 Last class we talked about chord of contact. Last time we talked about the chord of contact, chord of contact and the equation for the chord of contact was t equal to 0. We also talked about the chord bisected at a given point, chord bisected at a given point. In that case the equation was t is equal to s1 and of course equation of pair of tangents drawn from an external point that was t is equal to t square is equal to ss1. So these concepts remain the same as what we had studied in case of a circle. Before proceeding for today's session I would like to take up one or two questions based on these concepts which we have studied in the last class. So meanwhile people will join in also. So today my agenda would be to finish off Parabola because we have to start with ellipse. There is one simple concept I will not say simple concept but interesting concept which is left off and post that we can start with ellipse. And one more thing guys in maths we need to little bit such our normal classes okay three and a half hours seems to be slightly less. So since most of you have joined in here I would like to extend the class by 15 more minutes if you don't mind. Okay so four to seven forty five is what I want to keep it. Will that be fine with everybody? Okay okay thank you thank you great. All right so let's get started with the very first question while people join in. Question is find the locus of find the locus of the midpoints of the cords midpoints of the cords which subtend the right angle at the vertex of the Parabola which subtend a right angle at the vertex of the Parabola. Okay question is clear to everybody. So we have to find the focus of the midpoints of the cords which subtends a right angle at the vertex. So let me draw the sketch for the same. So there is a cord and this cord subtends a right angle at the vertex of the Parabola. That means this angle is a right angle. This is a right angle. So what is the locus of the midpoints of all these cords? What are the locus of the midpoint of all these cords? Okay Aditya, Aditya already got the answer. Well done. How about this? So those who have joined in first of all, good morning. We started with some questions which on the topics which we could cover in the last class but we did not have a time to solve questions on and those concepts where a cord of contact, cord bisected at a given point and equation of pair of tangents will take couple of questions based on the same and then we'll move on to the last topic of our chapter which is the co-normal points and post that we will be starting with ellipse but not in today's class, maybe the next class. Very good. Okay. So how do you solve this question? How do you solve this question? Where above? How did you solve this question? If you'd like to discuss, if you'd like to unmute and talk. Okay. So we have assumed that let these points be P and Q and he assumed that these points have parameters T1 and T2. Okay. Okay. By the way, I didn't mention that the parabola was y square is equal to 4x but now I'm mentioning it even though I have drawn the figure for the same. Okay. So let's say I say the parabola is y square is equal to 4x because that information is very important. Okay. So P has a parameter T1, Q has a parameter T2. Okay. Then what do we know about HK and these two parameters? They are the midpoint. Right? H comma K is the midpoint of P and Q. Right. So since H comma K is the midpoint of P and Q, can I say H is AT1 square plus AT2 square by 2 and K is what? K is 2AT1 plus 2AT2 by 2. Okay. In short, it is AT1 plus T2. Okay. Yes. Then what next? What did you do next? Then you use the fact that slope of OP into slope of OQ is going to be negative one because there is a right angle getting formed over there. Correct. So what is the slope of OP? Slope of OP is 2AT1 by AT1 square which is nothing but I mean, I'll cancel it out later on and same with this guy, AT2 by AT2 square. That's going to be equal to a minus 1. Okay. So AA, AA gone, one of the T's goes off and we know that T1, T2 will become a minus 4. T1, T2 will become a minus 4. Now what do I do? Now please note that T1, T2 are parameters. T1, T2 are parameters. We need to eliminate them in order to find. We need to eliminate them. So they need to be eliminated, eliminate them to find the locus. That's how our locus problems have been since our class 11 days. Right. Yes or no? Aditya, all set? All fine? Any questions, any concerns? Right. Okay. So now what I'm going to do is I'm going to write 2H by A as T1 square plus T2 square from your first equation. So let's me call it as the first equation. And from the second equation, from the second equation, I can say K by A is T1 plus T2. Okay. Now what I'm going to do is I'm going to I'm going to square the second expression and subtract it from the first. Okay. Let it be as it is and square this guy and subtract it from this. Now see, when you square this guy, you will end up getting T1 square, T2 square. In fact, I should write a minus here. Sorry about that. And you end up getting plus 2 T1 T2. Okay. And on cancellation, it gives you minus 2 T1 T2 and T1 T2 is minus 4. Correct. So in other words, I get 2H by A minus K square by A square is equal to minus 8. Okay. So let's further simplify this. So 2H by A minus K square by A square minus 8 minus 8 is equal to 0. Multiply throughout with A square. You'll end up getting 2AH minus K square minus 8A square is equal to 0. Now generalize this. Generalize this. Your equation of the locus will be 2AX minus Y square minus 8A square is equal to 0. Is it fine? Any questions, any concerns? Now there is one more technique to solve this question and that is through the concept of homosinization. Siddhartha is asking, is there any reason for squaring that term? Yes, because I wanted to eliminate T1 and T2 and thereby produce an equation which has directly relationship between H and K. And I can only do that when I square this term because I want to cancel out T1 and T1 square and T2 square, Siddhartha. That's the reason I squared it up. Okay. Yes. So I was talking about one more method to solve this problem and that is through the concept of another method to solve this second method and that is through the concept of homogenization which I will be taking when I do pair of straight lines with you. Homogenization. Okay. This I will do when I do pair of straight lines with you. Pair of, will be discussed in pair of straight lines because it has got a lot of prerequisites. You have to understand the general equation of general second degree equation. When does it represent a pair of straight lines passing through the origin? So that's actually called a homogeneous second degree equation. So all those concepts will come into picture which I will take separately as a chapter. Okay. So meanwhile, any questions, any concerns in this solution do let me know. Dear all, I would again like to reinstate that the locus type of questions are very, very important when it comes to serious exams like J. Locust questions are asked in and out in most of the cognitive exams. Okay. All right. So without much waste of time, we will now move on to our next concept and that concept is the concept of co-normal points. We will come back to more questions. Don't worry, but let us complete the theory that is required for us to complete. Questions is an ongoing process. So we'll keep on doing questions. No worries. Okay. Now what are co-normal points? Okay. So co-normal points, let me draw the diagram for the same. In fact, I will make a diagram and keep it also. So co-normal points are those points from where normals drawn, normals drawn are concurrent. So what are co-normal points? Co-normal points are those points. Let's say I call them as A, B and C. Okay. If I draw normals from these three points, they will be concurrent. That means let's say I draw a normal from here. Okay. I draw a normal from here. Okay. What is the meaning of normal from there? That means this line is perpendicular to the tangent drawn at this point. So this is perpendicular. This is perpendicular. And one more I have to draw. This is going to be perpendicular. Okay. Now why only three points? Why not four points? Why not five points? Why only three points? Why maximum three? Okay. See, try to recall your equation of a normal. Equation of a normal is given as y is equal to mx minus 2am minus amq. Okay. Now this is basically a line. Let's say it passes through a given point. Let's say I call this point as h comma k. So this normal, let's say, I mean normal at any of the given point, let's say A, B, C, any one of the point. If let's say it passes through alpha comma beta, sorry, h comma k, can I say h comma k will satisfy this equation? Okay. In fact, let me call this point. By the way, everybody knows that this point coordinate is minus 2am. Sorry. This point coordinates is am1. I can choose the coordinate as am1q minus 2am1. Everybody knows about this. Okay. If not, then I'll just quickly discuss it. See, when you, when you used to draw, let me draw another figure. When you used to draw a tangent at a given point whose parameter is let's say t. Okay. So let's say you're drawing a normal at this point whose parameter is t. What used to be the equation of this normal? Equation of this normal used to be y plus tx is equal to at plus at cube. Correct. And let's say this line has a slope of m. Okay. If line has a slope of m, we know that m is going to be negative dx by dy at that point. And negative dx by dy is going to be negative 2at by 2a. That is nothing but negative t. So the parameter at that point and the slope of the normal at that point are related by this relation. They're negatives of each other. Right. So we can say t is equal to negative m. And that's why this point can also have or also set to have a coordinate of minus a, sorry, not minus plus a m square minus 2am. That is what I have used over here. Correct. Okay. Now please, please pay attention. Please pay attention. Let us say, let us say I call that parameter as, I call this point as having a coordinate of this. So I'm putting my m as m1 everywhere. Similarly for b, it will be same thing. For b, it will be m2h minus 2am2 minus am2 cube. Okay. I'm assuming that this point now has a coordinates of am2 square minus 2am2. Similarly, c point will also satisfy because that is also passing through the same point. Let me call it as b. So it will be k equal to m3h minus am3 minus am3 cube. Okay. So c point is, let's say, am3 square minus 2am3. Now why it cannot have more points because, see, if you see these are all, these are all basically satisfying the fact that this equation, this equation is a cubic in m. Correct. This equation is a cubic in m and it could only have three roots. So you can maximum allow three m1, m2, m3 itself. You cannot have more, you know, m values. Right. So that is the reason why in such case, we can only get three co-normal points at the maximum. Are you getting my point? Now I'm not denying that these points cannot be coincidence. Two of them can be coincident. One can be separate. Okay. So the idea here is to understand that since the normal equation to a parabola was a cubic in m, there could be max three values of m and hence three coordinates possible because as you can see, the coordinates of these particular points are functions of m. So hence only three such points, three co-normal points at the max. I'm saying at the max because all the roots may not be real. Okay. So at the max, three co-normal points can be possible. Okay. So the same definition which I gave you right now, I can just rephrase it and say that from a given point, let's say p, you can draw at the max three normals to the parabola. Are you getting my point? So from a point, you can draw at the max three normals to the parabola and the feet of these three normals are called the co-normal points. Okay. So here let me write it down. So a, b, c are called co-normal points. Okay. Is the idea behind what is the co-normal point clear to everybody? Any questions related to co-normal points? By the way, if this point is outside the parabola, right, there can only be one normal drawn. By the way, I'll show that particular fact to you. Don't worry. I'll show that fact to you. Not to worry. I'll show you through GOGBRA also. Okay. So before I proceed, let me take a confirmation from you. What is co-normal points? Is it clear to everybody? Again, I'll repeat it from an external point. So from a point, from a point, let's say p, there can be a maximum of three normals, three normals drawn to the parabola as the normal equation is a cubic polynomial in M and the feet of and the feet of the normals are called the co-normal points are called the co-normal points. Is it fine? Great. So now we'll do some analysis on this. There are a lot of things that we need to understand about it. So let us start with our equation. I always say I'll take a pic and I keep forgetting. Let me draw the situation again. Sorry for forgetting that fact. So I'll just draw once again. Let me take a pic first because I'll keep using it and I'll keep drawing it every time. Yeah. Now let's say again I start with the same diagram which I have created in the last page. So let's say I'm drawing three normals. So these are three normals drawn from the given point h comma k onto this parabola. Okay. So can we say that first of all y is equal to mx minus 2am minus am cube being the normal. This will be satisfied by this will be satisfied by the point h comma k. Okay. So when I substitute h in place of x and k in place of y this is what I end up getting. Okay. This is nothing but so let me just bring it as a cubic equation plus m 2a minus h minus k equal to sorry plus k is equal to 0. Okay. So am cube and these two terms I have also brought to the left hand side. It will become 2a minus h m plus k equal to 0. Okay. And let's say the roots of these equations are m1, m2, m3. Okay. So first of all by using Vita's relation, by using Vita's relation can I say the sum of the okay. What will be the sum of the roots here? Who will tell me? Look at the equation and tell me if this is a cubic polynomial equation in M whose roots are m1, m2, m3. What would be the sum of the roots? Some of the roots will be 0. Okay. What is the reason for this? There is no m square term. Okay. So as you can see in this entire equation the m square term is missing. By the way everybody knows Vita's relation. Right. Vita's relation is basically a relationship between the coefficient of a polynomial equation with the roots of that equation. Right. So I'll just write it down in case anybody is not aware of it. So let's say if this is, I'm just taking an example of a cubic equation. So let's say this is a cubic equation whose roots are let's say, whose roots are let's say alpha, beta, gamma. Okay. So Vita's relation says that the sum of the roots will be minus b by a. Okay. Sum of product two at a time will be c by a and the product of all at a time will be minus d by a. So this is your Vita's relation. Please recall, you may have done this in your class 10th polynomial chapter. I don't know whether it was excluded or not, but you should have done this. Okay. So in the light of the same a is a for us, b is zero for us. Now treat your x as your m. Okay. So b is zero for us. So zero by a is going to be a zero. So some of the roots here will be zero. Okay. What about product two at a time? m1, m2, m2, m3, m3, m1. That will be equal to c from this relation c by a. So here your c is two a by h, two a minus h. So two a minus h by a. Okay. And product of the roots, m1, m2, m3 is going to be minus d by a. In this case, minus d is going to be minus k, minus k by a. Is it fine? Are these three expressions clear to you? Any questions, any concerns? Okay. Now from here, certain things come into picture. What are the things that come into picture? Let us try to look at it. What is this fact? Okay. Just ignore these terms like equation and all. Okay. So this is just look at the theory part. I didn't want to write it. So I took a snapshot so that I save my time also your time also. So the first corollary says the algebraic sum of the slopes of the three concurrent normals are zero, which is true from this equation. Isn't it? Yes or no? From this equation, it is very obvious that the slopes of these three normals, let me call, let me name their slopes on the figure itself. Let's say this has a slope of m1, m2, m3. So the algebraic sum of the slope, what is the meaning of the word algebraic along with the sign? Okay. As you can see, m1 and m2 are negatively sloped, m3 is positively sloped. So when you add them all along with the respective sign of course, then you will always get a zero. It will always get a zero. Okay. Now, my question to you all is, is this true? Is this true for any parabola? Is this going to be true for any parabola? What do you think? Yes, no, maybe. Or is it true only for y square is equal to 4x? Okay. Shruti is saying yes. See, I do understand the fact that I can take any parabola and I can have a, you know, a co-normal points based on that. So is it going to be true for any parabola? Is that is the question to be answered? Okay. So what I'll do is I'll take up a, you know, exercise on this. Let Geo-Gebra tell me whether it is possible. So I'll just pull out a parabola equation, which is not the normal one. Okay. So let's take up this parabola. I'm picking a generic case of a parabola. Okay. Okay. We'll see that. I'll see that. Okay. So let's go to our Geo-Gebra. Geo-Gebra, let it confirm, guys, what is going to happen? Okay. So here, I'm going to take a very, very nasty case of a parabola. So 3x plus 4y minus 4 whole square equal to, let's say 20 times 4x. I just, I'm just picking up a, you know, generic case, any generic case. I just picked up one of the examples of that. Okay. So as you can see, there is a parabola, which is basically created over here. Okay. Now I have to, now I have to choose up three points on this. So let me choose three points on this A. Okay. Let's say B and let's say C. Okay. Now, I don't know whether they're conormal points. I will try to make them conormal. Okay. So let's see how I make them conormal. So first of all, let's make tangents at these points. Okay. And I will make normals also to these points. Okay. I hope you can see the normal. Okay. Now I'm going to slightly move these points so that they become concurrent. Right now you can see they are not concurrent. Okay. So let me, let me shift them. There you go. Okay. We all see now that they are concurrent at this point. Let me just use my marker for the same. So at this point, they are concurrent. Okay. I just shuffled the point and made them concurrent. Good enough. Okay. Now, let us find out the slopes of the normals, which I have drawn. Okay. And let us try to add them up. Okay. So let's do that. By the way, if I just write down, oh, sorry. Where am I choosing? There is an option of finding the slope. Yeah. This is the slope. Slope of this line. Okay. I think they have written the slope. Slope of this line. They have written the slope and slope of this line. Okay. Is it, is it okay? I hope their slopes are being shown to you. So M1, M is the slope of this very first line. It is very high slope. It is almost 65. Okay. M2 is the slope of this line, which is actually negative 3.47. And M2 is the slope of this line. What do you see? Do they add up to zero? Does it add up to zero? Does 65 minus 3.47 plus 0.52, do they add up to zero? The answer is no. That means this concept of the slopes of the normals or the conormal points adding up to zero is true for the case which I have taken, y square is equal to 4ax. So please do not generalize it. Please do not generalize it. In many of the books also, they have written, you know, all these faulty stuffs. Right? So please use your analysis. Okay. These books are also written by some mathematicians. They may not, they may have slipped or overlooked or that would have been an over prediction from their side. Okay. Anyways, so coming back to this, the answer to this is it is only applicable, it is only applicable to whatever case I have taken. Okay. It's applicable to this case. Okay. It may be applicable to several other cases, but you cannot generalize it. That is what I am trying to say. No, no, no, no. The concept of conormal points is true in general. Arjun, see when I made conormal points, I took some generic parabola. Right? The idea is not the concept. The idea is the conclusion from that concept. Correct. Okay. But if the origin is shifted, then it will work. Origin is shifted that doesn't make it a standard case. So if you want to make your vertex as the origin, you, I mean, you may not say that the sum of the algebraic slopes will become zero. Zero. It is true for y square is equal to 4x and may be true for other cases also, but in general, it is not true. That is what I'm trying to say. Okay. Now coming to the second part of the corollary which says that the algebraic sum of the ordinates of the feet, now see here, ordinates of the feet of the three normal points is zero. So it is very obvious that these three points, ordinates, what are the ordinates? Let me write their coordinates. First of all, let's say AM1 square minus 2 AM1. This point coordinate is AM2 square comma minus 2 AM2. And C coordinate was AM3 square minus 2 AM3. Correct. Now the corollary is saying or this particular theorem is saying that the sum of their ordinates will be zero. So let me write down the sum of the ordinates of ABC, which is nothing but minus 2 AM1 minus 2 AM2 minus 2 AM3. Will it be zero? Yes, it will be zero because because M1 plus M2 plus M3 is zero. Correct. We just figured out from this Vita relation. Okay. So this first relation says M1 plus M2 plus M3 is zero. So this will be zero. Yes, definitely true. So both of them are speaking the same things in different tones, in different languages. Okay. So whether you say the sum of the algebraic sum of the normals slope is zero or you say the sum of the ordinates, the algebraic sum of the ordinates of the feet of these three normals or the co-normal points is zero means the same thing because it all evolves from M1 plus M2 plus M3 is equal to zero. Is this fine? Any question so far? Any question so far? Okay. Now this entire game, please understand, has come from this equation of the normal. And this equation of the normal is very, very pertinent to y square is equal to 4x. It is not the normal equation to any parabola that you would like to answer. But many times the results that we get from a very specific case, they can be generalized. They can be generalized. But in this case, I am not able to generalize the fact that the algebraic sum of the slopes of the normal is equal to zero or the sum of the algebraic sum of the y coordinate or the ordinate of the co-normal points is zero. That I cannot say. So please do not make that conclusion, blind conclusion. The next property which I am going to take up that is very important, very interesting as well. By the way, don't take care of the numbers. The numbers may be misleading. You must be thinking where is corollary 3 gone? Okay, so just ignore the numbers. Now read this particular statement. Everybody read this particular statement, especially NPS Rajaji Nagar will be able to make more sense out of it. HSR, Kormangala and YPR, you'll have to wait for the next class to happen on application of derivatives. But don't worry, that will not put you in a disadvantageous position here. I will talk about this as if everybody has equally aware of the situation. Now see the corollary says if three normals drawn to any parabola y square is equal to 4x from H, K point are real and distinct, are real and distinct, then this condition must be satisfied. That means if I go and check the previous figure, let me upload the diagram first of all. Yeah, so if you want all the normals to be distinct and real, okay, let's say this was one of the normals, this was another normal and this was another normal. If you want them, if you want your slopes to be distinct and real, that means M1, M2, M3, which are actually the roots of the equation are real and distinct. Now this is a question to, this is a question to NPS Raja Ji Nagar students. In a cubic polynomial, when do you have all the three roots distinct and real? Do you remember the condition? Just say yes, no need to give me the condition. Let me name a person. Kinshukh, Kinshukh will tell me, Kinshukh, do you remember the condition Kinshukh? Either this is saying is yes. Okay, anyways, so those who have not attended this part, it's very simple. All of you please understand this. Okay, if you have a cubic polynomial, so let me take this in this instance itself. So our equation was what? Our equation was a m cube plus 2 a minus h m plus k equal to 0. I hope I'm writing the right equation. This was our equation of the normal after substituting the point h comma k, a m cube plus m 2 a minus h plus k equal to 0. Now all of you please pay attention. Let me call this as a function of m with special attention to HSR, YPR, and Kormangala students because you people have not done this concept officially with me. This is a concept which we normally do during the application of maxima and minima. So this is basically based on the fact of the nature of roots of a cubic polynomial equation. So all of you please listen to this very carefully. Okay, let us say if you differentiate this particular function, okay, now it's in terms of m, so I'm differentiating it with respect to m. And let's say when I differentiate it, I get two roots for this, let's say alpha and beta. So what did I do? This cubic polynomial, I differentiated it and put it to 0. Of course, cubic polynomial on differentiation will become a quadratic polynomial. Quadratic polynomial will add the maxive view to roots, correct? Now I'm not claiming that these two roots are both real. I'm not claiming that they are distinct. Let's say these roots are alpha and beta. Now just answer one simple question of mine. If you want your this equation to have three real distinct roots, how many places should the graph of f of m So this is your m, f of m. How many places should it cut this graph? You will say, sir, three places it must cut because the cubic polynomial, right? Now any how it is shaped, let's say I take a rough shape like this. One place it cuts, two place it cuts, three place it it's cut. Okay. So I'm basically making this graph cut three distinct points in order to have and let's say these are your three roots. So let's say this function f of m equal to 0 has got three real distinct roots. Then the graph of f of m with respect to m will cut the m axis at three locations. Okay. Wait, wait, but now I'm coming to that. I'm coming to that. Okay. Now, if you wanted to cut at three locations, let me draw once again. I think I made it pass through y axis, but it may not. So let me let me make it like this. Yeah, that will be good enough. So let's say this is your m one location. This is m two location. This is m three location. Okay. Yes. Now coming to the analysis part, but enough whatever you have said. See, if you wanted to cut at three locations on the m axis, that means these points that you see, these are your maximum minimum points, they must be the roots of the derivative of the function equated to zero. Am I right? So these two must be your alpha and beta points. Correct. Right. So first thing that you will notice here is that alpha and beta have to be real and distinct themselves. Okay. Real and distinct themselves. That means this quadratic equation discriminant must be greater than zero. Okay. That means discriminant of discriminant, discriminant of f dash m must be greater than zero. Okay. Now that is not, that is a necessary condition, but that is not a sufficient condition because many people will say, sir, what if it happens like this? Here also alpha and beta are separate alpha and beta are distinct, but it only has one real root then. Right. So this is necessary condition, not a sufficient condition. Okay. Now what is the necessary, what is the next condition which will make it sufficient? Then you realize in this case f alpha and f beta would have negative signs. As you can check, f alpha, f beta would be opposite signs. That means their product will be negative, which will not happen if your case was like this. In this case, let's say this is alpha, this is beta, f alpha, f beta would be of the same sign. Now many people say, sir, why do you say same opposite sign? Why didn't you say f alpha is positive, f beta is negative? Right. Because what if my parabola, what if my cubic polynomial was like this? Okay. Then I cannot say f alpha is positive and f beta is negative. There are opposite, there are opposite sign I can say. Right. Because see, here you know, a is positive. Right. In y square is equal to four a x, a is always kept as positive. I hope everybody knows that we had done this in class 11th. Right. In class 11th, I had categorically emphasized on the fact that in y square is equal to four a x or x square is equal to four a y or whatever, a is always kept as positive because it is considered to be the distance between the vertex and the focus. Right. So here is positive. So you can take this case, the yellow case, you know, but even if it is not the yellow case, even if it is a white case, this fact will still hold true. That is what I am trying to say. Okay. So these two conditions, these two conditions, if they are met, then can I say, then can I say, if these two are met, then can I say f of, f m equal to zero will have three real distinct roots. Is everybody convinced with this or not? So if that particular cubic polynomial I differentiate, put it to zero, that quadratic equation I get has number one, two real distinct roots, alpha and beta, that means the discriminant is greater than zero. And alpha and beta are such values for which f alpha into f beta, the original f, the cubic polynomial f, is multiplying to give you a negative value. Then can I say that this cubic, this particular cubic, which I have shown it with the curly braces here, will have three real and distinct roots? Yes or no? At least this part is clear to non-Rajaji Nagar students. Sir, why didn't you teach us that, sir? When you come to that stage, then only I will teach you. So as of now, you are not at that stage. You are just doing, you just finished up with monotonicity and you started with this. So Rajaji Nagar, people are slightly ahead. Okay. Any question, any concerns? First, get it sorted out, then only I will move forward. Clear. Okay. Good. So now let us try to apply it to this particular scenario. So this is my, this is my f of m. This is my f of m. So now let me execute the process. This was just a theory part of it, a bit of theory. So now let me execute it. So now let me differentiate this guy. So 3 a m square, this will be 2 a minus h equal to 0. As you can see, m has gone off. Okay. m has gone off. Okay. Now this is just a pure quadratic. Right. And it's roots you can easily figure out. So let me write down the roots. The roots, let's say are alpha. You can say alpha is h minus 2 a by 3 a plus and beta is minus h minus 2 a by 3. Okay. You don't have to go and find discriminant because here the roots can only be found out easily. Isn't it? Yes, I know. So are they distinct? Is the first condition met? Are alpha and beta real and distinct? Yes. First condition is met. Any questions? Okay. Next thing that you have to apply is f alpha into f beta must be negative. This is the main condition which will help us to get to this particular result. So I have to reach this result. See, this is my target. Okay. So this target will be achieved through this inequality. f alpha into f beta will be negative. Okay. Now as of now, we can see that alpha and beta both are negatives of each other. So you can say beta is negative alpha just in order to write less. Okay. Just in order to write less. Oh, yes. That's a very, very important thing that Aditya has brought into picture. Now here, all of you please understand it. If you want this, then h must be greater than 2a. Yes, definitely. So I'm assuming my h is greater than 2a. Okay. My h is greater than 2a. Then only my roots of alpha and beta will be, will be, will be real numbers. Okay. h must be greater than 2a. Okay. Now I actually missed that point out because I wanted to, okay. Let me show that as well. Show that as well. See, if you want your roots to be m1, m2, m3 to be real. Okay. It is very obvious that they must satisfy this condition. Okay. And if they must satisfy this condition, then m1, m2, m3 square minus 2, m1, m2, m2, m3, m3, m1. This must also be, I'm so sorry. This must also be greater than 0. Okay. Now this term is already a 0. This term is minus 2 and the product of 2 at a time is 2a h, sorry, 2h minus a by a. So this should be greater than 0. That means 2h minus a should be less than 0. Sorry, 2h, no, h minus 2. This is 2a minus h. Yeah, sorry. Yeah. So 2a minus h must be less than 0. That means h must be greater than 2a. So this condition is what is to be stated overhead. Okay. Sorry about that. I missed this telling you this part. Okay. All right. So coming to this particular fact, please note this down also very important. So if you want your coordinates to be real, at least this condition should be satisfied. Okay. See ideally you should read it like this. If h is greater than 2a, if h is greater than 2a, let me write it other way around, if m1, m2, m3 are real, are real, then h should be greater than 2a. The other way round is not true. The other way round is not true. I'll show you how. Okay. This way is may not be true. Okay. I'll show you. I'll show you in some time. Just note it down. As of now I will show you the geometrical way of analyzing the whole thing. Okay. Anyways. So this is to answer Aditya's query that shouldn't h be greater than 2a. Yes, h should be greater than 2a Aditya. And that comes from this fact. Okay. And we'll be using the fact over here as well. Thank you for reminding me that. Right. So now here f alpha and f beta should be less than 0. So let me write my beta as negative alpha. Okay. Now, what is f? This is your f term. This whole thing is your f, isn't it? This whole thing was your this whole thing was your f, isn't it? f of m, you can say that. Okay. So in that, I'm going to put my alpha and minus alpha. So let us see what happens. So it'll give me a alpha cube plus 2a minus h alpha plus k. And if I put a minus alpha, I will get minus a alpha cube minus 2a minus h alpha plus k less than 0. Okay. So I hope f alpha f minus alpha is clear in this expression instead in place of m, I have put alpha and minus alpha. Okay. And I multiplied it. So this gives me an expression like this. Now, don't worry, this looks to be a big expression, but you can easily simplify it because here you can see something like you can call k as your x, you can call this as your y. So it is like y plus x and minus y plus x. Okay. So you can use your x minus y, x plus y formula. So that will give me something like this. k square minus a alpha cube plus 2a minus h alpha the whole square less than 0. Correct. Any questions, any concerns here? Let me pull it down over here. So it becomes k square. Take an alpha out from this particular expression, you'll get a alpha square 2a minus h alpha square less than 0. Correct me if I'm wrong. Do let me if I'm missing out anything. Now, what is alpha square? Now, all of us, let us go back and revisit this term. Yes, revisit this term. So alpha square is actually h minus 2a by 3a. Correct. So let me replace that. Let me replace this in our alpha square term over here. So it'll be k square minus a times h minus 2a by 3a into 2a minus h into alpha square. Alpha square is h minus 2a by 3a. Does it make sense? Please convince yourself before we move on. Please ensure whatever I have written, you have clearly understood it. And a clear on the chat box would let me know that you have understood it. So please write clear if you have understood this. Oh, yes, yes, yes. This has a square term. Alpha cube became an alpha because Siddhish, I took an alpha common from here, common from here outside the square. Got it? Yeah. Okay. Any questions? All right. So now what I'll do is this a and this a goes off. By the way, 2a minus h, if you take common, it will make 2a minus h 1 minus 1 by 3, which is 2 by 3. Okay. Square. And this is h minus 2a by 3a less than 0. So let us send everything to the right side. By the way, before you send everything to the right side, I'll just take a snapshot of it because I'm out of space. Can I go to the next slide? Oh, it's too big actually. Yeah. Now, all of you please pay attention on simplification. This gives you k square minus 4 by 9. You can write it as h minus 2a also. Doesn't make a difference because you're anyway squaring. So what happens is that you can actually club this guy with this fellow h minus 2a and make it as a cube and make it as 27a. In other words, k square is less than 4 times h minus 2a cube by 27a. In fact, 27a k square is less than 4 times h minus 2a the whole cube. And this is what we wanted to prove. Let me put the, let me put the corollary once again on the screen. Yeah, this was the corollary. This is what we wanted to prove. So if there are three distinct normals, then 2a, sorry, 27a k square should be less than 4h minus 2a the whole cube. This is very, very important. A lot of questions in j advanced are framed on this. By the way, this concept is asked mostly in j advance, not in j main, but we never know. We never know. Is this fine? So this particular concept has a small prerequisite which I already discussed with you. So non Raja Ji Negar students, I hope you have made sense out of this whole analysis because most of you have done maxima and minima to a certain extent. Okay. Now I would like you to see something very interesting. Let me, let me make a locus out of it. Okay. First of all, let me equate it. Okay. And I will show you something very interesting on GeoGebra and let's make a locus out of it. So if you make a locus out of it, it becomes 27a y square is equal to 4x minus 2a cube. Okay. It will actually show you a curve. I would like to show this on GeoGebra to you. This page is completely corrupted. So I'll just discard it and I'll open one more. Yeah. So let us take y square is equal to 4ax. Okay. a being one, let's say I don't want to complicate my scenario. Okay. Now the same curve which I have shown over here. Let's keep a as one and try to sketch it. So it's 27 y square, 27 y square. Okay. Equal to equal to 4x minus 2 whole cube. 4x minus 2 whole cube will come as a bracketed term like this. Yeah. Like this. Okay. Now all of you please pay attention. Very interesting thing I'm going to show you here. Very, very interesting thing I'm going to show you here. Now, if I pick up a point, let's say here. All right. All of you can see the point A on the screen is my camera blocking your view. No. Yeah. You can see that point. Now what I'm going to do is I'm going to create a cubic in x treating my x as m. Okay. Now see everybody what I'm going to do. y is equal to x cube. In fact, yeah. Okay. x cube plus 2a minus h. So 2a is one minus h. Now h is the x coordinate of a. So I'm doing some coding over here. I hope you can make sense. Okay. x plus k. k is the y coordinate of a. Okay. On your screen, you are seeing a cubic polynomial generated. Right. Now what is this cubic polynomial? Let me explain you how I have framed it. See normally the cubic term that we write is a m cube, a m cube plus 2a minus h m plus k equal to zero. Right. So what I did here, I made my a as one and x as, so a is one and your m is x. Okay. So I made something like this. Sorry. H is nothing but h is the x coordinate of this point. So this point a is h comma k. So that is why I wrote in my code x of a. x of a means it'll take up the, it'll take up the h coordinate, x coordinate of the point a. Okay. And k is nothing but the y coordinate of a. So you can see in your formula, in your code here, I have written y a because we picked the y coordinate of a. Okay. Now this cubic should have how many roots if you have, if you are satisfying 27 a k square lesser than, what was that rule? 27 a k square lesser than, lesser than 4, 4. What was it? 4h minus 2a the whole cube. Right. So what I've done, I've taken a point a which is actually in that zone. So this zone that you see, this zone that you see is a zone where this inequality will be satisfied. It's very obvious. Why? Because if I take it on the curve, it will be equal to, if I take towards the origin, as you can see, origin will not satisfy this inequality. Correct. So origin will be when it is greater than 0. Correct. So this zone I've taken a point a and I have framed a cubic in x, which is actually corresponding to the cubic in m. And that cubic diagram you are seeing right now on your screen, how many roots that this cubic has? 1, 2, 3. Are they real and distinct? Yes. Are you getting my point? Is this clear? Is this analysis clear to you? A little bit of mathematics and coding I've done over here, just a little bit. I'm sure all of you can understand it. Yeah, that I will do, Siddish. That is the main thing I'm going to show you, but I just wanted to show you that if I take a point h, k which satisfies this particular scenario, then the cubic polynomial in m will have three distinct real roots and these are the three distinct real roots. Aditya, first you have understood the scenario. I will show you the implications by shifting that point h, k. Don't worry. Is it clear? Okay. Now, one more thing I would like to show you that this is your 2a point. This is your 2a, 0 point. This is your 2a, 0 point. So, h, k is basically satisfying the fact that h is more than 2a. Okay, h is to the right side of that 2a. You can say the vertical line x equal to 2a and that's why we are getting three real roots over here. But other way around is not true. I will show you in some time. Is this clear? I mean, I'm not getting the clear from you people. Yes, no. If you want to ask something, you can ask me again. Should I explain what I did again? Could you repeat the situation? Okay, sure. I'll repeat the situation. See Aditya, what had we figured out in the previous discussion that if any point satisfies this condition? Okay, let's say this point h, k. First of all, is this point satisfying this condition? Yes or no? Any point here if I take? Correct. That is satisfying the condition. Great. So, what I did was I took a point, orbit point, this satisfied this condition as your point a in the curve. Okay. And that cubic polynomial which you have learned in terms of m, which was this, okay. I basically made a curve out of it by removing this zero, basically putting it as y. Okay. And replacing your m with x's. Okay. So, this is what I have plotted here. Got it now? Now, if you wanted to solve this as zero, which means you want to see where is this cutting the x-axis. So, in this diagram, I showed that it is cutting the x-axis and hence the root of that particular cubic polynomial in m will be three distinct values. That is what I am showing here. It is a showcase of the particular concept that you have learned. Now, it is clear? Everybody? Everybody? Okay. Now, the interesting thing is let me move this point. That is what many people want to see. So, when I move this point, you will see that this cubic polynomial will start changing the nature of its root. Okay. How? Let's say, first of all, first of all, if I move it anywhere in this particular zone, it will still have three roots. You can check. Okay. But I should not go out of the parabola. The moment I go out of the parabola, despite having h greater than 2a, it will not have three real roots or all the roots will not be real. That is why I said h greater than 2a is a necessary condition, but not a sufficient condition. Okay. So, till the point, you are within the parabola and within that particular bracket. Okay. I'll just call this curve as a bracket curve. Okay. So, till you are within this zone, you'll see anywhere I take it, anywhere I take it, it will have three real and distinct roots. Are you able to see it? Okay. Exactly at a, if I put, see what will happen. It will have three real and equal roots. Three real and equal roots. By the way, it has got slightly corrupted because of shifting. Yeah. It is showing that it has stopped just before zero, but it is actually reaching at zero. Yeah. Now the graph is problem. From move, that starts, you know, having some pixel glitch. Yeah. Is this fine? Any questions? Okay. Now if I take this point on this particular curve, see what will happen. You'll realize that two of the roots have become same. See here. That means two of the roots, let's say this was your M1. This is your M2 and M3. They have become the same. So, when you are exactly on the curve, you will end up getting two of those three real roots equal. Are you getting my point? So, let me move it. Let me move it. You will realize that when I move along this particular curve, you will always have, you will always have two of the roots equal. Okay. Are you seeing that? Okay. Here also if I move two of the roots will always be equal. I mean, it is moving from my hand. That's why it is appearing to go. Yeah. This is how it'll appear. Is this clear? So, here is the next corollary, which I would like to, you know, put it over here. In fact, it is actually like a question. I'll put that over here. Yeah. So, see here, this question basically tells you the same thing. So, you have actually solved the question geometrically. So, this question shows, says, show that the locus of the point says that two of the three normals drawn from them to the parabola coincide is this condition, which is actually this curve. Okay. So, if my h comma k is on that bracketed curve, you would realize that your cubic polynomial in M will have two roots equal and one will be distinct for sure. Okay. That is what is basically seen over here. So, the fact that it is touching here shows that one root is separate and two roots are equal. Are you getting my point? Okay. Now, many people say, okay, if I have to solve this without making a graph, what condition will you write here? Here, the condition that will come will be f alpha, f beta equal to zero. The one which had, which we had taken in the discussion as less than zero, that will now become equal to zero. Why it will become equal to zero? Because what will happen if you have two roots such that the graph has got two real equal roots at this point, then let's say I call this as alpha and call this as beta, f of beta or f of alpha, one of them will be zero because it will be touching the x axis. Okay. And that's why their product will be zero. That's why their product will be zero. Are you getting my point? So, the less than zero which we had taken, see, we had derived this whole thing by using what? f of alpha into f of beta less than zero. That less than zero will become equal to zero if you want those three real roots not to be distinct from each other, but two of them to be equal to each other. So, this is when all of them are distinct. This is when they're real, but two of them are identical. That means equal roots. Are you getting my point? Does it make sense to you? Okay. So, this is what is primarily happening. Now, if I go beyond this particular curve, let's say I come in this zone, you can see that one root will be real, two roots have become non real. In fact, if you go outside the parabola, then definitely, you will only get one real root, two will be non real. So, that is why we say that from any point outside the parabola, from any point outside the parabola, you can only sketch one normal. Okay. So, let me write that down also. Very important point. Let me write important point. The important point is, let's say I take three positions. One is p, one is q, and other is r. Please note this down. From positions like p, which is outside the parabola, outside the parabola, only one normal can be drawn to the parabola. Getting the point? From positions like q, which is on the curve, 27a y square is equal to 4x minus 2a the cube. Three normals with two of them coincident. That means the slope of two of them will be the same. Two of them coincident can be drawn to the parabola. See, I'm pinning it down so that whenever you are turning back your notes and trying to see it, you are able to recall this stuff. Now, when I say two normals, even let me take it to this position. Let me take this guy to this position. Inside the parabola, but within this. So, here also you can only draw one. So, see here, only draw one. I'm just moving it in this space. See, only one way, it will cut the x-axis only at one position. See here? Okay, so let me add that thing also. Let me take p, q, r, s positions. Okay, so p and s. s is within the parabola, within the parabola, but outside the curve, 27a y square is equal to 4x minus 2a the whole cube. Okay, within the parabola. Yeah, you can only draw one normal. Okay, outside the parabola, but inside the bracket. So, we'll check that out. Very good. Outside the parabola itself means it will not be, it will only be one. So, I'll just take it. Outside the parabola, but within the, yeah, as you can see here, you are getting three roots. Absolutely. Outside the parabola, but within this. So, let me write that case also. So, I think this will be a more generic case. So, let me just remove this and just write it like this s position. Okay, and this I will take it to be here. This I will take it to be here. Okay, so at p position, so I think p position will be covered under this. P and r positions will be covered under the same. Right, p and r positions will be covered under the same. So, p and r positions which are within the, within the curve, 27a y square is equal to 4x minus 2a the whole cube. There can be three real and distinct normals. Okay, so as you can see here, I have kept it outside. Let me just, yeah, so I have kept it outside here. Okay, so if it is outside, but outside that curve, then basically you'll end up getting only one real root. Okay, again, I'll repeat the whole procedure. The procedure is, if it is outside that bracketed curve, this curve. Okay, it will definitely have only one real root. Correct, whether outside or inside. Okay, if it is on that particular curve, if it is on this particular curve, it will have three real roots of which two will be identical or two will be coincident. Okay, and if it is within it, whether outside or inside doesn't matter, it will have three real and distinct roots. Is it any questions, any concerns? No, in this case, if it is outside, but within that curve, within that curve, it will have three real and distinct roots. Is that fine? Any questions? See, it is outside the parabola, still it has real and distinct roots. Could you take A behind it? Yeah, behind means this position or this position? There, yeah, so only one. Only one it'll have. It is only cutting at one. Don't think like if there are three coincident roots, because if I put it on A, then it will, that will become a cube, pure cubic x cube on A. This is a minute difference. If you have kept the point here, then this root will be, then it'll have three real roots. It will have a point of inflection. So this will be an inflection, but the moment you drag it a little bit over here, let's say you keep it in this position, then this will become more straightened off like this. That means only having one real root. It's a normal only in its first place. It intersects the parabola. No, it can be normal at the other place also. Your question is a good question. Let's say I keep a position like this. So you can always draw. No, let's say there was a point like this. Okay, here. Let's say I'm just drawing some orbit positions. So here it may be normal. Here it may be normal. And here it may be normal. Are you getting one? Don't treat these points into your consideration. Treat these points. That can be possible? Yes, exactly. As long as it is in the bracketed curve, it will have three. Okay, front of has a question. So does anything happen if the point moves on the parabola? Okay, that's a very good question. Now you people have started analyzing the situation from geometrical point of view. Okay, nothing happens. It can still have. If it is within that curve, it will still have. But the moment you go outside the curve, gone. Okay, whether on the parabola or anywhere else. Till it is within that given bracketed curve, it will have three. Okay. So guys, we had a very good analysis on this. And I wanted to cover up one small concept actually. All of you please pay attention. There is a question which I'm going to basically write it down in terms of a diagram. I will not, sorry. Why am I drawing it when I already have the curve? Pay attention. Let's say this is our y square is equal to 4 a x parabola. Okay. And these are my three co-normal points. Okay, three co-normal points. So a, b, c are three co-normal points. Are three co-normal points. If I try to make a circle passing through it, if I try to make a circle passing through it, I may be not be able to make a very perfect circle out of them. But I'm just trying to make a circle out of them. Okay, let's say number one, prove that this circle, prove that the circle passes through the vertex, passes through the vertex of the parabola. Okay, that's number one. Number two, find the equation of the parabola. Sorry, find the equation of the circle. I'm sorry. Find the equation of the circle. Okay. Now, in the interest of time, I will only solve the question. Shutech, didn't we see the case that normals drawn at two points intersect at the same point on the parabola? Didn't we see that? Whose coordinate used to be negative of? Okay, we'll discuss it in more detail. Let us finish this question first. Okay, let's not waste everybody's time. In fact, we will talk about it through our geometrical analysis of it. Okay, so let us try to solve this question. Let us try to solve this question. The first part of the question says that prove that the circle passes through the vertex of the parabola. And we also want to know the equation of the circle. Now, how do I solve this question? See, let us say my circle is x square plus y square plus 2 gx plus 2 fy plus c equal to zero. So let the circle equation be this. Okay. And let us say these co-normal points, these co-normal points in general, in general, the co-normal points or the points which basically satisfy this particular circle are of this nature, a m square comma minus 2 a m. Okay, let's say this point of the parabola satisfies this equation. So when I say that, when I say that in place of x, I can put a m square in place of y, I can put minus 2 a by m minus 2 minus 2 will give you minus 4, I believe minus 4 f a m. Now, clearly, if you see this, it is actually a m square coefficient will be, let me write it down, m square coefficient will be 4 a square plus 2 a g and minus 4 f, or you can say m times, m times minus 4 a f plus c equal to zero. Okay. So this is a bi quadratic, this equation is a bi quadratic. In m, what are the meaning, what is the inference from the fact that I am getting a bi quadratic polynomial equation in m, when I substitute any generic point, it means there will be four roots. Right. That means there will be four points. Okay. And let's say these four roots, these four roots are m1, m2, m3, m4. Correct. So m4 is basically the parameter for, you can say negative of the parameter of the fourth point, which I am guessing is going to be my origin. And that guess is actually correct. Why? Because if you see there is no cubic term over here. Right. Since there is no m cube term, can I say by beta's relation, this thing should be zero. Correct. And we already know that m1 plus m2 plus m3 should be zero because they are, they are conormal points because they are conormal points because ABC are conormal points. So what I'm going to do is I'm going to use my previous theorem that we had seen, a previous corollary which we have seen. So basically let's say this point was having the coordinate minus, sorry, this point was having the coordinate a m1 square minus 2 a m1. This point b was a m2 square minus 2 a m2. Okay. This point was a m3 square minus 2 a m3. And this point where it is passing through, I don't know where it is passing through. I'm just taking a guess for and I'm just calling it as a m4 square minus 2 a m4. Okay. So these are the four points where this circle is basically cutting the parabola. Three of them will be definitely the conormal points because you're making a circle pass through the conormal points. Okay. So m1 plus m2 plus m3 is definitely zero. What does it mean? So if you put this as zero, what does it mean? m4 is zero. And if m4 is zero, this point automatically becomes zero comma zero. This point automatically becomes zero comma zero. That means it must pass through the vertex. Correct. So it means it means the fourth point D, which will be a m4 square minus 2 m4 will be zero comma zero. Correct. Yes or no. That means the circle passes through the circle passes through origin or the vertex. Let me not write origin. Let me write vertex. Is this fine? Any questions? Any concerns? Now, will it be true every time? What do you think? Will it be true every time? Even if I take a generic case of a parabola, can I say if I make a circle pass through the conormal points, feet of the conormal point, it will always pass through the vertex? Guys, no areas has to be given very, very carefully. Okay. So this is a task for you. All of you will open GeoGebra and try to figure it out. Okay. I don't have much time to show it. Okay. Please figure it out and let me know the answer in the next parabola class. Okay. The last part of the question, find the equation of this circle. Okay. Find the equation of the circle. Now everybody, please pay attention. I'm writing this curve once again. Let me take a snapshot because I'm falling short of place over here. So I'll do it in the next slide. Now, all of you please pay attention. We just now figured out that M4 was 0. So let's say the roots of this equation is M1, M2, M3 and M4. Now, since M4 is equal to 0, that means one of the roots is 0. One of the roots is 0. It implies that 0 should satisfy this. That means C should be 0. That means that circle which I will be getting will have no constant term. Correct? Yes or no? Correct? So if the circle doesn't have any constant term, even this equation that constant term C, I will take off. I will not provide it to the... Okay. So I'll just keep it like this. No C anymore. Okay. So automatically M equal to 0 comes out as a root of this. Right? Makes sense? Okay. Now, if I drop this root, if I drop this root, let's say M equal to 0 is the solution for this, the three other roots, M1, M2, M3 should actually come from this guy. Am I right? So M4 was the root of this guy and M1, M2, M3 would be the root of this guy. Am I right? Yes? No? Maybe? Correct? But we also know that... But we also know that... Let me write it down. M1, M2, M3 are roots of... By the way, this expression has an extra A. Okay. So please allow me to drop that A also because there is A everywhere here. Okay. So this A I will drop off. Okay. But we also know that M1, M2, M3 are roots of Am cubed plus M2A minus H plus K equal to 0. Yes or no? Yes or no? So can I say that these two equations should actually have been... Should actually have been the same cubic equation? So these two should have been the same cubic equation because they are the same set of roots. Okay. So if you see coefficient of M cubed is already matching. So coefficient of M should also match. Right? In other words, you get your G value as... Correct me if I'm wrong. In fact, 2G value as minus 2H minus 2A. Correct? And minus 4F is your K. Minus 4F is your K. That means F is equal to... Or you can say 2F is equal to... 2F is equal to minus K by 2. Is it fine? Any questions here? So can I now write down the equation of a circle like this? So the equation of the circle that will be passing through those three conormal points drawn from H comma K will be X squared plus Y squared plus 2GX, right? Plus 2FY... Oh, sorry. 2FY equal to 0. So this is your equation of the circle. Now, many people ask me, sir, do we have to remember this as a no need? As I told you, these all concepts are basically covered in J main, J advance, mostly J advance, I would say, not in J main. Yes, HNK will be known. Yeah. So HNK will be known. Had I not shown it? HNK? No. No. Okay. So these three conormal points, they pass through, let's say here and there, H comma K. See, if H comma K is varied, conormal points will keep on changing. So I need H comma K. Okay. So H comma K will be given to you. And from that point, you're drawn three conormal points. Conormal points are not sacrosanct. It keeps on changing, depending upon your HK, because your cubic polynomial itself has an HNK in that. So if your cubic polynomial parameters are changing, HNK are involved in it, will not the roots of M1, M2, M3, or will not the roots of that cubic polynomial change. But still there will be conormal points only. So there can be millions of conormal set of points created, depending upon your point from where you are sketching it. Okay. So guys and girls, we will stop this chapter over here. We have done this chapter to considerable depth. And the next class that you're going to meet, sorry for taking extra 10 minutes of your time. I know all of you would be waiting to have your breakfast. Next class, we'll start with ellipse.