 The next concept that we are going to talk about is the concept of co-normal points. Having learned about the equation of the normal, now we will talk about the concept of co-normal points. Now in order to understand what are co-normal points, let us understand this fact that in general, four normals can be drawn to an ellipse, not more than four, can be drawn to an ellipse from any point. Please remember that all the four normals need not be real. So let's say I assume a point h,k inside the ellipse. Okay, let's say I assume this point h,k. From this h,k, I can draw four normals. I am again repeating it is not necessary that all the four normals should be real. First of all, how do I prove this fact that I can draw only four normals max to max from any point h,k onto the ellipse? How do I prove this fact first? Okay, so in order to understand this, by the way, what are the co-normal points? The points or the feet of these normal a,b,c,d, they will be called the co-normal points. So a,b,c,d are the co-normal points. So co-normal points are nothing but they are the feet of the normal drawn from a point h,k onto the ellipse, right? So is there any question with respect to what are co-normal points? Now sometime I am going to prove that we can actually draw four co-normal. We can actually draw four normals from a point h,k to the ellipse, not more than that. So let us prove that first. Now we all know that the equation of the normal to an ellipse is given as a square x by x1 minus b square y by y1 is equal to a square minus b square. So now what I will do is, I will assume that these points have eccentric angles, alpha, beta, gamma and delta. So the eccentric angle of these points are let us say alpha, beta, gamma and delta. So now what I will do is, in general let us say the eccentric angle is phi. So in general, let us say the eccentric angle of these points of the co-normal points is phi. So when I write it like this over here, I can write it as a square x by, x1 could be written as a cos phi and y1 could be written as b sin phi is equal to a square minus b square. So I can further simplify this to a square x by cos phi minus b square y by sin phi is equal to a square minus b square. So what I am assuming that, I am assuming the opposite way round, I am assuming that from a general eccentric point phi I am drawing a normal. So this will be the equation of that normal. Now this normal is also passing through h comma k. So I can replace my x with h and y with k. I can replace my x with h and y with k. Now guys here I will take the help of complex numbers. I will take a help of the complex numbers. I will say let z be a complex number which is given by cos phi plus i sin phi. That means in the Euler form I am writing it as e to the power i phi. So of course 1 by z will be cos phi minus i sin phi. So it is clear that 1 plus 1 by z is 2 cos phi. That is cos phi is going to be half of, I can write it as z square plus 1 by z. And similarly sin phi would be z minus 1 by z into 1 by 2 i. That means it can be written as z square minus 1 by 2 y z. Now what I will do is, in the equation of the normal, in the equation of the normal which I have written in terms of cos phi and sin phi, I am going to make the replacement with z square plus 1 by 2 z. And I am going to change sin phi by the way this a and a will get cancelled. And here I am going to replace sin phi with z square minus 1 by 2 i z. And this is equal to a square minus b square. So please pay attention now I am going to simplify this. But for the interest of time I am going to directly write down the result of the simplification so that you all don't waste time knowing that what is important is some properties which we are going to discuss with respect to the conormal points. So when you simplify this, you are going to get a bi-quadratic equation in z. This is a bi-quadratic in z. Bi-quadratic in z. Clearly suggesting that there are four values of z1, z2, z3 and z4 which are going to be the roots of this equation. That means there can be four eccentric angles because each one of them is e to the power i phi. So I can say e to the power i alpha, e to the power i beta, e to the power i gamma, e to the power i delta would be basically be the roots of this particular bi-quadratic equation in z. Suggesting that there can be four eccentric points. There can be max to max four eccentric points which are going to be the feet of the normal drawn from h, k on to the ellipse. Hence only four normals can be drawn, not more than that. Only four normals can be drawn. Is that fine guys? Any question with respect to why four normals only can be drawn from a point external to the ellipse on to the ellipse? Okay. So for the purpose of giving you some properties I will rewrite this particular expression once again. You don't have to memorize it because it's very very lengthy procedure to derive these equations. It's just for your conceptual clarity that I'm going to write this. Okay. Now I'm going to ask you few questions. I'm going to ask you few questions. The first question is prove that the sum of the eccentric angles of the conormal points is going to be an odd multiple of pi. Prove that the sum of the eccentric angles of the feet of these conormal points or of these conormal points is going to be an odd multiple of pi. So what I want to say here is that if you add the eccentric angles over here alpha, beta, gamma and delta their addition will be going to be an odd multiple of pi. Can you prove this? If you're able to prove it, please send me the snapshot on my personal ID, personal WhatsApp. Alright. So guys, if you look at this equation here what are the product of the roots? Z1, Z2, Z3, Z4. Okay. Be careful that there is no Z squared term over here. So you can write zero Z squared. So remember, sum of the roots is minus B by A then C by A then minus D by A then E by A. So E by A would be this. E by A would be this. That's going to be minus of one. Right. Now, Z1 is e to the power i alpha. Z2 is e to the power i beta. Z3 is e to the power i gamma. This is e to the power i delta. That's going to be minus one, which means e to the power i alpha plus beta plus gamma plus delta is equal to minus one. Right. Now, basically it says cos of alpha plus beta plus gamma plus delta plus i sine alpha plus beta plus gamma plus delta is equal to minus one plus i zero. So that can only be possible if your this is equal to minus one and sine of alpha plus beta plus gamma plus delta is equal to zero. So this can only be possible if your sum of the eccentric angles happen to be an odd multiple of pipe. An odd multiple of pipe. So this is how we can prove this. Any questions with respect to this proof, please feel free to type in in the chat box. If it is clear, please type CLR. Alright. So second property is again if alpha, beta, gamma, delta are the eccentric angles are the eccentric angles of the co-normal points eccentric angle of the co-normal points then prove that sine, take any three of them let's say alpha, beta, gamma so alpha sine alpha plus beta plus sine beta plus gamma plus sine gamma plus alpha is equal to zero. This should be very simple for you to prove since you have already taken the previous case while proving this you will be able to realize many more properties as well. Guys, again let me tell you if je has to ask this question they will actually ask this in the je advance part where they will first tell you what is co-normal points. They will not expect you to come prepared with this theory. Je is most interested in locus whether you are clear with locus if at all some extra concepts are to be asked they will always frame up a paragraph based question explain you what is co-normal points and then probably ask you follow up questions on that. So look at this again we will recall back the previous equation so again for the purpose of using it I will recall it for you so this was the equation that we had written in terms of z in terms of z again I have not written it but zero z squared should be here because z squared is not present and that's actually the key point for us this is the key point for us if you realize that z squared term is absent that means product of all the roots product of z1 z2 will actually be zero that means e to the power i alpha e to the power i in fact I will directly write it e to the power i alpha plus beta e to the power i alpha plus gamma e to the power i alpha plus delta e to the power i beta plus gamma e to the power i beta plus delta e to the power i how many have written 5 the 6 one is left e to the power i gamma plus delta it should be zero correct now when you express it as a polar form you would realize that summation cos alpha plus beta plus summation i sin alpha plus beta is actually zero plus zero i right summation means understood right two angles taken at a time which clearly indicates that these two individually are zero zero right so again this is some extra information which I got from the problem however I have to prove only the thing which is given to us so how do I prove this so let me write down these angles once again sin alpha plus beta sin alpha plus gamma sin alpha plus alpha plus delta sin beta plus gamma sin beta plus delta sin sin gamma plus delta equal to zero okay now we all know that alpha plus beta plus gamma plus delta is an odd multiple of pi so let's say pi okay it's an odd multiple of pi so can I say that can I say that alpha plus now look at the terms which are not required if you look at the proof we only require these terms this beta plus gamma and gamma plus alpha gamma plus alpha this we don't require this term this term and this term okay so what I will do is first of all can I write alpha plus gamma term as pi minus beta plus alpha plus delta can be written as this alpha plus delta term as pi minus beta plus gamma correct so sin of alpha plus delta is sin of this term which is actually sin beta plus gamma okay so these two terms these two terms are same this term and this term are same these two are same similarly I can say that beta plus delta that is this term is equal to pi minus alpha plus gamma so sin of beta plus delta is equal to sin of alpha plus gamma so this and this are the same term this and this are the same term and similarly you can say sin of gamma plus delta will be sin of alpha plus beta so the first and the last are the same term that means the same expression can be written as twice of sin alpha plus beta sin alpha plus gamma sin beta plus gamma any three angle you can take okay so dropping a factor of two from both the sides hence proved is that clear guys any question with respect to this