Very good here. So three minutes over. I'll give you one more minute for the people who are about to crack this about to finish this. Okay, let's. Yeah, let's invite everybody now to guess it. If you have not been able to solve it, I would request. I think four of you have responded. Three of you. Please put forth your response. I think just two of you, I guess. Man Shah and who else? Yeah, just Man Shah. We are waiting for Man Shah. Man Shah take a guess if you are not able to solve it. Put it on the chat box. Man Shah, can you hear me? Yes, sir. Give a response on the chat box so that we can start the discussion for the same. I have already replied in the poll. You put it in the chat box also. Okay. Okay, good. So everybody has responded. All of you leaving Tirupan have said B. Okay, let's check whether B was correct or not. See, very simple. First of all, we are aware that the equation of a chord whose midpoint is given is T equal to S1. So let's say I want to find out the locus of the midpoint of the score. So I am assuming the look at the midpoint to the H comma key. So T as you all know is nothing but xx1 by a square plus yy1 by B square is equal to or you can say minus one is equal to S1. S1 will be h square by a square k square by B square minus one. There is no point writing minus one because minus one is going to get cancelled from both the sides. Okay. So this is the equation of this chord AB. So this is the equation of the chord AB. And this is under the assumption that h comma k is the midpoint. Okay. Now we all know that if the chords, if the tangent is drawn at the end of a chord meet at 90 degree, that means this C must lie on the director circle. Okay. Director circle is the locus of all such points from where tangent drawn to a particular conic are at right angles to each other. Okay. So for such a conic, the equation of the director circle is given to be this. Okay. So can I do one thing? Can I assume that the point C is now this is your R square. This is treated like R square. So let's see the R cos theta comma R sin theta. That means I can assume C coordinates to be this cos theta, this sin theta. Okay. So in light of this, if I have to find out what is the equation of this chord of contact drawn from C. Now I'll use the equation of chord of contact. Code of contact equation is T equal to zero. So code of contact will come out to be X under root of a square plus B square cos theta by a square plus Y under root of a square plus B square sin theta by B square equal to one. Now ideally both of them represent the same equation. Both of them will represent the same equation. So now you can start comparing their coefficients. So if you compare their coefficients, so let's say I take this coefficient, divide by the coefficient of this. Correct me if I am wrong, if I am missing out on anything, you will get a square B square cos theta by H is equal to under root of a square B square sin theta by K. And on this side, on the other side, you end up getting a square B square by a square K square plus B square N square. Okay. So I've done this mental calculation over here. I hope things are on the right track. Now try to eliminate theta here. So if you do that, your cos theta will come out to be, from here your cos theta will come out to be H times a square B square by under root of a square plus B square times B square H square a square K square. And sin theta will come out to be a square B square K. Why? Same thing under root a square plus B square times B square H square a square K square. Now we know the very famous Pythagorean identity that cos square theta plus sin square theta has to be a one. Okay. So if you do square of this and square of this and equate it to one, this is what you should be seeing and correct me if I am wrong. You should be seeing a to the power four B to the power four and you will in fact be this thing you can call it as as a lambda. So let me write it like this. So you will have lambda s square plus K square is equal to one. Okay. So lambda square will be a to the power four B to the power four. So this is nothing but X square plus Y square. And on the other side you will have a square plus B square. And you can call this back to back to X and Y. Okay. So whichever option says it that option would be the correct option. And I think that clearly matches with there's a square here. So that clearly matches with option number B. Anybody who did it in a shorter way or it took some special values of A and B and started working on it. Anybody who did that? We can assume a chord whose point is A by 2 common, B by 2. So even the end points of that chord will be perpendicular to each other. The tangents are drawn from the- See, whatever point you are assuming, make sure the tangents drawn there will meet on the director's circle. Yes, sir. If you can assume that point, then you can directly put it in the equation itself and check. Yes, sir. Alright, so let's move on to the next question. I hope everybody was clear about it. Well done. I think most of you got this right. This is from, I think, pair of straight lines, I think. Again, we'll have around three and a half minutes for this. Put your response on the chat box. Question is if the pair of lines given by x square plus 2x y plus A y square equal to 0 and A x square plus 2x y plus y square equal to 0 have exactly one line in common. Then the joint equation of the other two lines is which of the following. Guys, there was an assignment practice sheet shared with you on Saturday. I hope you're done with it. Some of you are still doing it. No worries. You can take today's time to complete it. But I've shared one more with you and that is supposed to be done by 8th, end of the day. Please take these sheets very, very seriously and solve them within time bound manner. That's the only way to success in J advance. So at the end of the class, please do let me know what was the status of the sheet which was shared. I think Richard wanted some more time. Others do let me know. Okay, Mansa, very good. Mansa, Mansha. Take some time for me to get that pronunciation correct because we had a student Mansa for two years. Yes. So that pronunciation is still left in me. Guys, time is going to be over in next 30 seconds. So I would request you to put forth your response. So I have received it from Mansa and Thrippan. Excellent. Very good. Yes Pratham, Richard, Shamik. Okay. Shamik is left. Okay.  Let's check whether B was correct or not. All right, now both these lines basically are homogeneous equations, right? Suggesting that the components or the lines which make these will be the ones which pass through the origin. So let this be the common line. Okay, so let this be the line in common. So when you put Y as MX over here, you will get 1 plus 2M plus AM squared equal to 0.  Please don't write unnecessary things in the exam because just wasted your time. Okay. Now from here, I would like to see these are two quadratics in M squared. Either you can use a direct formula or your direct understanding of the fact that these two quadratics in M have exactly one root in common. Read the question. They are saying exactly one root in common or you could eliminate M squared. So how would I solve this? I would just eliminate M squared. Let's say I multiply this with an A. Okay, so let me multiply this with an A and subtract it. So when I subtract it, I get A squared minus 1 plus 2M A minus 1 is equal to 0. Now this leads to A plus 1 plus 2M is equal to 0 because A cannot be 1. Because if A is 1, this will no longer have this Y squared term in it. Now here, your M becomes negative of A plus 1 by 2. Let me make this back to your original expression. Now let's do one thing. From here, I can say A is 2M negative 2M minus 1. Okay, put it here. I'll get negative 2M minus 1 plus M squared is equal to 0. Okay, plus 2M is also there. So 2M minus 2M will get cancelled off giving you M as plus minus 1. Okay, if M is plus minus 1, what will be the value of A? So if M is 1, A value will be minus 3 and if M is minus 1, A value will be 1. But if A becomes 1, both these equations will become same. That means they will become a perfect square. But in that case, we will treat this as if there are two coincident roots and both are same. So we will rule this condition out. We will take this condition into account. Even though I understand, you would be saying that it becomes a perfect square having only 1 root M. But in that case, we can basically say that there are two coincident roots of minus 1 and both are common. So both the set of lines will become common in that case. So this is ruled out. Any question regarding why this condition was ruled out? No? All well? Okay. So now when A is minus 3, A is minus 3. Let us look into these lines. So x square plus 2xy minus 3y square. How would this get factorized into? Can I factorize this? Yes, of course we can. x square plus 3xy plus xy minus 3y square. So you can take x common, x plus 3y, take minus y common, x plus 3y. So basically you will have x minus y times x minus 3y equal to 3. Similarly, if I look into the other equation. If I look into the other expression, we have negative 3x square plus 2xy plus y square. Again, this can be factorized as plus 3xy minus xy plus y square equal to 0. Take 3x common, I take minus 3x common, so it's x minus y, take minus y common, x minus y. So you'll have 3x plus y times x minus y equal to 0. So x minus y is the common line for us. So what I'm going to do is, I'm going to make up a line for, make up a line using these two. So you get y plus 3x times 3y plus x equal to 0. This clearly gives you 3y square. This will give you 9 plus 1, 10xy and 3x square equal to 0. So any option which says, this should be your right option. Let us figure out which option says it. Oh, I can clearly see B is the option which works. Any other shorter route to do it? Please discuss for the benefit of all the students. Anybody who got a shorter way to do it? Sir, did it using the sum and product of this? It's easier to find the common one. Sir, can you hear? Yeah, I can hear you. Sir, I did it using the sum and product of routes. I took alpha, beta and alpha, gamma. So then it becomes easy in one step. You can multiply two, you get alpha directly, then you can find out. Okay. Good. Okay, so let's take another one. Question number three. The question is the equation of the circle for which the straight lines given by this are normal. And the size is just sufficient so that it contains the circle. Okay. This looks to be an easy question. So I will not give you more than three minutes for this. Time starts now and put forth your response on the chat box to me. Okay, Shamik. Okay, Man Shah. Last minute. This is not a difficult question by any standard. So you should be able to answer it in a shorter duration of time. Okay, Richard, waiting for Trippan here and Pratham. Oh, okay, no problem, Richard. So Richard Parik has joined in. Board practical. Was it the actual board practical or just a mock? So mock. Okay, here. Pratham and Trippan. Okay. So some of you have said C, some of you have said D. Okay, let's check. So first of all, let's find out what are the component lines involved over here? So x square plus 3 y plus x y plus 3 x, right? If I take, if I take these two together and these two together, I get x x plus 3. Okay, and I take y x plus 3 equal to zero. So I get x plus y times x plus 3 equal to zero, which implies that the two component lines, which are normal are this. Now please note, normal means they are passing through the center of the circle. So one is a line which is x plus 3. So basically it's a vertical line. Okay, and another is a line x plus y. So that would be a line like this. Okay, anyways, so I come to know that the center has to lie at the intersection of these two, which is clearly negative 3 comma 3. Negative 3 comma 3. Okay, no doubt in finding the center. Okay, now whichever option you think gives you the center as negative 3 comma 3, that is going to be the answer. So B and D are anyways ruled out immediately. Second is it must contain this circle. Now this circle itself has a center at minus 1 comma 3. So minus 1 comma 3 means it is on the same level as this somewhere over here. So minus 1 comma 3. And if you see the radius, the radius would be under root of 1 plus 9 minus 9. Correct, which is 1. So let's say it is a circle of this nature. Okay, just sufficient to contain it. So when this circle itself has a radius of 1, this has to have a radius more than that, which clearly indicates that C would be your answer. But anyway, if you want to solve it properly, this has to be 2 and this is 1. So it has to have a radius of 3, thereby indicating that this should be 9. And hence C is anyways the correct option. So people who went for D option, two of you, what mistake you made? Should I name the people? Tripan and Pratap, careless mistakes. KMS mistakes is the last thing you should be doing in a J-advance example. Okay, next question. The question is locus of a point P. Three normals drawn from which on the parabola, y square is equal to 4ax, are such that two of them make complementary angles with the x axis. So it's a concept based on co-normal points on parabola. You have to give me the locus of P. Hopefully three and a half to four minutes is what I can give you for this question and give me a response on the chat box. No response from anybody so far. Let me give you one more minute. Okay, very good answer. Mansha, oh my God, what is wrong with me? Sorry, Mansha. Guys, Bakka, Bakka, last 30 seconds. Okay, Ruchir, okay, Tripan. Okay, so Tripan wants to change his answer to okay. Ruchir Singh, okay. Okay, so I'm getting a mixed kind of response. All the options have been touched upon by you guys, A, B, C, D, all of them. Okay, now all of you listen to this. Probably this is a concept which you may have forgotten even though it was done in the class. So co-normal points are basically those points, right? From where if you draw normals, they will be concurrent at a point. Also has to say that if I choose a point, okay, from there I can draw. Now I'm not saying all of them will be real, right? There may be one real and two imaginary. There may be three real also. So these points, A, B, C, D such that the normal drawn at them meet at a point. They are called co-normal points. Now one of the very important properties of co-normal point is, if let's say this normal has a slope of M1, this normal has a slope of M2 and this normal has a slope of M3, then the sum of their slopes is zero. Now why? We'll try to understand it. First of all, when does a line y equal to Mx plus C becomes a normal to be parabola? y square is equal to 4ax. What is the condition for normality? What is the condition for normality? Anybody knows it? I've forgotten it. C is equal to negative 2am by minus amq. Yes, C is equal to negative 2am minus amq. Very good, Trippan. Remember that. So now if you see this, this normal has to be satisfied by the point h comma k. So in place of y I put a k, in place of x I put an h. Which clearly leads to a cubic polynomial in M. Something like this. Now this doesn't have an M squared term, so there is no M squared term. Which clearly indicates that the sum of the roots must be zero. And what are the roots here? The roots here are M1, M2, M3. The ones which I've shown you in the diagram. So cubic equation means there can be three possible normals drawn from a point onto a parabola. No, I'm not claiming that all of them would be real. Two of them may be imaginary, one of them may be real or three of them may be real. So that really brings to this fact. So this is very important. If you remember it, you really save a lot of time. You really save a lot of time in the exam. Now coming to the equation, the question says, two of them make complementary angles with the x-axis. So let us say, if M1 is the slope of the other, and let's say this is tan of theta, then M2 is actually tan of 90 minus theta. Okay, let's say, and M3 be M3. Which clearly means M2 is one by M1. So this relation is anyway is known to us, apart from the fact that M1, M2 plus M3 is equal to zero. Now, let us try to get the locus of the point P here, which is h, okay. Now I'm explaining and doing everything. So it may take a little bit time. So don't think that this process is lengthy because normally I would, I would not do all these things in my exam. I will directly come to the final, this expression and this expression. Apart from that, we can see from this equation. That put up the fruits two at a time. So M1, M2, M3, M1 minus BC by that should be equal to 2A minus h by A, correct. So here I can do one simple activity. I can replace M2 with, I can replace M2 with one by M1. So this will become a one anyways. And I can take M3 common from it. I'll get M1 plus one by M1 is equal to 2A minus h by A minus one, which actually becomes A minus h by A minus h. Okay. Anyways, then second equation I can write is the product of the root is, product of the root is going to be minus K by A. So product of the root is M1, M2. M3 is equal to minus K by A, which clearly leaves with M3 being equal to minus K by A. Okay. Now all of us, let us try to understand from here. From this equation, I can say M1 plus one by M1 is actually negative M3. So this guy here, I can replace with negative M3. So negative M3 square is equal to A minus h by A. And from here, I have M3 minus K by A. So I'll put this here. Okay. So that will leave me with negative of K by A the whole square is equal to A minus h by A. Now this is going to be your locus. Let us try to simplify this. So on simplification, this gives you K square is equal to A minus sign. I'll flip the h and a position, okay? Which means y square is equal to A x minus a. So whichever option says this, that would be your option. So y square is equal to A x minus a. Y square is equal to A x minus a is clearly option number C. And C was only given by Krippan and Mancha. Others who made a mistake. I hope there is a lot of learning hidden in this question. Locusts. Locusts is simply a concept which is loved by JMM exams. Sorry, J advance exams. So ensure that you are well-versed in your art of finding locusts. It's an art actually, I would say that. Because you know the process. Well done Mancha and Trippan. Can we now move on to the next question? I hope you know your mistakes now. Okay. All right. Time for the next one. Tangents to the hyperbola drawn from the point alpha, comma, beta are inclined at an angle of theta and phi to the x axis. If tan theta into tan phi is 2 and the hyperbola is 3x square minus 2y square is equal to 6, which of the option do you think is correct? Put forth your response on the chat box. This is a very easy question. Should not take you more than three and a half minutes. Max to max. Okay Pratham. Very good. Okay. Trippan. Okay. We are waiting for the rest four of you. Which is Singh by the way. Which is Parik waiting for your answer. Mancha, Sharmik and Hia. Okay Mancha. Okay Sharmik Hia. What's your response? Okay. Now basically it's a case of pair of tangents. Right. So there's a hyperbola from a point. Okay Rujal. So from a point alpha, comma, beta you are drawing a pair of tangents. Basically let's say these are the slopes of these two tangents. The question is basically saying that m1, m2 is equal to 2. Okay. So first of all, what is the equation of a pair of tangents drawn from an external point to the hyperbola? It is t square is equal to SS1. Okay. This is the equation of pair of tangents. So now this will normally give you a general second degree equation like this. A x square, B y square, 2 h x y plus 2 g x plus 2 f y plus c equal to 0. Correct. Now if let's say this corresponds to a pair of straight lines. Okay. Now what gives you the product of the tangents? Now please note that the product of the tangents information is hidden in is hidden in A and B. Right. So it's actually your A by B. This everybody knows. All of you know. So basically if you write your y as mx plus c. Okay. You would realize that A would be like your constant term and this would be like your m square term. So we know that in a quadratic expression like this, which will be coming out in terms of m, the product of the slope is constant by the coefficient of m square, which is A by B in this case. So now we have to figure out this A and B from this guy t square is equal to ss1. So what is t square? What is going to be t square? So t square is going to be 3 alpha x minus 2 beta y minus 6. Okay. So square of this will be t square. What is ss1? Ss1 will be 3x square minus 2y square minus 6 into 3 alpha square minus 2 beta square minus 6. Okay. So from here I need to just pull out the ratio of the coefficient of x square by the coefficient of y square. Now you don't have to expand it to get that coefficient. You can directly get it if you basically segregate out the coefficient. So what will be the coefficient of x square? Coefficient of x square will be 3 into this. By the way, this is the constant term. So it will be 3 into this and from here I will get minus 9 alpha square. Okay. Fine. What is the coefficient of y square? Coefficient of y square will be minus 2 into this and from here I will get minus 4 beta square. This ratio is given to us as 2 in the question because m1, m2 is given to us as 2 in the question. So from here we have to find out the relationship between alpha and beta. So let's try to simplify this.  So I'll be getting minus 6 beta square minus 18 upon plus 4 beta square and minus 4 beta square will get cancelled off. So I'll get minus 6 alpha square plus 12. Okay. By the way, 6 also we can remove from here minus beta square minus 3 by minus alpha square plus 2. Take it to the other side minus 2 alpha square plus 4 is minus beta square minus 3. So that gives you beta square as 2 alpha square minus 7. I think I've seen this option. Yeah, that's option number B. Option number B. I think it will all be a B as the answer.  So shomik, hiya and ruchir parik. Where did you go wrong? Where did you go wrong? I hope you are clear where you went wrong. Okay. So do all your mistakes here only. Don't do this mistake in your day advance exam. Okay. My teacher always used to say there is always a element of misfortune and bad luck in quota of everybody. It's better you exhaust that quota before the exam. So whatever is the misfortune, whatever is the bad luck, do it when the stakes are low. Okay. Let's move on to the next one to some multiple choice questions. Okay. So let's take a multiple option correct question. Sir, we can take point slope form and then use condition of tangency. It comes out faster. Okay, good. More than one options are correct here. Please ensure you have marked all the options. So from common sense, only A or B could be correct. Both of them cannot be or both of them could be wrong also. But A and B together cannot be correct. C and D, of course, they are all different things. So let's figure it out. Okay, mancha. Guys, one more minute I can give you for this. Last 30 seconds. Okay, here. Okay. So mancha, pratham, and here I have responded. Others, four of you. Guys, please back up. Or if you're not able to solve, you have to take a guess because already three to four minutes have been given for this question. Ruchir, both Ruchir. Okay. So I'm getting one person is okay. Okay. Ruchir Singh has only given one option. Okay. So most of you have said BCD. Mancha has said AC and Ruchir Singh has just said an A. See, guys, first of all, the centroid. This is very easy to figure out because we know that ortho center centroid and circum center lie on a line, which we call as the oiler's line, right? And the centroid divides ortho center and circum center in the ratio of two is to one. Correct. So this is three comma four. This is one comma two. It's very easy to figure out what is the centroid coordinates. So that's two into one, two into one, three into one will give you a five five divided by three and two into two four four plus four eight eight divided by three. So that means C has to be pukka pukka correct. In fact, all of you have Mark C. There's no problem with that. Next thing is, I don't know how many of you realize this fact that the mirror image of the ortho center about this line PQ will actually lie on the circum center. Now I can easily prove this. This is very simple. See, basically what I'm trying to say is that what I'm trying to say is that this angle and this angle are same. That means this two distances are also same. How does same very simple. If you look at this arc, arc h dash q, it basically subtends this angle over here, which is actually 90 degree minus C. And if this is 90 degree minus C, then this angle must also be 90 degree minus C. Angles are printed on the same mark are equal. Second thing is if you see this is 90 degree and if this is angle C, then this also has to be 90 degree minus C. So these two angles are same. Hence, let me call this as M and P H M and P H dash M are congruent. Okay. So can I get H dash by taking the mirror image of H about PQ? Now, if you recall, we had done a formula for finding the mirror image of any point about a line. If you have forgotten that formula, I'll just take one minute to recap that. So let's say if alpha beta is a point and there is a given line to you AX plus BY plus C equal to zero and you want to know what's the mirror image. Okay. So let's say mirror image is, what do you call it? Let's say X5. Okay, let's recall it as X5. No, let me not call it X5. Let me call it as PQ. Okay. So let's say PQ is the mirror image of the object alpha beta about the line AX plus BY plus C equal to zero. Then the formula that we know is, what is the formula? P minus alpha by A is equal to Q minus beta by P is equal to negative 2. Negative 2. A alpha V beta C by A square plus B square. Okay. So this is a direct formula for finding the mirror image. By the way, just to remind you what is the foot of the perpendicular for that you just drop this 2. I hope you remember these formulas. We have been using it time and again and frequently to solve questions. So from our given situation, let us try to find out what is the mirror image of the ortho center, which will happen to lie on the circum circle. Okay. So this is 3 comma 4. So let me call this point as P comma Q. So I'll use the formula over here. P minus 3 by A. If I'm not mistaken is 1. Q minus 4 by minus 1 will be equal to negative 2 times 3 minus 4 plus 7 by A square plus B square, which is 2. Okay. So from here, if you figure out the values of P and Q, they will come out to be this is 10 minus 6. So minus 6 and this will come out to be minus 3. So P will come out to be minus 3. And this will come out to be 10. So Q will come out to be 10. Okay. So this point is minus 3 comma 10. Now this will clearly help you to get the radius. So for radius, it will be 4 square plus 8 square under root 4 square plus 8 square under root, which is 16 plus 64 under root, which is root 18. So it is obvious that option number A has to be correct. So B will be wrong and D will be wrong. So option A and C are correct. And the only person to get this right is Mancha. Well done, Mancha. Very good. Now did you guys need more time to do it? Was that the case? Because I'm not very happy with the way you have responded in this question. Just one person getting it right. Ruchir Singh, why do you have to guess? What happened? Ruchir Singh. I wasn't able to do it in time. What was the problem? I didn't know the property. The reflection of the ortho-center will be in a circle. Rupal. Sir, I didn't get that. Okay. Next question. If a variable tangent to the circle, x square plus y square is equal to 1, intersects the ellipse at point p and q, then the locus of the point of intersection of tangent at p and q is, what is this question? Okay. Is a parabola. Okay. I was thinking where is the equation? Okay. Is a parabola with later sector 4? It's a parabola with focus as 2 comma 3. It's an ellipse with eccentricity root 3 by 2. And it's an ellipse with eccentricity greater than half. Okay. This is I think more than one option correct again. So please solve this 3 to 4 minutes is what I can give you. I think it should not take you that much time. Give me a response on the chat box. Okay. Very good. Monsha. Yeah. It's very obvious that if C will be correct, then D has to be correct. Weissa versa need not be. Okay. Very good. Guys. Last minute. Good trippin. Pratham. Very good. Ruchir Singh. Ruchir Parik. And here. Ruchir Singh also. Both the Ruchir and here we are waiting for you. Okay. Let's discuss. So it's very obvious that let's say there is a tangent. Let's say there is a tangent to this circle. And let's say I call the tangent as x cos theta y sin theta equal to 1. Okay. You can treat this tangent to be the chord of contact generated by this point of intersection. H comma K. Okay. So treat it as if you have drawn a pair of tangents from this point H comma K and this variable tangent to the circle behaves like the part of contact. So we know that the equation of the chord of contact is T equal to zero. So it'll be H x then it'll be 2 y K equal to 4. Now both these equations are same as per our question. Both these equations are same as per our question. So I can say cos theta by H is equal to sin theta by 2 K is equal to 1 by 4. In other words you eliminate theta from here by using the fact that cos theta is H by 4 sin theta is K by 2. So using the fact that cos square plus sin square is going to be 1. It's very simple. H square by 16 K square by 4 is equal to 1. Which means x square by 16 y square by 4 is equal to 1. So this is clearly an ellipse. Okay. And A is more than B so as eccentricity will be under root of 1 minus B square by A square. Which is under root of 3 by 2 in this case. Very very clearly. Which means option number C is correct. And because C is correct, D is also correct. So C and D are the only right options. Everybody got this right. Well done. So without much mist of time we'll move on to the next question. Okay. So here's a question. Equation of a line that touches the curve y is equal to x mod x and x square plus y minus to the whole square is equal to 4. Yes. Yes. Any success for anybody? Okay. I don't know whether it's a single correct or more than one. You have to figure it out. I think a simple construction of the entire situation will tell you how many answers should be there. So I've got a response from Mansha here and Richard Parik so far. And all of them have given only one correct option. One correct answer. Okay. Only one. Okay. Okay. Pratham. We are just waiting for trippan. Okay. So trippan gives two options. Okay. Fine. So we all know that this guy x mod x is basically x square when x is positive and negative x square when x is negative. Okay. If I just have to sketch it, it is just going to be two parabolas, one opening upward and other opening downward, other half opening downward. Okay. And there happens to be a circle whose center is at zero comma two and radius is also two. So there's a circle which is like this. Let me use my tools. Now it is very obvious from this diagram that there could be two tangents possible. One would be like this. Okay. And the other could be like this. Now, let's try to figure that out what is going to be the answer for this. So I think there should be two answers coming out. Let's try to verify it. If I take the case where x is positive, that means I consider y is equal to x square and the circle. Okay. Now we know that for y equal to x square, what is the condition for tangency? What is the condition for tangency for x square is equal to four a y. So when I say y is equal to mx plus c is a tangent to this kind of a parabola. What is the relationship between mc and a? Who will tell me? Who will tell me? What is the relation? c is equal to minus a m square. Absolutely correct. Okay. So now in this case, your a happens to be one fourth. So c should be minus one fourth m square. Right. In other words, you're claiming that y minus mx plus one fourth m square is a tangent to the parabola of course and also to the circle. It's a tangent to the circle. So you can use the condition that if something is a tangent to the circle, that means the distance of the center from the tangent must be equal to the radius of the circle. Correct. So the center here is zero comma two. Okay. So when I put that, I get two. This is zero plus one fourth m square mod upon under root of one plus m square. This should be equal to two. Okay. Let's have to solve for it. So let's write it as on expansion. This will give you four. This will give you one, six, seven to the power four plus m square is equal to four plus four m square. Okay. I hope I'm not missing out on anything. Do let me know if I'm missing out on anything. So this will give me one by 16 m to the power four as three m square. Now m is zero. That is less likely. Okay. So m square. I'm going to cancel from both the sides because I acknowledge the fact that m cannot be zero. So m square is equal to 48. So m happens to be four root three. Okay. Plus minus four. So four root three. I can at least see over here. Let's try to verify with the constant part of the constant part is fine. You can go ahead. So why minus four root three X plus one fourth of m square m square is 48 48 by four is 12. So that clearly matches with option number B. So B. Let me check again. Yes. B is the right option. So anybody who has marked with B is definitely correct. Now this is half the problem done because I speculate that they cannot. They can be another tangent to the other arm, which is Y is equal to negative X square. Okay. Now let me repeat the process. So when you have X square is equal to minus four a Y and you have a tangent Y equal to m X plus C to eight. What is the condition of tendency? What are the conditions for tendency? C is equal to m squared. Okay. So why minus m X minus. Again, your A is one fourth. Okay. Again, using the same fact that the distance of the center from the tangent should be equal to the radius. So zero comma it will give me something like this. So this mod is equal to two under root 1 plus m square.  Four four getting canceled. Oh, sorry. Yeah. So this will be, this will be giving you this equal to five m square. So m is equal to plus minus root 80. Okay. So root 80 is like four root five. Okay. So four root five, I can see sitting in my options. Okay. And four root five. Now, why not minus four root five? Can it be minus four root five? No, because here you are considering that your tangent. Let me ask you this question. We have two answers coming out. One is four root five and other is minus four root five. Okay. Only one of them is possible. Which one of them is possible and why? Yes. In the third quadrant, if I'm making any tangent, it has to have a positive slope. Why? Because divide by dx is minus two x and you are in the third quadrant. So your x is a negative quantity. So your slope has to be positive. So you cannot have minus four by root five. So whichever option says minus four by root five that anyways would be ruled out. C and D will be ruled out, but I have some hope with A. So let us try to figure out. So this guy will be y minus four root five x minus one fourth of 80, which is going to be I think 20. So four root five x plus 20 is going to be the answer. So option number is definitely correct. So A and B are the right options. The only person to get this right is trippin. I mean, get this completely right. Okay. So a quick recap of our condition of tangency for our standard parabola case. So let's now move on to the next question. So we'll take this one. It's a paragraph based question. I hope this figure is large enough for you to read it. As of now, I'm interested only in the 19th answer. So please solve the 19th one. So the question is there's an ellipse. Those semi measure access is of length to semi minus of length one. It is slipping between the coordinate axes in the first quadrant while maintaining contact with both x and y axis. What is the locus of the center of the ellipse? What is the locus of the center of the ellipse? This is a one minute question, but I can give you two minutes if you want. That's correct. Oh, sorry. I should not say that. Okay. Very good. Given the answer, as I told you, it's a very simple question should be solved within one minute. Just saying. Okay. Very good. Who's left? He is left. Yeah. Okay. So most of you have gone with D for dog. Okay. See here, we must all understand that these are your coordinate axes are like tangents to this ellipse. And because they're meeting at right angle, basically origin must lie on the director circle. Okay. Now, one thing is very interesting about director circle and you would be knowing it also. Basically, if you have an ellipse, okay, x square by a square plus y square by p square equal to 1. I just give you an example of this ellipse. Okay. So this ellipse, let's say has the center at origin. Okay. Now, let's say this is your director circle. Okay. Now, if I start rotating this ellipse, will it change the equation of the director circle without changing the origin without changing the center? So let's say if I start rotating this ellipse about the center, will it start giving you different director circle or director circle will remain the same? Absolutely. Director circle will remain the same. So director circle of an ellipse basically depends upon the A and B values and the position of the center. That's it. Okay. So if the center is at some point h comma k, then the equation of the director circle will be simply this. That's it. Okay. Whether the ellipse is rotating or not, it doesn't make any difference provided it maintains the same center and it maintains the same value of A and B. The director circle equation is not going to get changed. Okay. Now, if you see here, in this case, our center is changing, but we know for the fact that the director circle here, this point always lies on the director circle. Correct. So wherever is the center, the distance of the center from alpha comma beta must always be under root of A square plus B square. Okay. So this clearly brings to the fact that under root of alpha square plus beta square should be under root of a square plus B square, which is in this case, two square and one square respectively. So alpha square plus beta square has to be five. And if you generalize this, the locus of the center has to be option number T. Okay. But my main aim is our main ambition for you guys is to solve the question number 20. So now I would request you to put your response for question number 20. Okay. Very good. Mancha, Ruchir, very good. So three of you have already responded with the same option, so it looks to be correct. Okay. Tripan has also responded with the same option. Ruchir Singh also with the same option. Awesome. I hope you're not communicating among each other. Okay. Who's left? I think Hia is left out. Hia and Shahmiknaal. Okay. So only Hia is left now. Hia, you're waiting for you. So everybody has said B, so I think it requires no discussion. Let's discuss it anyways. So first of all, if the ordinate of the center is one, the abscissa has to be two. Correct? Because alpha square plus beta square has to be equal to five. So if ordinate is one, okay, then this has to be R2. Okay. Now what are the question? The question is a ray of light passing through the origin and the center of the ellipse. That means passing through the origin. So let's say this is zero, zero, and this is two comma one. So there's a ray of light passing through it. Okay. What this ray does, it hits the tangent at the vertex of this parabola. Now if you look at this parabola, it is basically y minus five the whole square, if I'm not mistaken, and you will have negative four x, 25, and you have to have a minus 10. Okay. And remember, if you have any case of a shifted parabola like this, plus minus four a x. Okay. The vertex tangent, the tangent at the vertex is always obtained by tangent at the vertex is always obtained by putting capital X equal to zero. Okay. So if you remember, I had basically discussed this with you when I was telling you the very general equation of any parabola. So when you say why, why is actually nothing but the distance of the square of the distance from the axis, square of the distance of the point on the curve or the parabola from the axis. Okay. And this four a basically represents the length of the latter sector, the length of the vector. And this X basically represents the distance from the distance of the point from the tangent at the vertex. So when you equate this to zero, you directly get the equation of the tangent at the vertex. So tangent at the vertex is X equal to 10 here. So what is happening? There is a line X equal to 10. So that has to be a vertical line. So let me just sketch it. So there is a vertical line X equal to 10. Now what happens? This ray hits here and goes back and cuts the Y axis somewhere. Okay. So let's say this is your Y axis. Y axis. So this is your Y axis. So where does it cut the Y axis? That is what we need to figure out. Right. So how will you figure this out? How will you figure this out? First of all, this line is Y is equal to X by 2. Okay. Yes or no? So this point here is your 10 comma 5 point. Right. So from the symmetry of the figure, we can say that we can say that this angle and this angle are the same. Correct. So if this jump, what is this jump? This jump is a 5. Correct. So this will also be 5. Yes or no? Because it is an isostasis triangle. So you have to jump 10 to reach this point. So you will be reaching the point 0 comma 10 where the reflected ray will hit the Y axis. So option B is definitely going to be correct. Is that fine? Most of you got this right. Well done. So I was expecting that also. So let's move on to the next question. This is a column match. In fact, you already know the instructions so there's no point showing you the instructions. Now give your response together. Don't be like A is matching to let's say R and then you wait for some time to see B option. So give everything together. Okay. So whatever is, do you think is matching to A, B, C, D. Write all of them together and then press an enter. Okay. Don't give your answer in bits and pieces because if you answer one then other person will answer in between. So I will lose track on what you had actually answered. Actually the list part was needed only for this question. Tripan, sorry. I lost touch. Which question you were referring to? Previous one. Okay. Okay. Yeah. Okay. No issues. Should we discuss it now? Are we all ready? I got only response from here so far. Okay. Pratam. Good chair. Parik. Good. Okay. So what about others? Mansha. Richard Singh. Shomic. Tripan. Okay. And uptime has been given. So those who want to guess, you can very well guess it. So I'm giving you an option to guess in case you're not able to solve. Okay. Take an intelligent guess and we will take it forward from there. So I think we are just mating for Mansha. Who is not in the call. So we'll move ahead. Let's take this forward. See the length of the focal chord, if you remember, we had done that the length of the focal chord is 4 a cosecant square alpha, where alpha is the angle that it makes with the x axis. So if you have any kind of a focal chord, okay, and let's say it makes an angle alpha, then the length of the focal chord, let me call it as pq is given by this expression, correct. Now, as per the question, it is inclined at an angle of pi by 3. So length of the focal chord is 4 a is 1 and cosecant square pi by 3 will be 2 by root 3 the whole square. Okay, that's nothing but 16 upon 3. Okay. So length of the focal chord inclined at an angle is L. So this is L. So they want to find out 9L by 16. So 3L by 16 is 1. So 9L by 16 is going to be 3 for sure. Okay, so it is greater than only p and q. Okay, so A will match to p and q. A will match to p and q. Let me check, who all gave the answer for A, S, P and q. So Pratham gave, Ruchir Parekh, I do not know why you've mentioned everything. Shamik gave, Ruchir Singh has not replied anything. Yeah, Ruchir Parekh I'm talking about. Oh, this is a 916 by 11. Next one, the number of normals that can be drawn from this point to the parabola y square is equal to 4x. Now we know that y equal to mx minus 2am minus amq is a normal equation, right. And since you're claiming that this normal should pass through 6 comma 4 and A is 1 actually. So I'll just put the values. This is going to be 4. 6m minus 2m minus amq. So this results into mq minus 4m plus 4 equal to 0. Now everything depends upon how many roots this guy has. How many roots this guy has. So how will you figure this out? How will you figure out how many roots does this particular equation have? Application of derivatives. We had done this concept in application of derivatives. So what we do is, first we differentiate this, right. And we see that the roots of this equation comes out to be plus minus 2 by root 3. If this particular equation has two roots alpha and beta, let's say, so this quadratic has two roots alpha and beta.  Now, if it is a situation where the graph just takes a turn and goes up again, right. Then you would realize that the value of f of alpha and f of beta will both be positive. Correct? Yes or no? So let's try to check that first. So 2 by root 3 if I use, I'll get 2 by root 3 cubed minus 4 into 2 by root 3 plus 4. What does it indicate about this value? This is almost 8 by 3 root 3 minus 8 by root 3 plus 4. This seems to be clearly a positive value. Is it a positive value? Seems to be positive? Yes or no? Because root 3, if you do 8 divided by root 3, it is roughly going to be 4.6. Negative 4.6. And this is going to be roughly going to be 1.53. So this is positive. Similarly, if I put minus 2 root 3 whole cube and of course this is going to be plus 8 by root 3 plus 4. This is again going to be a positive. That means this graph is not going to have more than one real root. That's the only real root it's going to have. The other two will be imaginary. So the answer to this question is the number of normals that you can draw is only one. So option B will match to P. Nobody has matched B to P. Rest nobody I think has matched B to P. Excuse me, sir. Yes, sir. Sir, if the line isn't on the axis of the parabola doesn't that mean it and it's on the exterior of the parabola doesn't it mean it can have only one normal on the parabola? Yeah, you can say from the diagrammatic point of view also, but I'm coming from a mathematical point of view. What you are doing is you're trying to look at the sketch of the diagram and trying to figure out how many normals. But I'm being doubly sure by confirming it with the equation. The number of points where a real flight parallel to the axis strikes the interior of the parabola. Okay, this is seems to be a weird question, but it's very simple. I guess so sorry. So if your free of light comes like this, of course, it's going to pass through the focus and any point focus will become parallel to this. So basically only you'll cut at two points. So option number C will match to Q. There's no other. Okay. So next, the area of the triangle formed by the pair of tangents from zero comma two and the chord of contact. Okay, let's figure that out as well. So chord of contact equation will be t equal to zero. So t equal to zero means y into two is equal to four x plus zero by two. So y is equal to x is the equation of the chord of contact. chord of contact. Okay. Zero comma two will be here. Try to understand him. If you see a situation like this, okay, where this is y equal to x and these are your pair of tangents. Yeah. So we have to find the area of this triangle. Any shorter way to get this. So you can find the perpendicular distance on the base. Okay. Can I first figure out the point Q here will be where y and x values are equal. So x value will be four. So this is four comma four point. Okay. And this guy has to be origin. So you know the four points. You know the three points of the triangle. Correct. Yes. Why are you saying something? So you can take the perpendicular from zero comma two onto y equal to x. Okay. So we take a perpendicular and that value will be two by root two into half into base. Base here will be four root. So root two root two two one answer is four. So that matches with S. So D matches with S. So the answer full and final answer to this question is a matches to P and Q. P matches to P. C matches to Q and D matches to S. Is anybody who got all this combination right thinks nobody got this right. Nobody could get this right. So why is CQ sir? Why is C to Q? The number of points where the real flight parallel to the inside the interior is two points only. I showed you here and here. So but it never said it reflects. It strikes the interior of a parabolic reflector. Actually they want to say. Can we take the axis also and we just one point P and Q. Axis is not a part of the parabola. So that's also a ray of light parallel to axis. No that is not parallel to it. That is coincident with it. But sometimes they'll do something and then give up on something. In that position these two points will become coincident at the vertex. This may be a controversial question also. I can understand your dilemma. But let's try to solve it by the most you can say the common feeling that everybody will get. Not by a very special feeling. Sir I had a doubt in the previous question. Sir in that I think the B part where they asked about that point. So Xiaomi was saying that if it is not on the axis then it has to be only one normal. I didn't understand what that meant. Can you explain? See he's coming from a dynamic point of view. He's saying that if the point is somewhere over here then there will be one real normal. That's what he's... No I think he's trying to say if it is outside the parabola they will know. That's what he was trying to say. Okay okay guys. I understood. So another column match is equivalent to solving four questions. So I think definitely it will take time. I'll give you around eight minutes, seven to eight minutes to do this. Put all your responses together. Yes guys any response? Okay so I've got for Mancha. Very good Mancha. Last 30 seconds. If you want to guess your answers you can do that. So I think eight minutes are over almost. Okay which is saying guess it. If you're not really able to solve it guess it so that we can discuss and we can save our time and learn the process. Yeah take a guess. Okay let's discuss. See guys let's look at the A part of the question. It says that these three lines are concurrent. The condition for concurrency is this determinant should be equal to zero. Correct? So if you simplify this what will you get? So let's simplify this. So one will have two minus minus three which is five then we have negative two, two a minus three b. And then we have a negative one minus a minus b this should be equal to zero. So on simplification this will give you minus three a plus seven b plus five. Correct? This is equal to zero. Yes or no? Now they're asking you what is the least distance of the origin from a comma b? So they're asking you what is the least value of this? Okay now see here. Try to imagine that there is a plane or there's a line for that matter. Okay whose equation is this and there happens to be a point on this line. So what is the least distance of origin from this point is basically trying to ask you what are the least distance of origin from this line? Negative three x plus seven y plus five equal to zero which has happened to be the perpendicular distance. Correct? So perpendicular distance the least value will be five by under root of nine plus forty nine which is root fifty eight. Okay? So s root fifty eight has to be equal to five. So the first option will, a will match with q. So if your a is matching with q absolutely correct with respect to a. Well done. I think most of you matched a with q only. Anybody who didn't do that? I think Pratham did not match it. Who else? Mostly everybody who answered matched a with q. Next is n we are to fix points on a line. Okay? So let's say there's a line and we are to fix points on it. Let the locus of a point be such that p a by p b is two. Now p a by p b is two. None of you try to understand it. If there's a point which moves in such a way that p a is to p b is a value which is a number not equal to one. Then basically p will lie on a circle. Okay? And that circle, most of you know the name for it also that circle is called the a plonious circle. This a plonious circle has a characteristic that it will cut the line joining a and b at such points. Let me use m and n such that m is to a m is to m b is two is to one. A n is to n b is also two is to one. That means this is the case where m and n are harmonic conjugates of each other. I'm sure you would have heard of this name harmonic conjugates. Harmonic conjugates are nothing but they are two points which divide the joint of two points in the same ratio but one internally and other externally. Are you getting my point? So m will divide a b internally. So this will be an internal divisor. Okay? Internally in the ratio two is to one. And this guy will divide externally in the ratio two is to one. Okay? And this will be a circle. Now, let us read the question completely what they're asking us. So they're saying that p a is to b b be a curve cutting L at R and S. Okay? So they're calling this as R and S. So I'll change their name. Nothing in the name. So they're saying the slope of PR is minus half. The slope of PR is minus half. Okay? So this has a slope of minus half. What is the slope of PS? Now you know for sure that this is going to be 90 degrees. So this has to be negative reciprocal of that. So answer is two. So b will match to R. b will match to R. Anybody who gave R as the answer. b will match to R. Tripan gave B as R. Shawmic gave B as R. Ruchir gave Man Shah gave. Very good. Everybody was on right track. Awesome guys. Next. Let tangents add p and q to a curve this intersect at t. Okay? Good. If two comma one is a point says that sp into qs is 16. Then the length of st is less than. Okay? Now if you see this curve. It is basically y minus one the whole square. Is equal to I think four x minus one. Okay? Now this has got a value as one. It's a shifted case. It's basically a shifted case of a parabola. Where this parabola has been shifted. One unit to right. Oh sorry. One unit to the right and one unit up. Okay? So where does the focus come for this? So the focus will come at a point two comma one. So s is actually the focus of this particular parabola. Okay? Now what is the question saying? Let tangents at p and q to the curve intersect at t. Okay? So let's say this is p and this is q. The tangents intersect at t. Okay? All right. Now everybody please pay attention. Everybody please pay attention. Just a second. Now all of you please just try to recall. In case of a standard parabola. If you were drawing drawing let's say y square is equal to four x. Let's say this is your parameter t one. This is your parameter t two. If I draw a tangent at t one and I draw a tangent at t two. Okay? What is the coordinate of this point of intersection of this two? Now if you recall this coordinate was go up.  Correct? Now what is the distance of this point? Let me call this as p. Let me call this as q and let me call this as t. What are the distance of go up from the focus? Now you would realize that the distance s t is nothing but a t one. A t one t two minus one the whole square plus a t one plus t two the whole square under root. This boils down to under root of one plus t one square one plus t two square. Okay? Now this result should be known to you actually because this leads to a plus a t one square into a plus a t two square. Now if you look at the situation very closely you realize that s p distance is a plus a t one square and s q distance is a plus a t two square. So your s t is nothing but under root of s p into s q. Okay? This is something that a serious J aspirant should know. Now coming back to the question. We already know even if it is a case of a shifted nothing changes because shifting is happening for all the points it's not going to change the relation. So s p into s q is 16 so s t is going to be under root of s p into s q which is going to be four. Now we're asking you this option is less than which of the following. I think it is less than q. So c will match to q. It is less than t. So q t is the answer for option number c. Qt qt qt qt Bruchir Parekh has mentioned qt. Mansa has not mentioned qt. Bruchir Singh has mentioned qt. Shaumik just t. Okay. So be careful. I think Shaumik you missed out marking q. Last part of the question. Any questions here? Anybody? Till ABC, does anybody have any concerns? Do highlight. So does the length from the product of tangents for like circle as s one? Does it work for other coding source as well? Under root of s one? Not under root because different length but the product is just s one. Product of what? The product of the two tangents as s one. I didn't get that. If you are drawing a tangent, two tangents to a circle, the length of the product of these two is s one. That's what you're saying. Yes. Can you do that for any point? Oh, no, I don't think so. I'll work for any other point. Okay. Let the double ordinate P and P dash of this hyperbola is produced. Both sides to meet asymptotes at Q and Q dash. What are the product of P Q into P Q dash? Okay. So let me just make this diagram quickly. So let's say this is your hyperbola. Okay. Let's let's take any double ordinate like this. Okay. Let's actually extend it forward so that it cuts the asymptotes. By the way, this is a very crooked line. So let's say this is your P and P dash and it cuts the asymptote at Q and Q dash. They're asking you what is P Q into P Q dash. Now, if I were you, I will just take a case where my double ordinate is a lattice rectum. Okay. So I know that this point is going to be, by the way, if I take our normal hyperbola, then this asymptote is Y is equal to B by X and this asymptote is Y is equal to minus B by X. Correct. And let's say this is my line X is equal to A. Okay. Passing through the focus. So now this point point P is actually a comma B by, oh, sorry, a comma B square by. Okay. And this point Q, if I'm not mistaken, it is a comma. And if I put a over here, it will become B. So the distance P Q is going to be B minus B square by. Similarly, the distance P Q dash is going to be B plus B square by. So they're asking you what are the product of these two. So the product will be B square E square minus B to the power four A square. But I have to use my formula below that E square is going to be or you can write it like this. B square by A square is one plus E square or sorry, E square minus one, E square minus one. So I'm going to write this guy as minus B square times E square minus one. So this results into B square. That means the answer here is actually a fixed value. Okay, B square. So we'll see what is the B square for this hyperbola. B square for this hyperbola is clearly pi minus one. So D will match to S. By the way, this is true even for any other point you take. Okay. If you take a double ordinate at any point, I have taken a very special case of a double ordinate, which happens to be the latter sector. But even if you take any other point, P Q into P Q dash will always be the square of the semi conjugate axis. This relation will always be true. However, many people do not know this result because it is not a very useful, I would say, commonly seen result. So in case such a thing comes, you may have to spend some time deriving it out. But again, if you want to keep this in mind, you can very well do that. So your option D will match to S. D will match to S. Anybody who has, everybody has matched D to S. Very good. Probably because it looked like very similar to the given expression. But if I were a smart question center, I would have made these values resemble these normal numbers. Okay. So that the students get confused. Can I move on? Oh, sorry. Sorry. Can I move on? I think this was a very beautiful question. We learned a lot of subs. Let's take some integer type questions also. If this circle bisects the circumference of this circle, then what is C plus D by 10? Two minutes for this. Not more than that. Okay, Pratam. Very good. Anybody else? Very good. Mansa, Ruchir Parik, Ruchir Singh. Waiting for Trippan. Shomik. Yeah. Okay. Yeah. Very good. Let's say, let's discuss it. Shomik, you want to answer? So other than Shomik, everybody has answered with 5. Trippan has answered 3. Okay. See, if the white circle bisects the circumference of the blue circle, it is very obvious that the common chord will actually be the diameter of the blue circle. So this common chord will happen to be the diameter of the blue circle. So common chord equation is s minus s double dash equal to zero. So just subtract these two equations. You'll get the equation of the common chord, which is in this case, 6x plus 14y. Now, if this is the diameter, it must also contain the point which is the center of the circle, which happens to be 1 comma minus 4. So 1 comma minus 4 should pass through it. So 6 minus 56 plus C plus D is equal to zero. So C plus D is 50. So the answer is loud and clear. C plus D by 5 has to be, sorry, C plus D by 10 has to be a 5. Trippan and Shomik. Kiochi, Sada, Shomik. I didn't have enough time. If the normal to this hyperbola at a point T meets the curve again at D dash, what is mod of T cube into D dash? So I think tomorrow you have a physics session. Tomorrow you'll have a physics class. Okay, Trippan. Shomik seems to like this Samba 4. Okay, Shomik. What about here? Shomik seems to like this Samba 4. Okay, Shomik. What about here? Pansha. Just saying. Patham. Okay, I'll give you, I'll give you last 30 seconds to wrap this up. So those are just one minute. Okay, Manu sir. Normal to a rectangular hyperbola equation, nobody will remember. So half a minute we'll go in, deriving that. Okay, this is the equation of a normal at any point T for a rectangular hyperbola. Okay, now you may derive this also in the exam. You may remember it, but again, this is up to you how much pressure you can take of remembering. So I'm not putting any pressure on anybody to remember things. Okay, in case you end up remembering it well and good. If no, you can derive it also. You know how to find the equation of tangents and normal at any point to occur. We have done application of derivatives. Okay. Now this normal is also satisfied by CT dash comma C by D dash. Okay, by the way, we normally assume a point on this hyperbola to be CT comma C by T that everybody knows about. So now this should also satisfy this curve. Right, so this will leave you with CT dash T cubed minus CT by D dash. Is equal to CT to the power four minus one. Okay, CCC you can remove from everywhere. So this will give you T dash square T cubed minus T is equal to T dash T to the power four minus T dash. Now from such situation, always T minus T dash will always come out to be a factor. Always come out as a factor. Why? Why? Because it is very obvious that the normal drawn at the point will also meet the curve at the same point. So the equation will give you all possible scenarios. So T minus T dash will always come out as a factor. Which you can see very clearly over here if you bring this term to this side and this term to this side. So you'll have T dash square T cubed minus T dash T to the power four S T minus T dash. And here if you take our T cubed T dash common, you will get T dash minus T is equal to T minus T dash. That means T dash T cubed is equal to negative one because T cannot be T dash. So more of this is going to be more of this quantity which is going to be one and that's what the question setters has asked. So it was supposed to be just a one minute one and a half minutes question for a person who remembers the equation of a normal to a rectangular hyperbola at any parametric point or at any generic point. Given two circles, let the radius of the third circle which is tangent to the two given circles and to their common diameter is 2P minus 1 by T. Find the value of T. So let's have three to four minutes to solve this. And please give your response on the chat box. Yes, any success anybody? Mancha. So just one person Mancha has replied so far. What about others? Do you need more time? So one minute. One minute. Okay. Okay. Okay. So let's discuss this. If you want to take a guess, you can do that. I have got just two responses. It's almost around four and a half to five minutes gone. Okay. So Trippan has taken a call. Okay. Parik has taken a call. He has taken a call. Pratham and Shomik. If you're there, you can take a... Okay. Pratham also. So here if you realize that we are talking about this circle whose radius happens to be minus three by root two, minus three by root two. This happens to be the center. And the radius happens to be nine by two plus nine by two under root. It's actually a three. And for the other circle, it is minus five by root two, minus five by root two. And the radius, let's say I call it capital R. That's going to be 25 by two plus 25 by two. That's going to be a five. So given the position of the centers and one more thing you would notice that they do not have any constant term. Constant term, absence of constant term indicates that your circle is actually passing through the origin. Okay. So one circle is clearly like this. And the other one is clearly like this, both of them passing through the origin. And let's say this is their common diameter. So they talk about in the question about the common diameter. So let's say this is the common diameter. They also seem to talk about the third circle, which is tangent to the two given circle as well as their common diameter. So let's say I assume that circle to be something of this nature. Touching the yellow circle, touching the white circle and touching the common diameter. So far so good. So let's say this is the center of the bigger circle. This is the center of the smaller circle. And this is the center of our unknown circle. Let's say I connect these points. Now let's say this radius of this unknown circle is X. Let me name the centers also so that we can, you know, address them by that name. So let me call the bigger circle center as O1. This is O2 and this is O3. Now, can I say, all of you please listen to this statement of mine. Can I say O2 O3 square, O2 O3 square is, let me call this point as P for the timing. Can I say it is O3 P square plus the distance O2 P square. Can I say that? Yes or no? O2 P square is O1 O2 plus, now see it. This is the sense I can write it as O1 O3 square minus O3 P square under root. Correct? So basically this is what is O2 P square and this is nothing but O2 O3 square. Now I'm going to put our given expressions, capital R, small r and X. So this is capital R. This is small r. So we are going to put that expression. So O2 O3 square is capital R square. Oh sorry, my bad. This is minus, this is going to be small r plus X square. This is small r plus X square. O3 P, O3 P is X square.  So this is capital R minus small r. Correct? This whole thing was capital R and if you subtract small r, you will get O1 O2.  O1 O3 is capital R minus X. So the whole thing is capital R. Subtract this part X from it. Minus O3 P square is X square. Let's try to simplify this further. So this will give me X square to Rx and this X square will get cancelled off from here.  Okay, now inside here also if I open the brackets, I will end up getting, I will end up getting just capital R square minus to Rx. Is it fine? Let's square this also.  R square minus to Rx and you'll end up getting two times R minus R under root of R square minus to Rx. Okay, so this is going to be R square plus small R square minus to RR. And this is going to be R square minus to Rx plus to R minus R under root of R square minus to Rx. Ooh, this expression seems to be quite a handy one. Okay, so I have to solve for X here. I have to solve for X here. Anything else that can be taken care of? So this is two R. This is going to be plus four R. This is going to be capital R square plus to R minus R under root of R minus to Rx.  Okay, but let's try to simplify things. Oh, one, one thing I can do here is that I can see here. I can see here some kind of, yeah. Why did you write four Rx? Wouldn't it just be two Rx that time? Okay. So before this here, two R plus four R. One second. Did I do a simplification L over here?  Under root of R square minus two Rx. Right. So I took this one. Minus two R, small R minus two Rx. So it should be minus two R plus X. Oh, yes. Sorry for that. So two R square minus two RR minus two Rx. That's fine, Shavik. Yes, sir. Any questions here? Okay, so dropping the twos from everywhere. So this will lose a two. This will lose a two. This will lose a two. This will lose a two. This will lose a two.  So clearly R minus R can be removed off from everywhere. Oh, this is going to be plus. Now it cannot be removed. Somehow it is becoming too lengthy. Or probably I should have put the values and simplified it too. Okay. So I should have put the values. Now if I have to make a correction, I have to make a correction right at this step. Let me just put the values. I think that will make our life more easy. Rather than dealing with these values. Sorry about that. I could have put the values in. So in this case, our small R was three. So three plus X the whole square. Okay. And this is going to be two. This is going to be five minus X the whole square minus X squared. Okay. So that will make things more easier. So X square and this is two plus under root of 25 minus X squared. Minus 10 X. And this is nine plus X square plus six X. Okay. So nine plus six X is equal to two plus under root of 25 minus 10 X the whole square. So nine plus six X will give us four plus 25 minus 10 X plus four under root of 25 minus 10 X. So this will leave us 16 X. And I think minus 25 16 X 34. Oh, sorry, minus minus 20. Now drop the factor of four from everywhere. This will give you. Oh, sorry, this will give you something like this. So 16 X square minus 40 X plus 25 is equal to 25 minus 10 X. So 16 X square is equal to 30 X. That is what we get. So X is equal to 30 by 16, which is 15 by eight. So now as per the question to P minus one by P is equal to 15 by eight. That means 16 P minus eight is equal to 15 P. So P value is going to be eight. Okay. So if I had not wasted time, here we could have done this problem well within three to four minutes. So the only person who got this right was Mansha. All of you gave a different answer, which was wrong. So P is equal to eight is the answer. Okay. So we'll stop the session over here. I would like to know from you.