smile on it as you're starting to have a happy dream. You're starting to forgive.  These are all beautiful metaphors using again the symbols that people relate to. You know, a smile you can relate to, a calm serene forehead you can relate to that. Those are the metaphors. And he also does guided meditations where he's asking you to go down in your mind and you sink into your mind through the clouds and it's so beautiful.  So you've got this face going down and through these cheeks. The cheeks can feel the moisture of the clouds. You go down. It's beautiful. He just has all this imagery. And so that's another good way we use the images in a way that is guided by spirit to take us beyond the images, you know, to take us back to the state of mind.


Hello and welcome to Do Try This at Home, brought to you by the Intitutor Physics. My name is Mikey and this is my front room. We're making these films because we want to help parents and carers in the UK and Ireland and all around the world to get their kids and families excited about science without even leaving the house. In this video we're going to show you how to make a super simple rubber band bass guitar. I love doing this one because it's so simple and easy to do and uses just one household item. A rubber band. But if you have more than one you can help everyone do it at the same time. But you might want to demonstrate how to do it yourself first. Stretch the rubber band between your thumb and the forefinger of one hand. If your family have smaller hands you might want to find a few smaller rubber bands for them to do it. Then you want to get a little bit of tension.  If you play around with tension and less tension you can vary the pitch of the sound. You could even try and play some famous bass lines like my terrible rendition of Seven Nation Army. Once you have your family trying it along with you I've sent them to take the finger out of their ear but keep twanging and pay attention to the difference in what they can hear. So what's actually going on here? Well all sounds are vibrations and when you twang a rubber band it makes it vibrate. And you can hear this even without your finger touching your ear because the vibrating band causes the air around it to vibrate and the air carries the vibration as a sound to your ear. But when your finger is touching your ear it's much louder and that's because solid things like your finger are much better at carrying sound vibrations than gases like the air. So with your finger touching your ear you hear a much louder sound from the rubber band. So that's how you make a rubber band bass guitar. If you're not totally sure about the science don't worry just have a play around together have fun with your family and if you need to look it up together. Thanks a lot and see you next time.


What's up everyone? This is Lauren from Lauren Learns. I wanted to do another video today on my bread machine. This is not the first of this series because I have a secret love affair with this machine and as you can see it's pretty well loved and beat up a little bit but I got it for three dollars like three or four years ago I don't remember and like I said in my first video which I will link up here I just I make everything in it and so today I wanted to show you what we do when we want to make some homey and hamburger or hot dog buns. We're having some friends over later for dinner and I don't know this is just my twisted thinking but it's easier for me to just think I have all the ingredients at home I'll just make buns at home than just buy a bag at the store to me buying bread when I already have the ingredients seems wasteful. So as you can see behind me there are some hamburger buns waiting to be put in the oven and the recipe I'm about to share with you makes about I guess enough for like 12 good size buns. Okay so I realized after the fact that I could have showed you all the ingredients that I had poured into my bread machine. Do you love the lovely music behind me? I know it's good. Okay so for this recipe you're going to need three and three quarters cup of flour, one quarter cup of milk and actually I use our almond milk in it so you don't have to use dairy if you're not if you're not eating dairy, one quarter cup of sugar, one egg, two tablespoons of butter, one quarter cup of yeast that's way too much yeast don't use that much. What I meant to say was one that's funny one and one quarter tablespoon of yeast and three quarters of a teaspoon of salt and you just dump it all in the machine I could have showed you that but I don't want to insult your time because you know what that would look like and you push the dough cycle on any bread machine takes about an hour and a half and then once the dough cycle has finished you take out the dough and divide it and form into whatever you want so like today I did hamburger buns um otherwise this recipe makes great hot dog buns or hoagies for sandwiches and then we just kind of keep them around for the week I don't have to buy any bread anymore so it's pretty sweet you could also um obviously make this very customizable if you're not eating white flour or if you'd rather eat a whole grain flour um so yeah I will go ahead and list that down in the description below but I've also given it to you in the video so hopefully you can recreate this at home I highly recommend it it's very soft yummy delicious bread and your house will smell amaze balls so yeah have a great week glad to share another tip with you things I'm always learning things I'm always trying and I have to go as always go get messy have fun and learn


I'm gonna keep you tight.  All right. I went to England, yeah. Right. I'm going to go to Premier Reserves. What is this? What is... There's not a bun. No. Jota! I'm gonna steal it for you so you can open it. Grab the rope. Come on. To reserve. Here. Good. What's it on? No, it's over here, man. Yeah, I know what it is. Hey! Hey! How's it going, George? Good. He's good. He's rad.


The ability of fungi to adapt to different environments is known as metabolic flexibility. This is particularly important for candida albicans, a common cause of life-threatening infections in humans. The fungus can survive in various host environments by activating an enzyme called alternative oxidase, AOX, which is induced under certain conditions such as when the fungus is exposed to respiratory stress. Deletion of the AOX2 gene, which encodes the main AOX protein, reduces the fungus' ability to infect mice, suggesting that it plays a key role in its virulence. Researchers have identified several transcription factors that are involved in regulating the expression of the AOX2 gene, including RTG1, RTG3, 181, ZCF11, and ZCF2. These factors interact with the AOX2 promoter to control the expression of the gene and thus determine whether or not the fungus will be able to survive in a given environment. This article was authored by Shungla Liu, Pauline Basso, Saif Hossain, and others. We are article.tv, links in the description below.


You may have heard of the story of Suicide Mouse, a disturbing 9 minute and 4 second long lost episode of the 1920s, 1930s version of Mickey Mouse. The last 30 seconds are very unknown, but what is known is that there is Russian text on the last frame and the final half minute has caused a death. Well, it has been found that on 8.34, it cuts from the Mickey Mouse end screen to Mickey, who has no eyes and his smile is perked, frantically running in this warped void. Yes, between 7-8-4, the floating rubble the buildings have become turned into said void and the sidewalk moving in impossible directions seemingly merges with it, though very unnoticeable, because of how distorted the video has become. Mickey seems to be going faster throughout the duration. Back to the final moments, there seems to be a dawn floating and when the mouse approaches it, he stops running. He slowly but surely picks it up, turns towards the screen and puts the gun to his head, throughout the duration of him saying, real suffering is not known, 7 times, each time getting gradually louder. At around 8.54, the screen turns white, and a gunshot sound happens, this is pretty much what the guard did when he saw the ending. On 8.55, you can see that the buildings and the sidewalk are back to normal, but the mouse is lying down, only his shoes being visible.  On 9.1, it suddenly cuts to a close-up of. McGrawl mouse. It's occupied with a loud demonic screeching noise played for around half a second before it cutting to the Russian text for the remaining seconds. It's also worth pointing out that on the very first frame of the video, on the very left, you can see a syringe lying on the floor, and on 4.32, where the video is mostly nothing, McGrawl appears for a split second before disappearing, this time with no screeching. There are also random weird noises throughout 1.32.6, probably of doors closing opening or items being picked up. Overall, Suicide Mouse probably wasn't even made by Disney in the late 1920s, early 1930s. It's almost like some Russian bootleg that probably got exported to the U.S. at the time the original story was written. We may just never know.


Hello, and thank you for coming to KubeCon EU 2021 Wasm Day, where we've been spending time at the intersection of WebAssembly and everything cloud-native. A tremendous thank you to Chris A., Eleanor Vogel, the CNCF staff, and my fellow Wasm Day program committee members for putting on an incredible event. I'm Liam Randall, and for the last couple of years, I've been working on a distributed application framework for microservices that leverages WebAssembly called WasmCloud. And today I'm going to share our perspective on WebAssembly as the future of distributed computing. Let's jump right in. WebAssembly brings some powerful value propositions to any domain, distributed or not. People are excited about WebAssembly, sometimes referred to as Wasm, for a lot of reasons. And while the future of WebAssembly is simply dazzling, today it already brings a lot to the table. First, it's efficient and fast. It runs at near native speed. Second, it's safe and secure, not just sandbox, but it operates in a denied by default mode where we need to explicitly grant capabilities. Third, it's open and debuggable. Fourth, Wasm's polyglot, choose your own language and you can bring well-tested libraries to a new domain. Finally, it's portable. WebAssembly is a compilation target and it runs on lots of different things, servers, browsers, embedded devices. Heck, you can even put Wasm inside of other applications if you wanted to. Next, let's introduce the modern distributed computing environment. The Linux Foundation Edge has helped to put together this continuum, a view of the world we can use as a starting point. This view of the world helps us to frame the complexity of the modern distributed computing environment. There are many different device types with a variety of system architectures and capabilities, but these categories actually undersell the complexity a little bit. Because if we annotate this view just ever so slightly to include some additional context with some of the platforms that are both incumbent and common in the ecosystem, we get a more complete and complicated world view that we need to account for. For example, on the service provider and public cloud edge, we of course find Kubernetes in both regular and microsizes and we find applications that are hosted on top of Kubernetes. We should include developer environments in our ecosystem as well. And certainly, if we're including mobile devices, we should have phones and lots of browsers and smart TVs and all the rest of the IoT. Even the really small things with teeny tiny capabilities. So let's pair up this new environment with our current evolution of technology. Now, over the last 30 years, technology has continued to evolve over two primary dimensions, the formats in which we deliver applications and the orchestration of those formats into a specific environment. Now, this picture, all the green is provided by the developer building the apps and the blue is the platform layer. And while 15 years ago, the launch of the public cloud ushered in its own revolution, the launch of Docker in 2013 and Kubernetes in 2014 accelerated the pace. And now today we find ourselves on the cusp of yet another revolution with WebAssembly. As we can see over time, these formats have gotten simpler and more decoupled from the underlying environments. We're moving from a tightly coupled world on the left to a more declarative world on the right. And the developer responsibilities continue to be reduced further by decreasing the number of things that they're responsible for maintaining. Let's put up some comparisons and see what really stands out. If we start to line these technologies up side by side, we start to uncover the patterns that emerge in the ecosystem. Each progressive wave has further to covered applications from their underlying architectures. That is to say with each format, we've carried fewer and fewer assumptions forward into the landscape. And we see this reflected at both the where our formats can run and in the security boundary. With security in WebAssembly, we've moved to an entirely sandboxed approach and we'll review the deny by default explicit granting of capabilities shortly. And the where is part of the reason why we believe that WebAssembly will come to dominate distributed computing. In fact, let's take these two world views and put them together. By observing what we are already seeing across today's ecosystem, we have compelling evidence that Wasm today is on the path to becoming the critical technology leveraged to deploy software across the distributed continuum. As examples, starting on the far right, we see native execution of microservices on servers with frameworks like Wasm Cloud. In research, we are already seeing WebAssembly as a native orchestration technology for Kubernetes with Microsoft's Crosslet, the Kubernetes Rust Kubelet. We see Wasm inside embedded inside critical technologies. For example, both OPA and Envoy are leveraging Wasm to execute untrusted third-party plugins where previously they might have chosen Lua or JavaScript. On the edges, we see organizations like Fastly and Shopify using Wasm as a deployment platform to run third-party untrusted code. If that pattern feels familiar, get used to it. I think a lot more organizations are starting to recognize that all code should be treated as untrusted code. We see thousands of examples of applications running in browsers on mobile phones, on tablets, and in web browsers everywhere. Google Earth uses Wasm. So does Microsoft Flight Simulator. Microsoft even makes a toolkit that leverages WebAssembly in the browser called Blazor. The anecdotes are one thing. What are the real challenges of this distributed environment and how well is Wasm actually stacking up? Now, this is not an exhaustive list. However, today I will highlight five critical challenges to be solved in distributed computing. First, the diversity of system architecture, which is that as cloud-native community, we are no longer targeting just the x86 monolith. CPU, memory, functionality, capability, and OS are all invariable now. The second great challenge will be the diversity of application architectures. How will these things be connected in the near future? What design patterns will they use? How will they speak to each other? The third will be our approach to distributed security. Fourth will be offline use. This incredibly diverse ecosystem will be running loads of complex software. And finally, bringing machine learning to the end devices to make them intelligent.  Both the number and incredible diversity of devices connected to the internet continues to simply skyrocket as we head towards an estimated 50 billion connected devices in 2030, the enterprise, the enterprise integration tasks are compounded by a dozen popular and incompatible CPU architectures, each of which has its own relatively unique operational lifecycle from dozens of major manufacturers. WebAssembly is already well positioned to operate well in this complex environment. Wasm runtimes can even be optimized to adapt an application for specific circumstances. For example, the bytecode alliance is wasm time for servers and their wasm micro runtime or whammer for short has been optimized for smaller embedded devices. Surely these vast collections of distributed and connected ecosystem systems will be interconnected in tremendously different ways and tier applications, model view controller, pub sub, microservices and more. Today we already see WebAssembly running on embedded into and running around our software stacks. Wasm excels here as well, its near native performance makes it ideal for standalone embedded or other types of integrations. As a component, we already have many large scale examples of WebAssembly running in complex stacks. Now the security of IoT is a broad topic. However, the native security design of WebAssembly is a welcome addition to distributed projects running anywhere in the continuum. WebAssembly itself functions as a sandbox, helpful when you want to adapt or bring code to a huge variety of complex systems. And WebAssembly brings to the ecosystem a powerful approach to capability based security. If you think back not that long ago when you started a child application, it usually inherited the rights of the parents. So if I could write or read to disk, then so could my child process. In more recent years, however, nearly every industry has started to migrate to a capabilities based approach where an application must be explicitly granted access to a capability in order to use it. With WebAssembly, the combination of a sandbox and deny by default approach means that is more difficult to abuse and compromise your applications. Now it is easy to think that when we're connecting billions of devices to the internet that we can simply offload all of the functionality to remote API services and that these could be dumb endpoints on the end user side connected back to an ultra smart core or that the edge moves the capabilities closer to the users and minimizes any performance impacts. While these are two valid and common architectures there are many compelling reasons to believe that most devices will be both connected and still offer some level of functionality when offline or during intermittent connections. Real complex logic will need to live on the device because for a few reasons. First, latency and determinism. For example, driverless cars need to react instantly. Second, large data must first be processed where the data resides at the devices. Perhaps an ML model extracting metadata to be shipped back back to the public cloud. Third, privacy and security will keep logic at the edges. Some data shouldn't just shouldn't be shipped. Fourth, limited or deliberate autonomy. Some devices will need to be able to operate under all conditions without connectivity. Ships far out at sea. Rovers, drones, even refrigerators. Finally, regulatory. GDPR and CCPA are the start of things to come. Big tech is going to be further regulated in the future and this will impact the where and how in the design of our distributed software systems. What this means is that these 50 billion connected devices by 2030 will amplify our workloads. These devices may be small, they may be cheap, but they will be increasingly intelligent and complex. For the aforementioned reasons, web assemblies right once, run everywhere, security model and portability are sure to come into play. Finally, one overarching theme of IoT will be making all the dumb things in our life smart. Tuning and optimizing models is difficult enough without having to worry about the differences between hundreds of potential different deployment chip sets. Even today, incredibly compelling work is coming out of the Google TensorFlow.js team where they are compiling ML models for use in the web browser, running and leveraging the speed and performance of web assembly in the browser. And Andrew and Son over on the Intel team have already ported their Wazi NN machine learning framework to Wasenton as an experiment. I can imagine a near future where these models are run throughout the distributed continuum. I hope you have found today's talk about distributed systems interesting and informative. If you care about these topics as much as we do, I hope you'll follow along as we build an enterprise solution to help developers navigate these complex opportunities. Join us working on WasmCloud, our open source distributed application runtime for the enterprise. Or for more information on future enterprise offerings, sign up for our early adopters program at Cosmonic.com. Thank you very much and have a wonderful KubeCon.


Hello everyone, I welcome you all for today's class. We will briefly look at what we discussed last time and then proceed further. In the last class, we saw how the oxazolidinones were used for the formation of Z enolate and how the Z enolates are then allowed to react with the alkylating species. And the oxazolidinones are basically derived from different amino acids and the configuration of the substituents on the auxiliary such as this for example here topropyl group it determines the orientation of the incoming electrophile when the enolate is reacting with the electrophilic species. And we also saw how the aldol products are formed in the formation of the aldol product that we saw that the oxygen metal bond which is boron enolate that is metal here is boron how does that form its kind of intramolecular chelation to make this bicyclic system here. And when the aldehyde comes in contact with this particular bicyclic enolate chelated species then the oxygen boron bond breaks and the oxygen of the aldehyde comes in contact with the boron and therefore we see that Zimmerman-Traxler transition state and then we see the aldol product formed. And during that process we also looked at how the dipoles of the this enol part and the oxygen carbon part of the auxiliary they ripple each other when the aldehyde forms the chelation with the boron and therefore the orientation goes back and then that allows the specific orientation of the aldehyde where our group is equatorially oriented followed by then the attack of the enolate from the C phase leads to the product formation which is synaldol. Now when the synaldol is formed we had no surprise because that is what was expected but how do we get then anti-aldol for the anti-aldol formation we saw that if we use extra Lewis acid then that extra Lewis acid does not allow the aldehyde to kind of break the enolate of the boron and the chelation remains intact as it is here and then the Lewis acid interacts with the aldehyde and disposes the aldehyde in such a fashion that the carbonyl of the aldehyde and this particular enolate they are away from each other this particular carbonyl group of the oxalidine oxazolidinone part they are away in the transition state and that allows the formation of the anti-aldol. Then more importantly towards the end we try to look at the enol silyl ether to be formed from a substrate like this and see what is known as Ireland Claisen rearrangement. So, I have taken up this Ireland Claisen rearrangement first instead of taking Claisen rearrangement because in Ireland Claisen rearrangement there is a very strong influence of the formation of the geometry of formation and geometry of the enolate and therefore I thought it was relevant that we do discuss it right along with this particular boron enolate part. So, in that case what we do is if we start with this and we carry out a deprotonation here. So, basically it is allyl ester which is then converted into the corresponding ester with the possibility of generating an anion here. So, you have 1, 2, 3, 4, 5 and 6 and therefore you have a possibility of a rearrangement which can occur in this fashion. So, this is how the reaction takes place and as you can see the conditions are very stringent conditions. The LDA reacts with this particular compound or the ester in the presence of tertiary butyl dimethyl salyl chloride but the solvent is THF and if the THF solvent is used what you can see that this is the new part of the enolate that is formed and the R group which is already present here they are anti to each other or a trans to each other. But if we keep everything else same but add extra HMPA which is hexamethyl phosphoric triamide which is having this structure you have ME, ME2N thrice PO. This is what is hexamethyl phosphoric triamide this is what is HMPA. So, this particular solvent has the nitrogens there are 3 nitrogens and they play a very crucial role and in that situation as you can see here the OX group and the R group are cis to each other. So, since now the geometry of the enolate has been achieved by different conditions they lead to different products as you can see here the R group and R1 group are cis to each other and here R and R1 group are trans to each other in the final product. So, the newly generated asymmetric centers are highly influenced in both the cases by the geometry of the enolate and therefore we need to understand why is it so. First of all we will see how this affects the product formations. For example, if we take a case like this kind of particular substrate here where the geometry of the double bond here is trans and then of course we have the corresponding ester here. In this case we have taken the geometry of the double bond as cis. So, if we start with the substrate like this and you carry out the reaction in THF as we saw in the last transparency and then we get the product which is cis product that is in THF. The same reaction if it is done in THF HMPA here we get the trans product here. So, the methyl orientations are trans to each other. This is not really a cis and trans in terms of the double bond but it is just a way of defining that the methyl groups and methyl groups are basically towards the same side. Similarly methyl group and methyl group here are opposite side therefore it is trans. It is not something to do with the double bond cis trans or so. So, we start with the trans double bond and we get a product which looks like this and if we do in THF HMPA the product is opposite to this. And with this cis double bond here if we carry out the reaction in THF we get this product and opposite of that we get in THF HMPA. So, there is a enormous amount of difference in terms of the solvent effect. So, as you can see in summarize here E double bond in starting material leads to EE transition state and leads to cis products. So, E this is E and double bond in starting material leads EE transition state. So, obviously this and this are away from each other. So, this is also E and therefore EE transition state leads to cis product. And if we take Z E double bond in starting material leads to ZE transition state. For example, here this is E and if it is THF HMPA then we get Z here and that leads to trans product. Now, you can see here that when the the rearrangement takes place here, this is the bond that is going to form here. And therefore this methyl is orienting up, this methyl is equatorial in the transition state. This is axial, this is equatorial. Therefore, axial equatorial on adjacent atoms will be cis to each other. And here as you can see that this is equatorially oriented. And this is also equatorially oriented. And therefore the two equatorial groups on adjacent carbon atoms in the cyclic transition state that will lead to the formation of the product would be trans to each other. Therefore, in the transition state as you can see trans product is formed here is going to be a cis kind of orientation of the two methyl groups. So, that is how it leads to the formation of the cis products here and the trans products here. That is what we are basically trying to refer to. So, it is very clear that with various kinds of product distribution that we can see with different stereochemistry of the double bond and different solvent systems we can then expect the specific cis or trans orientations of the methyl groups in the product methyl groups in this case, but there could be different substituents. So, we have to see why is this THF and THF HMPA this combination helps to make this particular enolate to be cis or trans or E or Z. Now, here as you can see that the orientation of the groups would be something like this where you have a carbonyl group here and of course you have the oxygen here and then you have this particular here you have a CH2 and you have a methyl group here and then of course you have a hydrogen here. Now, when the LDA that lithium diisopropyl amide comes into this, you are basically trying to have the anion from the N lithium here first chelates with this oxygen here. This is what is the is required here and then of course we have the, so something like this can be anticipated. So, lithium coordinates with the oxygen here and then we can write it here like this and this chelation allows the orientation of the hydrogen here to be in this direction of course. Now, at this stage the hydrogen can one the other hydrogen can be on this side or methyl can be here or vice versa that means methyl can be as we have seen it here it could also be methyl here or hydrogen here. So, in this particular case what is happening is now that in this such a transition state is required in THF where intramodular chelation is possible and then interactions of the carbonyl versus methyl and carbonyl versus hydrogen are there for important. Now, if we look at this transition state we can also look at and put it in this fashion and we can say that something like this is happening and we could put it in this fashion that we have a lithium here which is now chelating with this then you have a nitrogen here when you have isopropyl lithium diisopropyl amide and then you have a hydrogen here of course and then here there could be a hydrogen here and the methyl here is one possibility of course from here you will have this. So, this is the transition state that this is what I am referring it here. Now, as you can see that in this particular position when the hydrogen is axial there is not much of steric hindrance. But if we try and look at the put it here and instead of these two if we put the methyl group here and the hydrogen here then of course there will be steric hindrance. Therefore, this chelation which is allowing the orientation of the hydrogen to be in this direction leads to the formation of the E enolate. So, this E enolate is formed because now the double bond is being formed here and therefore the oxygen here enolate comes in here and the methyl are about opposite to each other. So, the intramolecular chelation with the LDA that is possible in THF makes the transition state look like this which is what is here in which the hydrogen prefers to be in the axial position and therefore, the methyl group is in the equatorial form and when the chelation occurs then the final product after deprotonation leads to the E enolate. Now, what happens when we have the HMP added to it when HMP is added to it this particular chelation breaks this particular chelation breaks and this particular chelation breaks here and the molecule remains free there is no chelation at all. So, there are two orientation that are possible one is like this here the and the other one is like this here. So, now we have two possibilities one the in the transition state when the proton is being abstracted from here the methyl can have orientation like this where you can expect the steric hindrance like this or like this where there is no steric hindrance there is no steric hindrance here there is a steric hindrance here. Therefore, this when methyl group is orienting towards this side it is upon deprotonation leads to E enolate on the other hand when the methyl group is on this side away from this large group here then of course, the deprotonation leads to Z enolate this is what happens because HMPA basically with the its nitrogen as I mentioned earlier it has three these nitrogen chelate with the lithium and therefore, the lithium is not available for the intramolecular chelation with the oxygen that we saw in the case of only THF as a solvent. So, this was the chelation that was possible in the absence of HMPA this chelation was possible but that chelation is absent when we add HMPA and therefore, the Z enolate formation occurs. So, this is how the effect of the solvent of HMPA versus no HMPA the formation of E enolate versus Z enolate occurs and that influences the geometry of the final product. Now, we look at the Claisen arrangement which is not really dependent on the enolates it is slightly different from what we have been doing it so far it does not have an enolate but we are dealing with allyl or aryl vinyl ethers that is what is important. Now, in this case what happens is it undergoes a stigmatropic rearrangement 3-3 stigmatropic arrangement if we have this aryl vinyl ether. So, this is the vinyl part here and this is the aryl part here if we take this and we number them as this is the bond that is to be broken and this is the here the bond will be formed. So, the bond that is being broken is numbered as 1 2 3 and 1 prime 2 prime 3 prime this is how it is undergoing cleavage at the 1 and 1 prime and the bond formation will take place. So, this is 1 prime and of course is 1 2 and of course it is 3 here and here is going to be 3 prime 2 prime and of course 1 prime. So, bond is being formed here and the bond is being broken here of course this is already a bond present here in fact. So, when this breaks of course we are writing in these two parts because there is a reason for that I will discuss in a while and once the bond formation takes place then of course you have enolization and that leads to the formation of the corresponding phenol. So, allyl vinyl ether lead to 2 allyl phenol formation and of course the transition state would look somewhat like this where the bond is being formed and the bond is being broken as you can see. The 2 double bonds are breaking here and the 2 single bonds are being formed here. Now, as we can see here if we look at the transition state with substituents around in non-aromatic substrates then we can see that the R2 group here versus R2 group here. Now here it is equatorially oriented and here it is actually oriented and the rest of the things are of course the same. So, now if we look at this particular part you can see that the this particular transition state where the R2 group is equatorially oriented is definitely more favorable. So, what here is the e geometry is basically what we are referring to is R2 versus this particular group and here it is R2 versus this. So, this is what is z and this is what is e. So, it is less favorable and that is more favorable, but we will look at it little more detail. Now, if we start with two molecules of this kind having an asymmetric center here and if these are chiral molecules but they have two different types of double bonds in which in this case it is a trans double bond in this case it is a cis double bond and since this reaction is concerted reaction. So, what we get is a product of this kind from this starting material and a product of this kind with this starting material and of course the chirality is also transferred in the product that means the products are also chiral because the starting materials are chiral. So, what it means that by changing the configuration at the allylic double bond this is the allylic double bond one can get different configurations in the products. For example, in this particular case the two methyl groups are beta oriented and in this case one is beta and the other is alpha oriented and of course as I mentioned since the reaction is concerted the chirality is transferred into the product. So, the geometry of the initial molecule is very important. Now, we are taking at looking at this slightly differently here is that we take without taking the absolute configuration into picture we take the allyl vinyl ether that the clasin rearrangement starting material like this for example then we can look at it in a slightly different way and see how does it happen like for example here if we start with EE that is this is also E and this is also E then the transition state would look like this where we have maintained the E here and we also maintain the E here when that transition state is formed and the C-C bond is being formed as then can see this methyl is pointing downward and this point this is pointing upward and therefore here we have these two and T to each other. On the other hand if we take the Z double bond here and the Z double bond here and then the transition state would look like something like this where you have this Z and this Z and therefore when the bond is formed here then we can see that the methyl group here is pointing upward the methyl group here is pointing downward. So this is how it is coming here. So EE gives the similar type of product as ZZ gives. So this is an interesting thing of course and therefore the geometry of the initial molecule becomes very important. In a similar fashion if we start with ZE as we have here Z and E then of course we can say that this is H H this is Z and this is H here and H here this is E and this is of course Z and this is E. So same thing happens and in the transition state as you can see methyl is down this methyl is down here, here this methyl is up and this is also up. So basically cis to each other. So they are cis to each other and here also they will be cis to each other it does not really matter up or down in this particular case because this is a racemic molecule. Now these clasin rearrangements and its variants are basically exothermic reactions as you can see 84 kilo kilo joule per mole and they are concerted pericyclic reactions now following suprafecial reaction pathways. So basically they are 3-3 the sigmatopic rearrangement which is what is the pericyclic reaction. But there is a substantial solvent effects that are observed. More polar solvents accelerate the reaction to a greater extent. Hydrogen bonding solvents give the highest rate constant for example water ethanol mixture gives 10 fold higher rate constant than in sulfolane. Sulfonyl is this kind of non-polar solvent and some Lewis acids such as trivalent aluminum accelerate the reaction. Now this part is something very very important and that is the reason why this radical pair or more so as ion pair type of rearrangements or the intermediates that are transient intermediates which do not affect the orientation of their double bond. But they are transiently available when the bond breaking and bond formation takes place and that has the reason that is the reason why these things are observed that the polarity of the solvent makes the reaction rate faster because of such ion pair intermediates that are involved in it. So we will stop it here and take some other aspects of Claisen rearrangement in the next class because there are say various alterations in terms of the Claisen rearrangement and they have a huge amount of application in organic synthesis. So please look at these carefully and go through the whatever content that I have discussed today and be prepared for the next class till then bye and have a good study period. Thank you.


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