 Hello, welcome to the Koenig section YouTube live session today So those who have just joined in Please type in your name in the chat box so that I know who all are attending this session All right guys, so today's session we are going to start with the fact that why at all Koenig sections Are known by the name Koenig sections, okay? so Koenig section is basically Given the name because they have been obtained by the section of a right circular double cone with a plane Right, so these Koenig sections they are obtained by They are obtained by the cross section of a right circular double cone a Right circular double cone by a plane Okay, so depending upon how these planes or how this plane passes through the right circular double cone Different Koenig sections are generated. Now. What is the right circular double cone a cone which has got? Let me explain this to you first a cone is basically Right the right circular double cone is basically a structure like this where this is called the axis of the cone So this line is called the axis of the cone, okay, and these two lines are known as the generators These two lines are known as the generators When the axis of the cone makes 90 degree with the base of the each cone We call it as a right circular double cone. Okay, so I will now show you How basically these Koenig sections are generated? Okay, so let us focus on Let's focus on this structure So I can as you can see on your screen. Okay, so basically this is nothing but a right circular double cone Okay, so I've shown it in a 3d structure now this right circular double cone if I cut it by a plane which is perpendicular to the axis Right as you can see here I've cut it by a plane which is perpendicular to the axis the cross section area from the top you can see is that of a circle Okay, so the cross section area that you can see is that of a circle Okay, if you cut it by such a cone Or such a plane which is making some angle Okay Then you realize that you start getting different types of Koenig's Okay In this case you end up getting a hyperbola Right, so these basics are all cleared in your mind And if there is any question regarding the basic of creating a Koenig section, please do ask In fact, I will quickly brush you through So when you cut this by a plane When you cut this by a plane Whose angle with the axis let's say is Let's say I call the semi vertical angle of this right circular cone as as alpha And let's say this angle is theta Okay, the angle of between the plane And the axis is theta if theta is 90 degree it generates a circle Okay, and If this plane passes through the meeting point of these right circular double cone the circle would be a point circle Okay If this theta is somewhere between 0 to somewhere between Alpha to 90 degree Alpha to 90 degree it will create an ellipse It will generate an ellipse Again here. I'm assuming that it is not passing through the meeting point of the generators Okay If theta is exactly equal to alpha That means the angle made by the plane is parallel to either of these generators The cross-sectional area left would be that of a parabola. Okay And if theta lies between 0 to alpha Then it would end up cutting both the Naps these are called naps Okay, both the naps Of this right circular double cone And you will obtain a hyperbola in that case If it passes through the meeting point if it passes through p if the conic section passes through Or if the plane passes through p if the plane happens to Pass through the point p Then in that case we call the conic generated as a degenerate conic Okay, and pair of straight lines is an example of a degenerate conic So first I will spend some time on the degenerate conic In order to explain you What is the general form of the equation of a conic section? So let's talk about pair of straight lines briefly What's a pair of straight lines? Pair of straight lines is basically a kind of a degenerate cone Where you obtain the Equation of this pair of straight line by multiplying two lines Okay, so if you start by multiplying two such lines which are passing through Let's say origin y equal to m1x and y equal to m2x Okay, if I ask you these are the lines which make that pair of straight lines Then the combined equation of the pair of straight lines would be y minus m1x times y minus m2x equal to zero Right, so this is the equation of a pair of straight lines, which is passing through origin Which is passing through Or intersecting each other at origin So both the straight lines are passing through the same point origin, okay If you clearly look at this expression It's very important for you to appreciate that This expression will be a homogeneous Expression in x and y Okay So we call this as a homogeneous function or homogeneous equation in two variables This is going to be a homogeneous equation in two variables Of degree two Of degree two So what is A homogeneous equation a homogeneous equation is basically a equation where you would realize all the variables are having the same degree For example x square y square x y they are all having the same degree, okay So any second degree equation, which is homogeneous in nature in two variables That would always represent a pair of straight lines passing through origin. This is very very important so always remember this Any equation of this nature a x square plus two h x y plus b y square Will always represent a pair of straight lines Passing through origin Now it is not necessary that both the lines would be real in nature It may happen that you may also end up getting non real lines as well, right You may end up getting imaginary lines as well that depends upon some relationship between h a and b that I will discuss with you later on Okay, now taking a clue from this You would realize that any conic that you see is going to be having a general formula or the general equation as a x square plus two h x y plus b y square Plus two g x Plus two f y plus c equal to zero Right, this is the general structure of any conic section that you would come across Right, which is called a second degree general equation of a conic section So even your straight lines as you can see here The the homogeneous its equation itself is basically is a second degree equation Even circle parabola ellipse will have these six terms appearing not or it's not necessary that all of them will appear together It may happen that some of them may be absent Now how why is this a second degree equation? Guys, if you if you see a very specific case from the example which I cited a little while ago the equation of the equation of if you see the equation of a cone I've taken passing through the origin as z square equal to x square plus y square, right? So this is the equation of the cone Okay And if you take any line any plane, okay, let's say I take this plane x plus y plus z equal to one Okay, and you try to eliminate your z from these two equations So from the plane equation and the cone equation if you try to eliminate your z Okay, you would realize that Let's say this was the equation of the cone Okay And let's say this is the equation of the plane any plane. Okay, so I'm taking just a simple example of a cone and a plane If you try to eliminate your z that means replace your z with one minus x minus y in this You yourself would see that When you square such terms You will end up getting some terms in x square Some terms in y square You'll see some term with x y You'll see some term in x you'll see some term in y and you'll see a constant, isn't it? So that is what we are trying to state in the previous slide that any conic section would have This kind of a structure. Sorry. This kind of a structure a x square Plus 2 h x y plus by square plus 2 g x plus 2 f y plus c equal to 0 Right so far any questions with respect to this Okay, so this is the general structure of any conic section Now depending upon your values of a b and a b h g f and c This conic section can represent various things Right, so now we'll discuss the condition to represent various conics condition to represent various conics The first condition that i'm going to talk about that's the very important condition if a x square plus By square plus 2 h x y plus 2 g x plus 2 f y plus c equal to 0 has to represent A pair of straight lines If it has to represent a pair of straight lines then An expression delta which is given as a b c plus 2 f g h Minus a f square Minus b g square Minus c s square will be equal to 0 that is In a simpler form the determinant made by these terms Will be equal to 0 Right now first of all i will prove how these condition comes So let us prove how this condition comes Okay So let us start with the fact that let A x square Plus by square plus 2 h x y Plus 2 g x plus 2 f y plus c equal to 0 represent a pair of straight lines Okay, let's say this represent A pair of straight lines Of course since this is not a homogeneous equation That means Their point of intersection of these lines is not origin because had it been origin These three terms would have been absent Right So let's say these pair of straight lines This pair of straight lines intersect each other at Let's say they intersect each other at x 1 y 1 And we are assuming that these lines are also not parallel to each other Okay Then do one thing then apply the concept of shifting of origin Shifting of origin 2 x 1 y 1 That means i'm assuming the point of intersection now to be my new origin Okay So when we do that all of you remember the concept of Shifting of origin where you replace your x with Where you replace your x with capital x plus x 1 and y with capital y plus y 1 So this was the concept which we had discussed In the beginning of the coordinate geometry chapter in class 11 So when you do these replacements in your equation Given to you you get something like this a x plus x 1 square Okay, b y plus y 1 square Plus 2 h x plus x 1 y plus y 1 2 g x plus x 1 2 f y plus y 1 Plus c equal to 0 Okay And if you simplify this you are going to see something like this a x square b y square 2 h x y And apart from that you will see terms like x 1 I'm collecting terms having capital x in it so 2 a x 1 I will see then from here I will get 2 h y 1 and I will also get term like 2 g Similarly if I collect terms containing capital y I will get 2 b y 1 I will get 2 h x 1 And I will get 2 f term correct And if I collect the constant terms I will get a x 1 square b y 1 square 2 h x 1 y 1 2 g x 1 2 f y 1 plus c equal to 0 Okay Now all of you please listen to this very very carefully If you want this pair of straight lines To pass through origin It is passing through origin Then it is very clear that it has to be a homogeneous equation And if it has to be a homogeneous equation That means this term This term And this term should be 0 Remember I discussed with you Any homogeneous equation Of second degree is basically representing a pair of straight line passing through origin And since I am shifting my origin to the point of intersection x 1 y 1 It means that I am assuming that These terms are going to be 0 This term is going to be 0 This term is going to be 0 And this term is going to be 0 Is that clear? Because if these three become 0 I will only end up getting this term Which is a second degree term And hence it will represent a pair of straight lines passing through origin So this implies we have a x 1 H y 1 plus g equal to 0 We have this is x 1 Yeah, this is h x 1 Plus b y 1 plus f equal to 0 And guys if you note your expression very very carefully If you note these expressions very very carefully you would realize that You can write this as x 1 into A x 1 Plus h y 1 plus g Plus This expression I am simplifying Plus y 1 into h x 1 Plus b y 1 plus f And the remaining terms would be g x 1 f y 1 plus c So what I did is I took an x common from this term And one of these terms I split it open as H x 1 y 1 plus h x 1 y 1 So I took a x 1 common from this term and I took x 1 common from One of these terms So this also I split it up as g x 1 plus g x 1 This also I split up as f y 1 plus f y 1 So one of the terms a x 1 h y 1 and g has come from taking this term This term and this term Similarly for y 1 if I take y 1 common from this term This term and this term I get this term I will get this term, okay, and the remaining terms that is G x 1 f y 1 and c I am writing it over here Okay, is that clear any question regarding this so far Now from 1 and 2 we already know that These two are 0 this is 0 This is 0 Correct that means I get a third equation from here That is g x 1 f y 1 plus c also equal to 0 So I get three equations altogether Like this So this is my third equation Now using these three equations I would now eliminate x 1 y 1 Okay, so from this you try to eliminate x 1 y 1 Okay, and how do I eliminate it if you remember The trivial solution concept of A system of homogeneous equation or a system of linear homogeneous equation The determinant If this determinant is equal to 0 Okay, then this system of equation. So guys what I'm trying to say is You know your h you know your x 1 y 1 and 1 right? They are not going to be be a trivial solution over here Okay, so if if the system of equation doesn't have any trivial solution We can say that the determinant formed by the coefficients Okay, so basically let me treat one over here if nothing is there So the coefficient of these variables would form a determinant which is equal to 0 right And if you expand this you are going to end up getting the desired result which is abc plus 2 f gh minus af square Minus bg square minus ca square equal to 0. Is that fine? Is that clear? So this is the condition Which we write in a very short form as delta Okay, so if delta is 0 your second degree equation will definitely represent a pair of straight lines Okay