 Thank you everyone for wishing me the teachers day. Thank you so much. Okay, so we are going to continue with ellipse I knew I had an agenda of taking two hours classes but unfortunately as I told some painting work is going on in my in my house so I will not be able to stretch beyond 930. So this room in fact where I'm taking it it's like half painted. Let's, let's get started. So today I'm going to start with the equation of normals. Okay, so we had already done the candy, we have already done the equation of tangents, we have done the point form, we have done the parametric form, we have done the slope form. And all those forms that I did I would like to reiterate once again that they were only meant for the standard forms. So today I'm going to start with again the standard form of an ellipse our x square by a square y square by b square equal to one ellipse. Okay, and we are going to learn the different forms of the equation of normals done to the ellipse. Thank you. Thank you so much. So let me just pull out my normal diagram I am sure I still have it. No. Yeah, I have it. Yeah. So let's say at a given point, at a given point. x1 y1. I'm drawing a normal to this ellipse. Okay, so we all know normal is nothing but it's a line which is perpendicular to the tangent at the very same point so if I draw a tangent at this point it is going to be perpendicular to the tangent at that point. Okay, so what is the equation of this normal let's try to understand very simple we already have done tangents and normals in our application of derivatives. So this should not be a difficult concept for us so if you differentiate it find d y by dx, you may also use your implicit differentiation formula. So let's say this is our implicit function. So d y by dx if you remember it is nothing but negative doh f by doh x by doh f by doh y. Right. So negative doh f by doh x will be to x by a square. And doh f by doh y will be to y by b square that is nothing but minus b square by a square x by y. If you're finding it at a given point you must substitute then only you'll get a tangent else you will get the gradient function. So please substitute the point in order to get the tangent else that will give you the gradient function. So this gives me minus b square by a square x one by y one. Now having known the point and having known the slope. So slope of the tangent or you can say slope of the normal is negative reciprocal of slope of the tangent. So that is going to become a square y one by b square x one. Okay, so this is the slope of the tangent. So let's find out the equation finally. So minus y one is equal to slope x minus x one. Okay, now we will basically simplify this a bit because we are going to convert it to a particular form. So let's simplify this. Let's multiply. Let's multiply b square on the other side. So you'll have b square y minus b y one. And let's keep. Let's keep. Okay, let's divide this why when down. I think it is b square. So the form that I would like you to note down on your formula list for the equation of the normal is a square x by x one minus b square y by y one equal to a square minus b square. So this is the form that you should be keeping in mind. This is called the point form because here you are trying to find the normal at a given point on the ellipse point form of the normal. Okay, so please note this down a square x by x one minus b square y by y one equal to a square minus b square. Later on, maybe not in today's class maybe in some later classes will be talking about equation of normal at a given point x one y one to a hyperbola. The hyperbola normal equation is very rhyming to this is just that b square will be replaced with a minus b square. Okay, so there will be a plus sign coming here and here. We will talk about it when we start that chapter. So this is the equation of this is the equation of the normal in the point form. Any questions? Could you show the diagram diagram is the customary diagram at a given point x one y one let me call that point as a P. I'm sketching a normal. A square x by x one minus b square y by y one equal to a square minus b square now one interesting mistake which people do here is this is the position of x one and x, but that is a mistake which you can easily catch because in an equation of a line you cannot have the variable in the denominator right so that mistake is easily a wordable if you are slightly attentive. So that mistake I'm just pointing out because many people in the past have done that mistake, switching the position of x x one and switching the position of why why when that mistake happens for many people. So, you can easily avoid that because a line equation cannot be of the nature something by x minus something by y equal to constant that will be a wrong equation of a line. Okay. Having taken the point form that is now move on to the parametric form of the equation of a line. So let's move on to the parametric form parametric form of the equation of a line. See in parametric form is just that. Let me just pull the diagram back again in parametric form we are going to learn what is the equation of the very same normal, but just that now the point is express as a parametric point you can say a cos theta B sin theta. Okay, so this point P has an eccentric angle of theta so what is the equation of the normal over here. So nothing different over here all you need to do is just take the old equation replace your x one with a cos theta replace your y one with also why one with B sin theta. So when I do that, I'll end up getting a square x by a cos theta minus B square y by B sin theta equal to a square minus B square. On slide simplification, your result will look like this a x seek theta minus B y cosec theta equal to a square minus B square. Now a very interesting question is framed on this particular fact. So everybody please note this down. So this is the equation in parametric form. All these equations which I'm giving you whether it is a point form or whether it is the parametric form they are applicable to a standard case of an ellipse please do not apply it to any non standard cases. Okay, of course you can apply it but only by accounting for the shifting and everything that you basically take care of when you are trying to solve a non standard case. So there you can apply but only along with the shifting concepts that you have already learned. Now, before going on to the slow form there's a small question I would like you to answer. If y is equal to mx plus C is a normal to, let's say our standard form of ellipse x square by a square plus y square is equal to be square. Then show that C square is going to be m square a square minus B square whole square upon a square B square m square. Okay, by the way, this result that you see over here, this result is also called condition of normality. Just like there's a condition for tangency, that is C square is equal to a square m square plus B square for a line y equal to mx plus C to be a tangent to a standard form of an ellipse. In the same way y equal to mx plus C to be a normal to a standard form of an ellipse. The condition for normality is C square is equal to m square times a square minus B square whole square upon a square plus B square m square. Everybody please try to prove this. It's very easy. And do let me know with a done that you are done with it. See, again, the process is very well defined. You have already got this equation know a seek theta minus by coseak theta is equal to a square minus B square, right. And we are basically trying to compare this equation with y equal to mx plus C. So let me write this equation like this mx minus y is equal to minus C. Okay, now we need to compare these two equations because I asked for the question the seek these two equations are the same because they represent the normal right. Yes. So, so what I have to do here I have to achieve this condition of normality by eliminating theta here. Okay, so let us do that very simple process. So, let's take the ratio of the coefficients so m divided by a seek theta minus one divided by minus by coseak theta rather I will write it just as one divided by B coseak theta and minus C divided by a square minus B squared. Okay, so from here you end up getting cos theta times m by a sin theta by B is equal to negative C by a square minus B squared. So from here cos theta is going to be negative AC by m a square minus B squared and sin theta is negative BC by a square minus B squared. Now we know our famous Pythagorean identity that cos square theta plus sine square theta is going to be one. So let us use that. Okay, so a square C square m square a square minus B square whole square, then B square C square a square minus B square whole square is equal to one. Okay, so first of all I would send the a square minus B square whole square to the other side. So you'll have something like this left. So here also you can just do a quick simplification nothing very challenging here. And there you go. A square is equal to m square a square minus a square minus B square upon a square plus B square m square and there was a square here also let's not forget that this is the condition for normality. This is the condition for our condition of normality. Is it fine any questions. Wasn't it easy. Yes, any questions. Yeah, it should have been done. Yeah, it was easy actually. Okay, let's have few questions one or two, and then we'll move on to a very interesting concept called the co normal points. Anything that you would know would like to know down from here please do so I'll be shifting my screen. So basically simple with the parametric form I ended up comparing y equal to mx plus C. Okay, so from here one more thing comes into picture that if I have been provided with a slope. If I've been provided with a slope, if M of the slope of the normal is given to you, then the equation of the normal can also be written as for a standard form of an ellipse, why is equal to mx plus minus m a square B In fact, it should come out as a mod but I've taken a plus minus, not an issue. Okay, so this can be called as the slow form of the equation of the normal. So we had a point form we had a parametric form now you have a slow form of the equation of a normal. Okay, let's take few questions. Simple one maybe I've already done a similar question a little while ago. But you can still try prove that the straight line Lx plus my plus n equal to zero is a normal to this ellipse. If this condition is met, just let me know with a done that you are done with it. See, they don't need to actually memorize the condition of normality you can proceed solving this question in the very same way as how I achieve that condition of normality. So please do not try to remember things which are unnecessary. So, solve it from the basics. See, you can start with the equation that we have discussed for the parametric form. Okay, so this was the equation the parametric form a x theta minus b y cosec theta is equal to a square minus b square now you're trying to claim that this line. This line is exactly the same. Excellent. Okay, so same way, compare the coefficients. So exit theta by L minus be cosec theta by m is equal to a square minus b square by minus m. Isn't it. So let's write, seek as reciprocal of course, let's write cosec as reciprocal of sign and use your famous Pythagorean identity to eliminate theta here. And sine theta would be very good and it'll cost theta would be minus na by L na by L by a square minus b square and sine theta will be nb by m a square minus b square. Do let me know if I have missed out on any expression here I don't think so I have taken care of everything yeah. So the fact that cos square theta plus sine square theta is going to be a one. So this is going to be n square a square by L square a square minus b square plus n square b square by m square a square minus b square whole square. That's going to be a one. Okay, so just simply take, simply take a square minus b square whole square by n square to the right hand side. And this is what we wanted to prove. This is fine. Any questions. Okay, with this we now move on to the concept of co normal points. So co normal points concept we have already seen in our parabola chapter that means from a point we can actually draw three normals to the parabola. We had also seen the condition where the parabola will be having a real form of the equation of the normal or how many real cases can be formed depending upon where is the point actually line. Okay. In case of an ellipse also will be learning a similar set of condition. Let me pull out the diagram. So from a point, let's say h comma k, from a point h comma k, you can actually draw four normals to an ellipse. So let's say I take the point to be h comma k here. Okay, so from this point I can actually draw. I can actually draw four normals to this ellipse. Okay, I'm just doing some orbit normals here. So maybe something like this, something like this. Now, first of all, how four normals. Okay, why four normals, why not lesser than that and why not more than that. Okay, let's try to analyze that first. Okay. So here a b c d, they will be called as the co normal points, co normal points. The feet of the normal, the feet of the normal drawn from this point h comma k to the ellipse. They will all form a co normal point. Okay, and of course they can be various such set of four co normal points right so this is just one example which I have shown you. Okay. So we'll first understand why four co normal can be drawn and why not more or less. Okay, let's try to look into the equation of the normal to understand that. The equation of the normal used to be a x seek theta minus by co seek theta equal to a square minus b square a square minus b square this was our equation of a normal, isn't it. Of course, the all the normals pass through h comma k so h comma k must satisfy, must satisfy the above equation. Let me call this as one. Okay, so if it does I will have a hc theta minus bk co seek theta equal to a square minus b square. Now, listen to this simple analysis of mine. This theta that you have in the question. This theta is a generic representation of the eccentric angle of the foot of the perpendicular right. So this theta this theta is a general representation general representation, the eccentricity. I should say eccentric angle I should say eccentric angle of a b cd. Correct. That means, let's say this is in reality alpha this is in reality beta this is in reality gamma this is in reality delta, but I don't know whether they will be four or not. So as of now, I'm not, you know, commenting on that, but I'm what I'm trying to say is that this theta is a general representation of the eccentric angle of the foot of the perpendicular. So the number of thetas will actually tell me the number of thetas will actually tell me how many foot of the perpendicular are getting formed. Am I right. So, the number of thetas that I get, they will indirectly tell me how many a b cd fgh I mean I'm just thinking from thinking from a generic point of view. So how many feet feet of the perpendicular is going to get formed that will be dictated to me by how many thetas I will get. Am I right. Yes or no. So what I'm going to do is I'm going to use a bit of complex number idea over here. How see. Let us say this theta I associate with a complex number. So let Z be. Okay. Can I do that. Can I associate. Can I associate the eccentric angles with a complex number. I can definitely do that. So cost theta minus I sign theta is one by Z. So if you add them you get to cost theta as Z plus one by Z. That means cost of theta is half of Z plus one by Z. Correct. Similarly, if you subtract them, you'll get to I sign theta as Z minus one by Z. That is to say that sin theta is one by two I Z minus one by Z. Now what I'm going to do is I'm going to substitute these two in our equation of the normal. Okay, so I'm going to substitute cost theta and sin theta in terms of Z. In the equation of the normal and see what is the degree of Z which I get. The degree of Z will be indicative of how many normal or how many feet of the normal points indirectly how many eccentric angles are given by this theta. So let's try to put that. So when I do that, I will end up getting something like this, a H by half, I can strike it like this. Instead of writing half that present I just combine it as that square plus one by two Z minus BK. And this will be Z square minus one by two I Z is equal to a square minus B squared. Is it fine. Let's do a bit of simplification over here. I think this will be to a H Z by Z square plus one and this will be minus two I BK Z by Z square minus one is equal to a square minus B squared. Okay, let's do a further simplification. To a H Z Z square minus one minus two I BK Z Z square plus one and LCM of this will be Z to the power of four minus one that I'm sending to the other side. Okay, let's simplify it. So here if I make a equation out of Z I will get a square minus B square Z square. Okay, I will end up getting minus two. Just correct me if I'm wrong I'm just collecting the Z cubed terms. Okay, so you get minus two a H minus I BK Z cubed. So I'm just collecting Z squared Z squared term is this anywhere no Z squared term okay so zero Z squared. Good. And what about Z term so Z term will be obtained from here and hit so it'll be to a H plus I BK Z and constant term will be minus a square minus B squared equal to zero. I've just simplified this. Nothing else I've just simplified this you can also verify that the simplification is correct. Anything that I'm missing do let me know minus a H, which one you're talking about. Proficient of which term here X Z Z term. No Z term will come on the right side it will come to a HZ and minus IBK Z. I'm bringing these two terms to the right side. Yeah, no issues, no issues. Okay. Now what does this fourth degree expression or fourth degree polynomial in Z tell you that there are four possible z so this is a fourth degree expression. Fourth degree expression implies that there will be four roots to it. Correct. And if there are four roots to it for each of the roots, there will be one. You can say argument, and that argument is nothing but your eccentric angles for those corresponding points that's alpha beta gamma delta. Okay, so let's say these are your arguments of arguments of your roots, Z 1 Z 2 Z 3 Z 4. So every argument basically signifies that there is a point on the ellipse where these normals are going to meet. In fact, indirectly speaking, that there are four points from where if you draw normals they will be concurrent at H comma K. So this signifies that there will be four co-normal points. There will be four co-normal points. Is this fine is this analysis understandable. There are a bit of complex numbers with coordinate geometry here. So this four point shows that there are four eccentric angles. See, when it comes to eccentric angles, basically I tried to compare it with argument. There are other ways also to do it. You know, many people convert cost theta and sine theta to half angles of tan and get a fourth degree expression in time. That's another way to do it. It's just like saying that if I have to have four angles. Okay. That means if I make those four angles alpha beta gamma delta to be your argument of four complex numbers and if I somehow show that it will be a fourth degree expression in Z, then I'm achieving the same thing am I not. That's the only purpose of doing it. So I was trying to relate eccentric eccentric angle to the argument. But let me tell you, this is not the only way to do it. Many people, as I told you, they write tan in terms, sorry, they write cos and sine in terms of half angles of tan, and they get a fourth degree equation. Now why half angles of tan because see this is very important to understand. If you get, if you just write it in terms of sine or cos or just a tan, you will end up getting two such angles for which tan will be the same. There by giving you a incorrect image that there were two distinct points not one. That is why when you convert it to half angles in tan, when you take it as a half angles, you have to lie between zero to pi and they will automatically different sign and they will tell you the reality. Right. Are you getting my point that is why many people converted to half angles of tan and do the process. But I find this one more convenient because related to this there are two properties which we're going to see in some time and we're going to prove that very easily when we use the complex number notation. Okay, so let us take a few properties based on the co normal points. Meanwhile, please note this down. I mean, don't have to remember this equation but just have just be aware of the fact that there are four co normal points for case of an ellipse. Unlike in case of a parabola they were, they were only three. So let's take, let's take a simple, you can say property based question. So it's a question also and it's a property also. Let's take this one. In general, four normals can be drawn to an ellipse from any point. And if alpha, beta, gamma, delta, the eccentric angles of these four co normal points, then please prove that alpha plus beta plus gamma plus delta will always be an odd multiple of pipe. This is a property of co normal points properties of co normal points, prove that the feet of the perpendicular, the feet of the perpendicular of these normals. The eccentric angles will add up to give you an odd multiple of pipe always. If you want, I can rewrite the equation once again that I had written in the previous slide. I think it was minus two h minus ibk right. Ibk, z cube, there was no z square term. So zero z square and two h plus ibk. Tell me how will you do this? The answer to that this question or this property is hidden in this equation itself. If you want to add the arguments of a complex number, what do you do normally? Tell me, go back to your class 11 days and tell me if arguments are to be added of complex number, what operation should happen between them? You need to multiply them. So if I ask you what are the product of the roots of this particular bi quadratic equation in z, what will you say? What are the product of a root minus b by c by d by minus e by a. So minus e by a. So let me just write it like this. Okay, so if all the roots product are required minus b plus minus plus, yeah, so it will be plus e by a. A here is also a square minus b square. Yes or no? So I'm basically using my beta isolation. So in beta isolation, the product of the roots of a bi quadratic equation is the constant term divided by the coefficient of z to the power four. Right. Some will be minus b by some of product two at a time will be C by some of product three at a time will be minus d by product of all four will be e by so that's minus one. Right. Now let us say each of these complex numbers z one is made up of is made up of such arguments which are the eccentric angles of those points. So the product will be what and if you want you can write it like this also not an issue Oilers notation. Okay, Oilers notation of complex numbers. So this is nothing but e to the power i alpha plus beta plus gamma plus delta that is equal to minus one. Okay. In short, you have written cos of alpha plus beta plus gamma plus delta plus I sign alpha plus beta plus gamma plus delta is minus one. Minus one is minus one is minus one plus I zero. So you are comparing two complex numbers here so which clearly implies that this is going to be this is going to be minus one and this is going to be zero. Now what is the only possibility which can give this. This can only happen this to can only simultaneously happen when your alpha plus beta plus gamma plus delta is an odd multiple of five. Isn't it. Isn't it for all multiple of pie cost will be a negative one and sign as we know that for any multiple of pie sign gives you zero. That's it. That's proved. Okay. What happens any concerns with the approach. See, normally eccentric angles are between zero to two pie. No. So alpha plus beta plus gamma can go from zero to eight by maximum. Yeah. All right. So next property, again, I mean it's based on the very same fact that you have derived. This property you can easily prove it by using the previous result itself. If alpha beta gamma are the eccentric angles of three points. So please note that they are just referring to three of the points. Okay. So alpha beta gamma are the eccentric angles of three points on the ellipse the normal set which are concurrent then show that or prove that this is going to be zero. The hint is, hint is use this and let me know once you're done. Done. Okay. I mean, it's not that that simple also. But yes, let's discuss it out. Front of says he is done. Okay, see everybody. Let us not forget that the coefficient of that square is actually zero. This is a very important part. Okay. Coefficient of that square is zero. What does it mean. It means that the sum of the product to at a time is equal to zero. Isn't it. That means Z1 Z2 Z2 Z3 Z3 Z4, let's say Z1 Z2 Z3 are the roots of that by quadratic equation in Z. Z4 Z1. I think Z1 Z3 I missed out and Z2 Z4. Yeah. That means this should be equal to zero. So what does it tell you that if you start writing your Z1 as e to the power i alpha Z2 as e to the power i beta Z3 as e to the power i gamma and Z4 as e to the power i delta. So can I say that e to the power i alpha plus beta e to the power i beta plus gamma e to the power i gamma plus alpha e to the power i delta. E to the power i delta plus alpha e to the power i alpha plus gamma and e to the power i beta plus delta this should be equal to zero. Because the coefficient of Z2 in the by quadratic was zero. Okay, so I'm just writing it down so that when you're referring to your notes, let's say little later on. Okay. So, what does it mean indirectly is that cause of alpha plus beta. I mean just letting, letting it down like this this will be zero and summation of sign alpha plus beta this will also be zero. In fact, from this particular concept these two properties also evolve. So please note down as a property itself that if you take cause of any two of the angles are some from those eccentric angles, which are the feet of the co normal points, the cause summation will always give you a zero. Similarly, sign summation will also give you a zero. Okay, now what I'm going to do I'm going to use this result to achieve my purpose. I'll see everybody sign alpha plus beta. Okay, let it be as it is sign beta plus gamma let it be as it is then sign gamma plus delta. Okay, gamma plus delta sign delta plus alpha sign alpha plus gamma and sign beta plus delta is equal to zero. I will not disturb this term let it be as it is, because it is there in my expression. I will not disturb this term I will let it be as it is because it's there in my expression. I will not disturb this term also. Where is, where is gamma plus alpha. Okay, there's no gamma plus alpha. Not possible. Yeah, I will not disturb this term also. Okay. So what I'm going to do is this term, this term. So there are six terms on all together right. Correct. So this term is also taken care of. Yeah. Now how do I take care of these facts. What will happen to the other three terms. Sign of no down sign of gamma plus delta is same as sign of an odd multiple of pi minus alpha minus beta why because alpha plus beta plus gamma plus delta was an odd multiple of. Pi, correct. So instead of writing gamma plus delta, I wrote it as 2n plus one pi minus alpha minus beta. Okay, and when you take sign of both of them, you will realize that this will automatically give you sign alpha plus beta. Right. Similarly, so this becomes this term becomes this term. These two are same terms. Correct. Similarly, you would realize that a delta plus alpha will be same as beta plus gamma. So this and this will be same terms. Correct. And this and this will be same terms. In short, you have written the three ticked terms twice here. Are you getting my point, which means you have written two times sign alpha plus beta sign beta plus gamma. And sign gamma plus delta two times, and you're claiming it to be zero, which means, which means drop the, you know, the two factor. So which means sign alpha plus beta. Sign, let me just drag it over here a bit. I think I have space. This is equal to zero. Hence proved. Okay, so from this tool, other properties also came up. That is cause alpha plus beta summation for all possible values of alpha and beta is zero sign alpha plus beta is also zero summation of that. And moreover, because of the fact that alpha plus beta plus gamma plus delta is an odd multiple of pi we also came into the fact that even if you take any three of the eccentric angles, let's say alpha beta and gamma in the case. And this term will also be giving you the zero expression. Is it fine. Anything that you would like me to repeat over here do let me know. So first I use the fact that since there is no z square coefficient or that square coefficient is zero. The sum of the product to what the time is zero, which led to these two results from the second result I wrote this expression in yellow. I did not disturb the terms which were already present in my given expression that is these three terms. I've ticked them on the top. Okay. And the rest three terms, they actually convert to the ticked terms only. So sine gamma plus delta is as good as sine alpha plus beta. Right, because of this hint which I gave you. Sine delta plus alpha is same as sine beta plus gamma. And sine beta plus delta is same as sine alpha plus gamma. And that's why this entire sum becomes twice of the given expression. And which is anyways going to be zero because this is zero. So the property holds true. Is it fine. Any questions. These feet of the perpendicular that means these go normal points. That's the third property. The feet of the co normal points. They lie on a curve. Okay, whose equation is, whose equation is a square minus B square x y. Let me write it down like this the feet of the co normal points. Concurrent at h comma can lie on a curve. A square minus B square x y plus B square kx minus a square h y equal to zero. Please prove this. Please prove this that the feet of the normal they lie on a curve which is given by this expression. Everybody please prove this. See the proof is very simple. I mean, just, it's just a, you know, observation that you have to do. So let me just pull out the diagram here. Okay, so let us say, let us say, this is my h comma k point. This is my feet of the perpendicular. I'm not getting. No worries. No worries. See, first of all, what is the equation of the normal that we know of a square x by x one. Okay. Minus B square y by y one is equal to a square minus B square. Okay. Now, as per this particular condition x one y one is the point. This is the, this is representative of the coordinates of the point. And this normal should pass through h comma k, right? So if you substitute your, if you substitute your h with x or x with h, y with a k, then this is the equation that I would be getting, isn't it? In short, what do you have written over here is a square a square h y one minus B square k x one is equal to a square minus B square x one y one. Okay. In short, you have written something like this a square minus B square x one y one plus B square k x one minus a square h y one equal to zero. Right. But can I say the same will be true. This will be true even if your point is x two y two x three y three and x one y four. Right now you started with this point as your initiating point you started with this equation. Then you made h comma k, you know, satisfy the equation. But the same will be true even if you have taken B as your starting point that means you would have started with your a square x by x two minus B square y by y two is equal to a square minus B square. And still h comma k will satisfy that equation of the normal, isn't it? Do you agree with me on this or not? This is very important. This will be true even for these three points other than x one y one. So x one y one is definitely satisfying it even x two y two will satisfy it even x three y three will satisfy it and even x four y four will satisfy it. Am I right? Are you convinced with that? Correct. So now just have a look at it even if you replace your x one y one with x two y two. Okay. And further replace it with x three y three, further replace with x four y four. Correct. Now what is the structure of this equation? This basically equation is trying to say that I'm so sorry. Yeah, this structure that you see over here, these four expressions that you see here, what do they seem to suggest? They seem to suggest that if you replace your x i y i's with your x y, you will end up getting a curve which is going to be satisfied by all the four points and that curve is this. So this equation will be satisfied by, let me write it here, this equation will be satisfied by all these points which are your feet of the normal. Correct. And this is the equation that we are basically asked to prove over here. Correct. So the feet of the co-normal points or you can see the coordinates of the co-normal points or the feet of the concurrent normals to an ellipse, they will pass through this curve. Okay. So this itself can come as a question to you in your comparative exams. By the way, this is given a special name. This is called the Applonian rectangular hyperbola. Applonian rectangular hyperbola. We use this word Applonian circle in case of a complex number last year. Correct. So a complex number which basically satisfies this locus condition, where k is not equal to one, then z will trace a Applonian circle. Right, remember, I don't know how many of you remember that last year we had done complex number. So basically the name of Applonian has come from a very, very ancient mathematician, Applonius. So just to honor his name, this was given, maybe he had done some work on this field, that's fine. So I'm not very sure about the history behind it, but this is what we call as the Applonian rectangular hyperbola. So the feet of the co-normal points, feet of the concurrent normals or the co-normal points satisfy this curve which is called Applonian rectangular hyperbola. Exactly, exactly. So this rectangular hyperbola will actually satisfy this four co-normal points will actually satisfy both ellipse equation and this rectangular hyperbola. Why don't we do one thing? We can try to figure it out, but of course we have to ensure I've taken a point h comma k. Okay, maybe we'll try it out, you know, offline. Some of you please try this out and see how does the diagram look like. Okay, so I'll just write it down, try it out on GOG. And share on the group if at all you are going to get these curves and so that everybody sees them. So you have to choose your point h comma k very carefully so that you can draw that four co-normal points. The idea to make co-normal points is choose your eccentric angles first such that there are some of the angles of the eccentric angle is coming out to be an odd multiple of pi. From there you sketch four normals, they will be concurrent at h comma k rather than doing it the other way around. Okay, next concept that we're going to talk about is very similar to what we have already done in circle. So in circles we had done concepts of pair of tangents. Okay, drawn from x1, y1. So if you remember, the concept was T square is equal to SS1. So this used to give you the pair of tangents drawn from an external point x1, y1 onto the ellipse. So let me just pull in our diagram quickly over here. So let's say from this external point x1, y1, you're drawing two tangents. So this pair of tangent equation is given by T square is equal to SS1. I hope you're all aware of what is your T expression. So if I talk about a standard case, okay, if I talk about a standard case, this is what we call as S. This is what we call as a S1 and this is what we call as a T. I hope you all are well aware now of, well aware of now the basic expression for SS1 and T. Okay, maybe a simple question we will take up on this. Let's say we have this ellipse. Okay. The question is, find the equation of pair of tangents, pair of tangents drawn from, let's say 0.4,3. Let's say 4,5 to be more, I'll say 5,6 to be more precise because I want to be outside the ellipse. Just to get a hands on on this formula, that's it. I don't have any other intention. Just to get a hands on on this formula, T square is equal to SS1. Let's just take this problem. No need to simplify it. Just write down the expression and leave it. I mean, you can always simplify it if the need be, but just write down the expression and just let me know by saying and done on the chat box. Done, Anusha is done. Okay, simple. So what is your T expression here? T expression will be X, X1 plus Y, Y1, okay, by a square minus 1. So whole square of this is equal to S and S1. S1 will be just replace your X with a 5 in the equation of the ellipse and Y with a 6 in the equation of an ellipse. I'm not simplifying it. Okay. So whatever equation you get from here, this will be a second degree equation which will represent pair of straight lines or pair of straight lines. Or pair of tangents. Is it fine? Any questions? Now, many times students ask me, sir, if I want to get the two tangents separately, then what will I do? One way is to basically factorize this second degree equation, which I will be telling you when I do pair of straight lines concept with you. Other would be you start the event. Once you know this point, you can start the fact that let the tangent be this. Okay. So once you start with this as a tangent equation, use the condition of tangency. Okay. Then use the condition of tangency. C square is equal to a square m square plus b square. You'll end up getting a quadratic in m. Okay. Solve the quadratic to get two values of m. Put it over here. You get the two tangents separately. I'll repeat whatever I've said. Okay. So please pay attention. If let's say the requirement be you want the two equation of the tangent separately. Okay. So what do we, what do you do for that? So you say, you assume that the tangent equation is y minus y one is equal to slope x minus x one. Remember y one and x one values will be given to you. So why one and x one values will be given to you. So this is the condition. This is the tangent. And this is the equation of an ellipse. You know your m, you know your a, you know your b, you know your C. Write down this equation in terms of, you know, a quadratic. This will give you a quadratic in m. Okay. So solve the quadratic. Solve the quadratic in m. You'll get two values. Okay. Those two values you put it over here. Those will be giving you the two tangents approaches here. How to find two tangents separately. Other concepts like equation of chord of contact. Let me just take up one more concept. Since we have discussed about equation of pair of tangents, director circle, director circle. So what is the locus definition of director circle? Director circle basically is the locus of points from where perpendicular tangents. Let me write it down from there. Two tangents can be drawn to the curve drawn to the curve, which are perpendicular to each other. Okay. So the very same diagram. Now I'm drawing it again. Oh, sorry. So if you have, if you have a point from where you're drawing two tangents, which are perpendicular to each other. Okay. Let's say this is my point P. Okay. Then the locus of all such points, the locus of all such points. Okay. That will be the director circle. And of course in this case also it will trace a circle only. So let me just draw a circular path around it. Now I would request you all to give me the equation of this director circle. Okay. So what is the equation of this director circle? So let's say this is a locus question which was given to us in let's say J main exam. Okay. So this is a standard case of an ellipse x square by a square y square by b square equal to one. And I want to find the locus of all such points from where if you draw tangents from where if you draw tangents to the ellipse. Okay. Any point you take it draw tangents to the ellipse that those tangents will be at 90 degrees to each other. So what is the equation of this director circuit? Please solve this and give me a response on the chat box. This will also help me to test your understanding of how to solve locus questions. Okay. Should we discuss it? All right. So let us say this slope is M. Okay. And it is passing through H comma K. So you can write down the equation of you can write down the equation of the tangent to be this. Okay. Or in short, y is equal to mx minus or you can say ms plus K minus mh. Okay. Now, first of all, you're claiming that this is a tangent. This is a tangent to tangent to tangent to our ellipse. Okay. Now, one important thing I would like to highlight over here. Many times people argue with me sir, why do you take H comma K in locus questions and then you substitute it with X and Y. See here, there's an already X and Y sitting in our tangent equation. And if you slide your H in K again with an X and Y that will create a lot of course. Right. So I think last week only I was teaching locus to your juniors. So in one of the batches, one student was quite irritated. He said that sir, why do you take H comma K and then replace it back with X and Y. In the last step, why don't you take X and Y in the beginning itself and save your time. So then I told you that maybe it is not going to hurt you right now. But later on when you do conics, there will be a lot of instances where X and Y is also used and H and K is also used. And then you will get confused. Okay. So it's very important that we follow the norms that is, you know, discuss with us. Anyways, if this is a tangent then condition of tangency must be fulfilled. So what is the C here? This is the C right. A square M square and V square that is not going to change. Now this is a quadratic in M. This is what I was basically talking about. If you write this as a quadratic in M, you will have K square minus 2KMH plus M square X square equal to a square M square plus B square. So this will be M square A square minus X square. Okay. And you'll have plus 2KMH plus B square minus K square equal to zero. Right. Now in this quadratic, you have two roots M1 and M2. Right. Let us say, and you want the, you want the two tangents to be perpendicular. So let's say if this is M1 and this is M2, and you want them to be perpendicular, clearly M1 M2 should be negative one. Isn't it? Very good. That's the right answer. So here M1 M2 is going to be minus one. M1 M2 is nothing but the product of the roots of a quadratic, which is C by A. We all know by beta isolation, this is equal to C by A. So this should be minus one. That is to say that B square minus K square is X square minus A square in short, X square plus K square is equal to A square plus B square. Now here you generalize your edge with X and K with a Y. That gives you the equation of the director circle to be this. Please note that this radius is even more than the semi major axis lens. So it is even more than that. It is under root of A square plus B square that is the radius. So radius of director circle is radius of director circle is and the center of the circle is same as that of the conic. Okay. So this is in general applicable to any type of an ellipse. Center will be same as that of the center of the ellipse, wherever it is, whether it is shifted, whether it is oblique and radius will always be under root of A square plus B square. So this is true in general. Is this fine. So before we move on, we will like, I will like to discuss with you. Equation of a chord bisected at a given point X one Y one. This concept is also same as what we learned for a circle and parabola, which is T equal to S one. Okay. So if you have been given a conic, if you've been given a conic X square by a square Y square by B square equal to one. And there is a chord whose midpoint is known to you. Let's say this midpoint is known to you X one Y one. Then the equation of this chord is given by equal to S one. And this, it is very easy for us to cancel out minus one and minus one, but just for remembering purpose. So that this is easy to remember, isn't it. So you can just carry that extra burden, even though it will get canceled from both the sides while you are trying to solve this problem. But this helps you to remember it T equal to S one. So I will not be going into the derivation, etc, because the result is more important for us. And we'll just take a question based on this. Again a locus question. The question says tangents and right angles are drawn to the syllabus. Show that the locus of the midpoints of the chord of contact is this curve. Beautiful question. Next when we meet we are going to talk about equation of the diameter and something related to conjugate diameter. I think that's the only concept which is left off conjugate diameter is also important. A lot of questions in J advance have come on that. So we'll talk about that. And then we can close this chapter and move on to hyperbola. Okay, so basically you're going to tangents from it, which are at right angles. Okay. This is your chord of contact. Okay. We have to find the locus of the midpoints. So let's say this is point P. Let's say this is AB. So we have to find the locus of this guy. Let's say I call this point to be Q point H comma K. So if you keep changing this P position on the director circle, obviously the chord of contact will also keep moving and the midpoint will also keep moving. And we have to find the locus of the midpoint. Okay. Let's solve this question in this way. Let's say this point is X1 Y1. Okay. Now, what is the chord of contact when you draw two tangents from X1 Y1? The chord of contact is T equal to zero. We have already seen it. So that means your equation of AB, equation of AB is going to be XX1 by A square, YY1 by B square equal to one. Now, at the same time, be very careful. X1 Y1 basically lies on director circle. We will use this little later on. As of now, we'll keep it in terms of X1 Y1. Same, if you find the equation of the chord whose midpoint is H comma K, the equation is T equal to S1. Even this will give you the equation of AB only, which is T equal to S1. And that is equal to HX by A square. I'm just writing T expression, KY by B square minus one that will eventually get cancelled. So that's not waste time writing that. Okay. Now here, these two equations represent the same things. This is also equation of AB. This is also equation of AB. Okay. So let's compare one and two, because both represent equation of AB only. So if you compare the coefficients, let's compare the coefficient of X by A square on both sides. So X1 by H will be equal to Y1 by K. Please note that other terms will get cancelled off. So there's no point writing it down. One by H square by A square plus K square by B square. So this is a vital step for us because after this our problem is almost solved. How? Your X1 is H by this and Y1 is K by this. And this must satisfy the director circle equation, which is just now seen as X1 square plus Y1 square is equal to A square plus B square. So we're just putting these values in place of X1 and in place of Y1 over here that will lead to H square by H square by A square K square by B square whole square. Similarly, K square by H square by A square K square by B square whole square equal to A square plus B square. Okay. So this is really close to the expression. Yeah, so almost done, almost done. So S square plus K square, you can just, you know, take one by this common out. Maybe after the class you can try sketching it also what kind of a curve comes out from it. Okay, and then you can just, you can just make it on GOG brand check. Okay, now generalize it. After this you can generalize it. Okay, X square by A square, Y square by B square whole square as X square plus Y square by A square plus B square. This is the desired locus. Okay. Now this chapter is still not completed as I told you a concept of diameter conjugate diameter will also talk about bit of consicclic points also. In the next class I will be definitely finishing it off and maybe I will also start with hyperbola whenever we meet next. Okay. Thank you class.