 So I'm now going to start with the diameter of an ellipse. Diameter of an ellipse. First of all, how do you define a diameter? How do you define a diameter? So diameter is nothing but it's a line which bisects a system of parallel chords. OK, so let us say I have a system of parallel chords whose slope is m. So let's say this is a system. I mean, system means a set of parallel chords. And these parallel chords, they have a slope of m. If I start bisecting the midpoints of these chords, I obtain a line which we call as the diameter of the ellipse. So this is called the blue line is called the diameter. Now, I have drawn the diagram in such a way that it passes through the center of the ellipse. But let's figure it out. OK, this is just a speculation, which I'm doing. So is there any, you can say, a robust way to show that this diameter will pass through the center of the ellipse. And let us say these are all system of parallel chords whose slope is, let's say the slope of all these chords is m. OK, first of all, I would like to know the equation of the diameter. One more thing I would like to tell you here is that the point where the diameter meets the ellipse, they are called the vertex of the diameter. This is called the vertex of the diameter. OK, these two are the vertex of the diameter p and q. Let's say. OK, now I would request you all to give me the equation of this diameter. Everybody, please work this out and tell me what is the equation of the diameter? Use your concept of equation of a chord bisected at a given point. Good morning, everyone. Good morning, Charan. Yes. If there is any lag or something, I think last class there was a lag in the session in case anybody is experiencing any kind of a lag, do let me know. It's better. OK, thank you. Thank you, Aditya. Yes, find the equation of the midpoint of. So basically this becomes the locus of the midpoints of a system of chords having slope m. So using this as a hint, using this, get the equation of the diameter and give me a response on the chat box. OK, let's say let's say the locus of the midpoints is basically I'm trying to obtain that by assuming, let's say the midpoint of one of these chords as h comma k. OK, now if this chord equation, I ask you, what will be the equation? So what are the equation of the chord whose slope is? Sorry, whose midpoint is h comma k? You say it's a simple t equal to s1, correct? And let's say this is our standard ellipse, x square plus y square, x square by a square plus y square by b square equal to 1. So t will be xh by a square, yk by b square minus 1. s1 will be x square by a square, k square by b square minus 1. OK, so minus 1, minus 1 goes for a toss. So this is the equation of the chord whose midpoint is h comma k. OK, now as per the question, this chord must have a slope of, this chord must have a slope of? Slope of how much? M, as per the given question, correct? And what is the slope of such a line? Slope of such a line is, see, if you write it in the general form, let's say I write it like this. In the general form, slope of any line is what? Let's say this is our lines. What is the slope? Minus a by b, right? So minus a minus a divided by b. OK, so this is the slope. And this itself is the locus equation. This itself is the equation of the chord if you generalize it here, if you generalize it here by replacing your h with an x and k with a y. So you end up getting m is equal to minus x b square by a square y. In other words, y is equal to minus b square by a square x by m. This becomes your equation of the diameter. And you can clearly see that it is passing through. It is passing through the center of the ellipse. OK, so please note that even though I am basically taking a case of a standard form of an ellipse, this property that this diameter, any diameter you take, diameter by setting the midpoints of any system of chords will have to pass through the center of that ellipse. OK, so this is true in general. So in general, you can say it will pass through the center of the ellipse. So note down the equation, y is equal to minus b square by a square x by m. So you can clearly see that if m is infinite, that means you are bisecting a system of parallel chords whose slope is, I mean, which is parallel to the y axis, then your equation of the diameter will be such that it will be y equal to 0. That means your axis, the major axis will become the diameter. Is this fine? Any questions? Any questions, any concerns? So y is equal to minus b square by a square x by m. Now, let us now also talk about conjugate diameters. Let us talk about conjugate diameters. So let me just pull out the, yeah. So what are conjugate diameters? So let us say there is a system of parallel chords. OK, I'm just taking a system of parallel chords, not much. And let's say there is a diameter which bisects the midpoint of these chords. OK, now what do I do? I start making chords which are parallel to this diameter. OK, let me make just a few of them. So what I'm doing, these gray chords that you see, they are parallel to the blue diameter. OK, and if you start connecting the midpoints of these gray diameter, if you start connecting the midpoints of these gray diameter, you will get another chord. Let me make it in pink. Let me make it in pink. Then this blue and this pink, this blue and this pink diameters, they will be called as, let me write it like this, they will be called as conjugate of each other. So they are a set of, or they are a pair of conjugate diameters. OK, so let me write down the definition of it, maybe from the definition and get a clearer idea. So what are conjugate diameters? So definition of conjugate diameter is two diameters, two diameters are said to be, two diameters are said to be conjugate of each other, each other, when each, when each bisects all chords parallel to the other. OK, now you can see that the pink chord that you see, that will bisect all the chords which are parallel to the blue one. And blue will bisect all the chords which are parallel to the pink one. So as you can see, the green ones are parallel to the pink one. So these are all parallel to this guy. And these are all parallel to the blue guy. So I'm showing them with double arrows so that you can understand them. OK, now I have a question. If you have understood this definition of conjugate diameters, I have a small question for you all. First note down, in case of any concern related to the definition, do let me know. So two diameters are said to be conjugate of each other, when each bisects all chords parallel to the other. Now let the equations, let the equations of the conjugate diameters be y equal to m1x and y equal to m2x. OK, so let us say the blue chord, the blue chord over here, I'm taking its equation as y is equal to m1x. So m1 is the slope of this diameter. And this pink chord, I'm taking its equation as y equal to m2x. OK, now prove that m1m2 will be minus b square by a square. Prove that m1m2 is minus b square by a square. Very simple. Once you're done with this, let me know with a done on the chat box. It's just one line. The one line you have to write, and that's it. The proof is done. Kinshukh is done. Read the definition once again. Two diameters are said to be conjugates of each other, when each bisects the chords parallel to the other. Clear? OK, now see. Let us let me focus on the blue diameter, correct? Now the blue diameter is basically bisecting the chord which are parallel to the pink diameter, correct? Pink diameter slope is m2. So all these green chords that you see, they have a slope of m2. So as per the equation of the diameter, we have learned that its equation is y is equal to minus b square x by a square m2, correct? Yes, no? And this is representing the slope of the diameter which as per my question is m1, correct? So m1 is equal to minus b square by a square m2. So what is m1m2? What is m1m2? Minus b square by a square done, OK? Very, very important. You can say property, OK? A lot of questions have been framed in GE Advanced exams based on this. So please note that the product of these slopes of two conjugate diameters irrespective of however is your basically your equation of the ellipse, it will always be minus b square by a square. So you can also write this in terms of the eccentricity. So eccentricity is what? Under root of 1 minus b square by a square, correct? So this is going to be e square minus 1 equal to minus b square by a square. So you can write it as e square minus 1 also, OK? Anyways, this result is good enough. Please note this down. Now try to understand, if you're Now try to understand if your ellipse was a circle, that means I take a limiting case of an ellipse to become a circle, that means your b square and a square will become equal. Then the two conjugate diameters would actually become perpendicular because slope product will be minus 1, right? So only when there is a special case where your ellipse becomes a circle, that means b tends to a, then the two conjugate diameters would be perpendicular to each other. So in case of a circle, the conjugate diameters would be perpendicular, but not in case of a general ellipse because b and a are not equal, OK? And please also understand, many people think that, we'll discuss this in some time. Many people think that the endpoints of the conjugate diameters, OK? We will take this as a property. Let me not speak about it right now. We'll take it as a separate discussion point, OK? So everything on this page is done. So can we move on to the next page? OK, done. All right. Now we look into some properties of conjugate diameters. Check if I have taken a snapshot instead of writing now. We'll take this property first. Properties of conjugate diameters. The first property is the eccentric angle. This is what I was actually trying to say. The eccentric angle of the ends of a pair of conjugate diameters differ by a right angle. Now let me make a diagram first of all. Let us say this is my pair of conjugate diameters. So I'll make it in blue and pink, OK? Now I've not drawn the chords just for the purpose of keeping the diagram neat. So you can say, let's say, p, p dash and q, q dash, OK? So p, p dash and q, q dash are conjugate pairs. There can be several set of conjugate pairs, OK? So this is one set of conjugate diameters I have taken. Now what is this particular property trying to say is that if this p has an eccentric angle of, let's say, phi, then this p will have an eccentric angle of pi by 2 plus phi. Can we prove it? Prove that if one vertex of the diameter p, p dash has an eccentric angle of phi, then the vertex of one of the vertex of the conjugate to that diameter will have an eccentric angle of pi by 2 plus phi. Can we prove it? Can we prove it? Let me take theta, and this is theta also, because phi I can use later on, so it doesn't matter. Yeah. Can we prove it? Please try it out. Not a rocket science, don't worry. You have already done one of the properties which will help you to do it just a little while ago, not even one minute ago. This is the center of the ellipse. Let me know once you're done. Anybody who's done, just say done on the chat box. OK, to be more precise, let's say Cp has a slope of m1. Cq has a slope of m2. Now use it. At least I gave you as a hint. So the two conjugate diameters have slopes of m1 and m2. Now go ahead. If you do it, you'll remember it. Else it'll be just another theory, and it'll be washed off. Aditya is done. Very good. See, what is P? P is, let's say, A cos theta comma B sine theta. And as of now, let's call this angle to be phi. Let's say I don't know this. As of now, this information is not known to me, and I'll call this eccentric angle to be phi. So as per this assumption, this should be A cos phi comma B sine phi. Now a simple concept I can use, that slope of Cp is m1, which is B sine theta by A cos theta, because C is at origin. M2 is similarly B sine phi by A cos phi. And you already know that m1, m2 is minus B square by A square. So m1, m2 will be B square sine theta sine phi by A square cos theta cos phi. This is minus B square by A square. In other words, you get tan theta tan phi is minus 1. So far so good, tan theta tan phi is minus 1. So can I write this as tan theta is negative? Or you can say tan phi is negative cot theta. OK, negative cot theta, you can write it as tan phi by 2 plus theta. And from here, we can say one of the possibilities that phi could be phi by 2 plus theta. So that's why one of the vertex of the conjugate diameters will have a centric angle of phi by 2 plus theta. Please make a note of this. Now, guys, let me clarify a confusion with people. Get OK? People tell me, sir, it seems to suggest that the two conjugate diameters are perpendicular, because the product of the slope, let the theta and phi are the angle of inclination of these two diameters. Then the product of their slope is minus 1. That means there should be a 90 degree over it, isn't it? Guys and girls, I would like to reiterate this. This is a blunder that you are making when you're saying that. Theta and phi do not represent this angle and this angle. Please note that. Theta and phi are the angles which are made with the auxiliary circle. So if there was an auxiliary circle over here, let me make a diagram out of it. Maybe I will use an orange color. See, please understand. When I say, yeah, this is my auxiliary circle. What is an auxiliary circle? A circle which is basically having the major axis as the diameter. So when you say this point has an eccentric angle theta, that means if I connect this center to the auxiliary circle and from there I drop up perpendicular, then this angle is theta my dear. So tan of theta is the slope of, let's say I call this point as m point, tan theta is the slope of Cm, not the slope of Cp. Get this right, OK? This mistake is done by many people. Similarly, when I say eccentric angle of q is phi, means if I connect this to the auxiliary circle, from there I drop up perpendicular. Sorry, my line was not that straight. It's difficult to. Yeah, so something like this, OK? This angle is phi. Get it right. Just like this angle was theta, this angle is phi. Now this angle between these two lines, this is perpendicular, that I agree. But the angle between Cp and Cq, that is not perpendicular. That is not 90 degrees. Are you getting my point? If that was 90 degrees, then I would have got m1, m2 as a minus one. But that is not happening here. OK, so don't get confused. This is the slope of Cm and this is the slope of, let me call it as m dash here. This point is m dash. OK, so this is the slope of Cm dash. So don't get confused between this and this. They are two different things. Is it clear? Any questions? That is why it is very important to understand the concept of eccentric angle, which many people don't know properly. OK, so in light of this, if p coordinate is A cos theta, can I say q coordinate will be minus A sin theta, b cos theta. OK, so if this is p, this is q. So what will be p dash and q dash? That also please tell me. Who will tell me what is the coordinate of p dash? Now, guys, angles, very important to understand. If theta is the eccentric angle of p, what will be the eccentric angle of p dash? Write down on your chat box. If theta is the eccentric angle of p, what is the eccentric angle of this p dash? People are scared to say. Yes, pi. Yes, you can say pi plus theta. Because if this is theta and this is c, please understand. This angle is also pi and the change in the eccentric angle will also be pi because if you bring this down till it hits the auxiliary circle and then extend it upwards, even this difference will be 180 degree only. So eccentric angle for this will be pi plus theta. So your answer will be A cos pi plus theta, which is minus A cos theta minus b sin theta. You can say it is going to be diametrically opposite, means mirror image about origin. So just change the sign of the x and the y coordinates of the p. Similarly, the q dash point, this point. It will be negative of this. So it will be A sin theta comma minus b sin theta. Please make a note of this. So these are your vertices of the conjugate diameters. So p, p dash and q, q dash are the two conjugate diameters whose vertices are as stated over here. OK, note this down so that we can now go to second property. Now, since most of you are here, I would like you to give me a suggestion. Yes, I also need your suggestion sometimes. How do you think such a voluminous topic like Onyx section can be completed only in class 11? Is there any way out for this? And you already know in class 11, many students are not willing. Like you people come on a Sunday, spend another one and a half hours studying. The present 11th, I mean, they are still not used to this long duration. For them, three and a half hours is only too much. So as 12th graders, in fact, as my passing out batch, what would be your suggestion related to how do I complete this voluminous topic in 11th? Any ideas? I would be happy to hear any good ideas coming from your side. Start from the beginning. Yeah, how do I face it out? See, by the time I just complete the school level conic section, your year is over. February end comes and your first-term exam, sorry, your 11th exam happens. Tell them that they definitely will do here. Definitely not open the chapter also. Yeah, trigonometry is still two and a half hours. But again, yeah, I have to engage them because I have to take 12th batch also, 11th batch also. So I don't want to split the three and a half class subject break and conic after that. That's a good suggestion, Raghav, actually. That means after every class, last half an hour, you do conic section. That's a good suggestion, actually. Correct, correct, correct, correct. They are also not done with the syllabus, no, Akash? Chemistry, you know how voluminous chemistry is also. It is equivalent to three subjects, physical, organic, and organ. Yeah, Gaiti, that is also a point, well taken, because people will start living in the last half an hour. But anyways, those who want to study, they will not be at loss. Yeah, so those who want to study, no one can stop them from studying. Those who don't want to study, no one can even study. Unfortunate. Next property. Okay, do it in the first thought. That's also a very good solution, Kinshukh. Thank you. We can't do anything right. Okay, Kinshukh and Raghav, thank you so much. And others also. Yeah, two, three, ten minutes break. Correct, Siddhant. We'll do that, we'll do that. Okay, now moving on to one of the next properties of conjugate diameter. Now read this property, it says the sum of the squares of two conjugate semi diameters is a constant and it is equal to sum of the squares of the semi-axis of the ellipse. In short, this is equal to square of the, this is also equal to the square of the director's circle radius. Anyways, this is very easy to prove. Please prove it. In fact, I'll make the diagram once again. So this property is very simple. It says that they have called it as D. So I'll also call it as a D. Okay. So they have said that CP square plus CD square is equal to A square plus B square. Now this is very simple. Given that we have already done the endpoints of, or you can say the vertex of the conjugate diameter ends. So let's do this. This is A cos theta. This is, okay, and this will become minus A sin theta, comma B cos theta because the eccentric angles, the eccentric angles for these points, they differ by pi by two. Eccentric angles differ by pi by two. Again, the angle between CP and CD is not pi by two. Okay. So now this is very obvious that CP square will become A square cos square theta plus B square sin square theta and CD square will become A square sin square theta plus B square cos square theta. When you add them, your result is very much in front of you. Okay. So this will give you A square and this will give you a B square. Fine. Any questions? Next property. Can you go on to the next property? Okay. Distance from the origin everybody knows, right? So I don't have to worry too much about telling you a distance formula. So CP distance and CD distance I've written and I've just added them up. Property number three. The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter which is conjugate to the diameter through the point. Okay. Read the property once again. The product of the focal distances is equal to square of the semi-diameter which is conjugate to the diameter through the point. Okay. Maybe when I draw the diagram you will be able to relate to it. See. This property says if there is a point P. Okay. Of course, let's say these are your foci S1 and S2. Okay. So the product of the focal distances, so what are the focal distances? Focal distances will be S1P and S2P. Okay. So S1P times S2P. This product is basically said to be equal to the conjugate. Now let's say I draw a diameter passing through P. Okay. So one diameter passes through P. That means P is one of the vertices of a diameter. And let's say the pink one happens to be the conjugate of it. Okay. So the blue and the pink ones are conjugate of it. Okay. So let me write it like this. This property says this product of the focal distances is equal to CQ square. Right. That is the square of the semi-diameter which is conjugate through the diameter through the point. Very easy. Please prove it. I hope everybody knows the focal distance of a point on the ellipse. If no, we can spend half a minute discussing it. Yes, half a minute, not more than that. It will not take more time. See, let me make a separate diagram. This is, let's say the director is corresponding to this focus. Right. Let's say if I take a point P here which is X1, Y1. What is this distance S1P? Let's say this is your AE comma 0. Okay. Now please note that this distance SP is E times PM. Okay. And this equation is X equal to A by E. So SP will be E times PM. Okay. Now here to here, it is X1 and this whole thing is A by E. So it is A by E minus X1. So when you multiply, it becomes A minus EX1. And also note that, let me call it as S1 for the timing. Yeah, also note that if you want to know its distance from the other focus, from the other focus, let's say S2, then S2P will be just a addition of it. Okay. And that's how your other definition of an ellipse comes out. That ellipse is locus of a point which moves in such a way that the sum of its distances from a fixed point is a constant. That means S1P plus S2P is a constant. And this constant is the length of the major axis. So this is another locus definition of an ellipse. I am sure I would have told this to you in class 11 when you were doing conic sections. If no, then please note it down. This is very important, knowing the focal distances. So yeah, so let's use this to solve this property. Done. Just say I'm done on the chat box. Done. Diatry is done. Very good. Okay, so let's take this point to be A cos theta comma B sin theta. Correct? So S1P, as per this notation that I gave you is A minus EX1. EX1 is this guy. This is EX1, right? And this is Y1. So X1 is A cos theta, correct? Similarly, S2P will be A plus E times A cos theta. Okay, now what do I have to prove? This is equal to CQ square. CQ square I just now discussed with you. It is going to be A square sin square theta plus B square cos square theta. Why is that so? Because as we already discussed, the eccentric angle of PNQ differ by pi by 2. So this is minus A sin theta comma B cos theta, correct? So CQ square will be square of this, square of this, okay? So which is what I have written over there. Okay, let's now simplify this and try to show that it comes out to be CQ square. Simple, let's use our A square minus B square or X square minus Y square formula, okay? So this becomes A square. Now E square, we already know it's, let me write it down. So E square is 1 minus B square by A square. So A square E square is A square minus B square. So I'll use that over here, okay? So this A square E square I'm going to substitute with. I'm just going to box it up. Yeah, so if you simplify this, it becomes A square 1 minus cos square theta plus B square cos square theta. That's nothing but A square sin square theta plus B square cos square theta and that's definitely matching with our CQ expression. Look here, CQ expression is this. So it's definitely matching with CQ square, okay? So hence this property is true. Now these properties normally, they come as a question themselves, okay? So that is why it is very important to treat these as a question rather than learning them as a separate entity that, okay, this is a property, I should remember it now. Treat this as a question because everything is, I know getting derived from your prior knowledge, okay? Let's take last but not the least property, but before that any questions you have, do let me. That's a very good suggestion. First one hour I will keep for conics and extend the class by half an hour more maybe. People will not be able to leave also, right? Thank you. Okay, next property, starting class. Yeah, that is also a good idea, 345. Yeah, yeah, correct, correct. 345 and ending it at the same time and doing the first one hour or 45 minutes with conics. That's a good suggestion. Okay, property number four. Read the property, very interesting property. It says that the tangents at the extremities of a pair of conjugate diameters form a parallelogram. Whose area is constant. And that is equal to product of the length of the axes. So this area that of the parallelogram is four AB. A few topics of clinic and we grow to the bush course also. Now bridge course anyways, I think, I should die was too long for you people, right? We didn't spend half the time with their current batch because their board was delayed a lot and it never happened. So in the bridge course, we hardly get time to complete properly, you know, graphs and calculus. So British course will not be, you know, right time to start conics because they will not be using it also. Conic is just, you know, math specific topic. Yes, get it. 345 would be a good idea. Yeah, definitely not. It was not better in offline. Online, we got more time, right? We could discuss more questions in online. In this case, see, you can do, you can do a small cutting of the corner. Then don't try to prove it's a parallelogram because you can easily prove it by finding the tangent at these points. Okay. So these point coordinates are already well known to us. They will be parallelogram. So don't worry about it. So it forms a parallelogram. Let's take it for granted. I'm more interested in getting the area. If you can prove that area is for AB, that is, you know, all I need offline batches. We, what do you use to do? In offline batches, we used to call them in the summer vacations in the school premises. So summer vacations. Madam principal would allow us to conduct classes and we would call them on Sunday on almost every day, especially during the, during the summer break, which used to be very long. And then we used to complete conic sections. Yes, exactly. That used to be the scene with the offline batches. Okay. Let's try to first, you know, brainstorm. How do you, how do you find this area? Okay. Now, just in the interest of time, I would like to show you this diagram. See, if you look at this parallelogram, C, D, Q, P, this parallelogram, let me shade it. Okay. Do you realize that the area of the complete parallelogram, let me name it Q R dash Q dash. Okay. That will be four times area of, area of the smaller parallelogram. That is what I have shaded C, D, Q, P. Okay. Do you all agree with that? As you can see, there are four chambers that they have made just to show you that. Correct. Now area of the shaded parallelogram that is your yellow parallelogram, C, P, Q, D, or C, D, Q, P, whatever. What will be that first of all? Okay. You'll say base into height. Okay. Now base here is CM. Sorry. DQ. Okay. And height is CM. Base into height. While CM is very easy to find out because you're just finding out the distance of the center from one of the points. So distance formula of a point from a line can be used. But how do you find DQ? How do you find DQ? Because Q position is not known, right? How do you find DQ? Anybody can suggest me that. CM is easy to find. Okay. Let me just write down the equation of tangent, tangent through D or at D. Okay. I'll just rewrite it. Let's say this is eccentric angle theta. So this is ascending angle pi plus theta. So it'll be minus A sin theta comma B cos theta. So your equation will be xx1 by A square, yy1 by B square equal to 1 and x1 will be minus A sin theta. So this will become minus x by A sin theta. So the distance of origin will be your CM. But how will you find DQ? So are they suggesting something? Q is intersectional direction formula. Can I say DQ is CP? Isn't it? So DQ is actually CP. So it is just CP into CM. Because it's a parallelogram. This also is a parallelogram. So DQ and CP are equal and CP distance is easy to find out. Because you know your point P is A cos theta comma B sin theta. And this is origin. It's a distance formula. Isn't it? Why is DQ CP a parallelogram? See, this is now a very good question. Why is DQ CP a parallelogram? Isn't this whole thing a parallelogram? Whatever I am drawing and this is basically nothing but it will be parallel to this line. This will also be parallel to this line. Same will be PP dash parallel to this line. And the diagonals, they will be equal. So this length will be equal to this length. This length will be equal to this length. This length will be equal to this length. And they are parallel also. Now here, let us try to figure out CP length and CM length individually. So CM length is mod of 1, which is actually a 1, need not write mod. 1 by under root of sin square theta by A square cos square theta by B square. In short, this becomes AB AB under root of A square cos square theta B square sin square theta. And what about CP length? CP length anyways is under root of A square cos square theta B square sin square theta because the sense of P from origin is this. Now when you multiply CP into CM. CP into CM. Let me just circle this also. Yeah. So CP into CM if you multiply it, what will happen? It will become AB under root of A square cos square B square sin square into under root of A square cos square B square sin square. So this will get cancelled off giving you AB only. And please let me remind you that AB is just the area of one fourth of this parallelogram. There are four such parallelograms sitting over here. So use this result over here. And you end up getting area as four AB. Okay, so this becomes your area of the parallelogram drawn by or formed by the tangents drawn to the vertices of the conjugate diameters. Is it fine? Any questions? So more or less the important part of this particular discussion is over. We will spend few times talking about few minutes talking about. Coincidently points. Some interesting questions can be framed on that as well. So let's move on to our next point of discussion. Coincidently points. What is basically coincidently points? The concept is very simple. Let us say there is a circle and there is an ellipse. Let's say this is an ellipse. And let me take a circle. Let's say these two, you know, geometric figures, they intersect. Let's say they intersect at these four points. PQRS. Okay. So the points where the circle and this ellipse intersect. Okay, please remember if they intersect, the minimum number of intersection point will be two or it will be four. Okay. If they intersect. Okay. If they touch, then also we can say there are two coincident points. Okay. So in case a circle intersects an ellipse, it will intersect it in two or four real points. Okay. And let us say the eccentric angles of these points are alpha, beta, gamma, delta. Now let me give this as a question to you only. Let's not, you know, give this as a theory. Prove that prove that the sum of the eccentric angles of the consicclic points. So let me write them PQRS, they are called consicclic points. So prove that the sum of the eccentric angles of the consicclic points is an even multiple of pi. There's always an even multiple of pi. Can anybody prove this? So if an ellipse and a circle meet at four consicclic points or maybe even two consicclic points, prove that the sum of the eccentric angles will always be an even multiple of pi. Please do not get confused with the conormal points. In case of conormal points, it used to be odd multiple of pi. So just take a very generic case, no issues. Take a very generic case. Okay. Take a generic case where it is a circle centered at origin. Okay. Will they always be diametrically opposite by the symmetry of the figure it is. But let's see if I just displace it also slightly here in their displacement if I do. Then also that property should hold valid. Maybe you're talking about a very special case. I could also displace this circle slightly here. Special case that will always come out. Okay, so in the interest of time, I know this will be slightly difficult for you. First, let me ask you a simple question here. If I connect alpha and gamma points, okay, I get a chord of a ellipse, isn't it? So what is the equation of chord PR? What is the equation of the chord PR? Or you can take any of the chords doesn't make a difference. Let me just take a chord like this. chord PQ. Let me take PQ doesn't make any difference, whichever two chords you take. Okay. And let me take a chord RS. So what are the equation of the chord PQ? Why I took PQ? Because you would remember alpha beta combination, which I gave you the in the equation of a chord. x by a cos alpha plus beta y by b sin alpha plus beta. In fact, x by a cos alpha plus beta by 2 y by b sin alpha plus beta by 2 is cos alpha minus beta by 2. Do you all remember this equation of a chord? We have done it before doing the equation of a tangent. Okay. Similarly, what is the equation of chord RS? You'll say simple x by a cos gamma plus delta by 2 y by b sin gamma plus delta by 2 equal to cos gamma minus delta by 2. Okay. Now, what I claim is the circle is a family member passing through the intersection of equation of this ellipse with the pair of these chords. So first of all, I'll create a pair of chords. Okay. How do you create pair of chords? Very simple. Multiply the two equations of the chords. So when you multiply it, it'll look like this. x by a cos alpha plus beta by 2 y by b sin alpha plus beta by 2 minus cos alpha minus beta by 2. If you multiply this with x by a cos, same thing, gamma plus delta by 2 plus y by b sin gamma plus delta by 2 minus cos gamma minus delta by 2. So this equal to 0 will give you a pair of chords. A pair of chords is basically the chord PQ and RS equation combined. Now, what is my claim? Listen to me. Listen to this claim. The claim is this circle that you see, this circle that you see is a family member or it's a curve passing through the meeting point of the ellipse and the pair of chord. Do you all agree? Because the circle is passing through the end points of the two cards as well as the ellipse. In short, the equation of a circle, the equation of the circle can be written as let's say ellipse equation plus lambda times pair of chord equation equal to zero. So that it is x square by a square y square by b square minus one plus lambda times this huge expression. Okay, which let me call this as a P. I will not be writing it again. Sorry. Okay. Now I have to choose a lambda. I have to choose a lambda in such a way that this equation should represent a circle. The equation should represent a circle in a circle you already know the characteristic of a circle is the coefficient of x square and y square should be equal and the coefficient of x y is zero. Right. So for it to represent a circle, please ensure two things coefficient of x square should be equal to coefficient of y square and coefficient of x y must be zero because in a circle there is no x y term. So what I'm going to do, I'm going to categorically pick up x y term from this equation and put it to be zero. By the way, this doesn't contain any x y term. So all x y term will be in P. So if you check your P term, your x y term will come when this fellow multiplies with this fellow. And when this fellow multiplies with this fellow. And the coefficient will come out to be let me just multiply and show you it will be one by a square. Sorry one by AB not a square one by AB cos alpha plus beta by two into sine gamma plus delta by two so I'm multiplying this fellow with this fellow. And now I'm going to multiply this fellow with this fellow. So that will give me one by AB. Sine alpha plus beta by two cos gamma plus delta by two. Okay, this should be zero first of all. The moment you put a zero you get the answer for your question. How see this is just as good as saying cos alpha plus beta by two sine gamma plus delta by two. Sine alpha plus beta by two cos gamma plus delta by two so as per our compound angle formula. This is a formula for sine of alpha plus beta by two plus gamma plus delta by two. Okay, now if sign of some angle is zero which means that angle should be a multiple of pi. So alpha plus beta plus gamma plus delta by two should be a multiple of pi. And being some integer, which means alpha plus beta plus gamma plus delta is equal to two and pi. Okay, where n is an integer, which clearly proves the fact that the sum of the eccentric angles of the corn cyclic points would be a even multiple of pi. That is what we wanted to prove this was our problem statement. Anyways, this is a slightly on a higher side. I mean, maybe this concept is going to be tested only in advance, not in JMA. Coefficient of x square and y square you have, it will come out to be the same. You can try it out, Aditya. So Aditya has a very valid point. So does this also ensure that coefficient of x square and y square will be the same? It will come out to be the same. If they're not coming out to be the same, that means the intersection is not happening at all. That means the points that you're talking about, they are not coincident at all. It will definitely happen. Aditya, check it out. Okay, so towards the last 20 minutes, we will take some questions. In fact, all the theory part we have already covered, there's nothing must to be covered. Just a small part is covered reflection property of an ellipse. This is for J advance. So people who are only aspiring for J mains, don't worry too much about it. It's only for J advance. Okay. So the last few minutes, I will spend on reflection property of an ellipse. That is something which is left off. Some questions have also been framed on that. Okay, so let's now move on to reflection property. Reflection property of an ellipse. Okay, all of you please pay attention to this. A very simple thing I would like you here to prove. Let's say there's a point P. Okay. So this P point, you connected to the two foresight. Okay, S1 and S2 are foresight. Okay, I would like you to prove a simple, you can say theory over here. Prove that the normal at P. This blue line is the normal at P by six this angle. Can you prove this? Prove that the normal at P by six the angle made by P with the foresight that is the angle between the focal distances. Anyway, I mean, if this was extended further, if this was extended further, okay, then the tangent at this point would have bisected the external angles. That means this angle would have been equal to this. Okay, anyways, I just want you to prove this part, the normal by six this angle. So this is equal to this. How would you prove it first of all, any, any idea to prove it. There's not going to be too much complexity of angles. Any safer way to prove an easier way to prove any convenient way to prove, maybe showing that and will give you some idea angles between that would be a angle by sector. I'm right. Okay, see guys, everybody knows that if there is an angle by sector. Can I say the ratio of S1P by S2P will be same as ratio of S2N by S1N. Okay, let's use this our angle by sector theorem by to go into too much of angles. Okay, if this is proved then means this line will automatically bisect this angle. Okay, reverse of angle by sector theorem you can say so. Now, now let us talk about, first of all, the equation of a normal drawn at X1Y1. Let's say our P point is X1Y1. What are the equation of the normal that we have learned. I hope everybody remembers the equation of the normal drawn to a point X1Y1. Hope you are practicing all the DPP is saying to you. So, for getting N1, so for getting N, you put your Y as zero, right. So, when you put your Y as zero, your X becomes a square minus B square X1 by a square. Okay, for Y equal to zero, so that means your N coordinate is this a square minus B square X1 by a square comma zero. Correct. Is it fine. What do you do next. So, basically you have been, you have found out CN. Okay, so CN length is now known to you. Another thing that you would all recall here that a square minus B square is a square E square, correct. We have already seen it in our previous discussion when you were talking about the length of the focal chord. Correct. Can I say this is this is a square E square X1 by a square that is nothing but E square X1. Yes or no. So this length is E square X1. Yes. So now finding, finding SN1 will be very easy or S1N will be very easy S1N will be what a E. This is a E minus CN. Am I right. That's nothing but E times a minus EX1. Similarly, S2N will be a E. This is your AE plus E square X1 which is a, sorry, a times. Sorry, E times a plus EX1. And we also know S1P. S1P is a minus EX1 and S2P is a plus EX1. Correct. Now you can clearly see from here that if I do S1P by S2P, I'll get a minus EX1 by a plus EX1. And the same result is obtained when you do, so you multiply with E on both numerator and denominator. So if you multiply AE on both numerator and denominator of this term, you will end up getting this. And this coincidentally is your S1N and S2N because you have just figured it out over here and here. So this property is holding true. That is your angle by sector theorem. Okay. Holds here. Oh, what happened? Holds here. Okay. So that means, that means this normal that you have drawn at P bisects this angle. Now this is a very interesting property which basically is related to reflection. That means if a ray of light comes from one of the four side, after hitting the elliptical surface, it will basically converge or pass through the other focus and vice versa. Okay. So if a ray of light starts from S1, after hitting the elliptical surface, it will pass through S2. Okay. So this is a property which comes out, the reflection property comes out from the fact that the normal bisects the focal distances. Okay. Similarly tangent will bisect the external angle over here. We can easily prove that as well. Is this fine? So last 10 minutes of our class, I will take questions, I will take one or two questions and then we will close this topic. Can I move on? Read the question. The question says a ray emanating from 0,6 is incident on the ellipse at point P with ordinate S. Okay. After reflection, the ray cuts the y-axis at B. Find the length Pb. What is P here? Oh, at the point P. Okay. Okay. So P is sitting over here. Sorry, I did not. Okay. Get the y-axis at B. Find the length Pb. And give me a response on the chat box. Most of the concepts in hyperbola will be a tweaked version of ellipse. Okay. So hyperbola can be done very quickly. So hyperbola, I think should not take more than two classes. Two classes for pair of straight lines and maybe two classes for. So maybe six more classes we'll have on Sundays. I think the ordinate that they have given is five. It is not S. Ordinate is five. Without that, you'll not be able to solve it because it could be so many other points. So that five, they have written it as an S by mistake. Yes. Is it taking so much time? See, first of all, let us try to analyze this ellipse. I think four into 64. This would be 64. And this will be 100. Okay. So this is x square by eight square y square by 10 square is equal to one. So basically it's a vertical ellipse. Okay. Vertical standard case of an ellipse. Okay. Aditya has given one response. This is a zero comma 10. This is zero comma minus 10. Eight comma zero minus eight. What is the assenticity of this ellipse? Assenticity will be one minus a square by B square because it's a vertical case. So a square is 64 by 100. So that's going to give me point six. Okay. So zero comma BE is zero comma six. And they're also talking about zero comma six. Oh, so this happens to be one of the four side. It's one of the four side. So basically it's saying that raise emanating from one of the four sides. So that's a good point for us because if it emanates from one of the four side, then after hitting the, you can say the elliptical mirror. After hitting the elliptical mirror, it will definitely pass through the other four side. Okay. So at least the point B is known to me. Why? Because B will be another four side, which is zero comma minus six. Okay. But the point where it is hitting the elliptical mirror that is something comma five. The ordinate is five. That is what the question center is giving us. The ordinate is five. I don't know about the abscissa maybe abscissa I can figure it out from the equation itself. So in this equation, if you put your y as a five, let's check what happens to. So this is x square by 64 is equal to one fourth. So this is three fourth. So x square is 48. So it's root 48. So you can take it as four root three. Okay. It could be other way around also it could like hit here also. It will be applicable. Okay. So minus will be there but distance will not change because these two triangles will be congruent. Okay. So, you know P, you know your B. So what is there? Find out the distance between them. That's all the question demands. So four root three square. 15 square under root. Oh, sorry. Not 15 square my bad. 11 square. Okay. So this is 48 121 which is 169 under root 169 is down to be 30. Option number D is correct. Okay. So one last question we'll take up. Oh, you have a test also to take right so guys will stop here. I don't want to extend your time you have a half an hour break to for your breakfast and all. So there will be a test sent to you please follow all the steps. Okay. Don't skip any step and then say said it is not opening for me and all those stuff. Okay. Don't give those kind of you have already taken so many tests. If at all you're stuck. Go to your class 11 Jay sorry class 12 Jay main folder 129 2021 test will be there that you have to take. But anyway, if you follow the link one by one starting from the first to the last. Okay. So execute every step which has been given in the instructions you will definitely reach the test so no need to you know, panic, don't jump any step. Fine. So all the best. 10 o'clock you can take the J main test. Need people also same. All right. Thank you class. Bye bye. Take care. All the best. Bye bye.