 Next is the concept of intersection of a line intersection of the line y equal to mx plus c With the standard case of a hyperbola With the standard case of a hyperbola again guys The concept is pretty much same as that we had for an ellipse. What do we do? We try to simultaneously solve these two We try to simultaneously solve these two Okay, by substituting y as mx plus c in the other equation So that gives you x square by a square minus Mx plus c whole square by b square equal to 1 and if you simplify this you are going to get a quadratic equation in x like this, okay and Just by deciding the nature of the roots of this quadratic equation you can actually make out whether The line is cutting touching or not even touching the given ellipse Sorry given a hyperbola, okay, so they can be three situations They can be a situation where it is tangent like this There could be a situation where it is cutting like this and They could be a situation where it is not even cutting like this Okay, so how do we distinguish between situation number a? b and c Just by looking at the discriminant of this particular quadratic equation So I'm just going to write down the result. I don't want to waste time simplifying this Okay, so if c square is greater than a square m square minus b square It will be situation number a that means it will cut the hyperbola If c square is equal to a square m square minus b square it would be situation number b That means it will touch the hyperbola and if c square is less than a square bs m square minus b square It would be situation number c that will not touch the hyperbola and please remember these conditions are very much specific to this hyperbola only Only to be applied on the standard case Only to be applied on this Okay, so if your hyperbola changes your situation will also change Your situation will also change Okay Out of this the condition number b is actually called The condition of tangency This situation is called the condition of tangency, right? And because of this you can actually write this Equation of a tangent if you know the slope as a square m square minus b square. This is called the slope form This is called the slope form Okay, so guys a quick analysis about this equation. There's something very important about this equation So when you say the slope form which is y equal to mx plus minus under root a square m square minus b square Please remember one thing that you cannot draw tangents with any given slope Your slope must be following this restriction. That means a square m square should be positive that means a Square m square should be greater than b square. That means m square should be greater than b square by a square or m Square minus b square by a square should be greater than zero, which means m minus b by a m plus b by a should be greater than zero That means m should either be greater than b by a or m should be less than Minus b by a Right in other words in other words Let me tell you what is b by n minus b by a slope. In fact, I know I'll just Take you to the Geo Jebra setup Let's say I draw a slope of y is equal to y is equal to b by a Slope line. So y is equal to 2 by 3 x or 2 x by 3 This is a y equal. This is a b by a slope and y is equal to minus 2 x By 3 So please remember the slope The slope of the tangent that you are going to draw Should be either Be greater than b by or less than minus b by that means Your line could have slopes of this nature. It could be like this It could be like this I'm just drawing the slopes. I'm not drawing the exact tangent So basically what I'm trying to say is that this zone is the zone in which your slope should like If your slope happens to be lesser than that, what will happen? It will end up cutting the The hyperbola Okay, for example, let's say I take a slope like this I can always shift this and make it a tangent like this Are you getting this point? But if you try to take a line with a slope this No matter wherever you shift No matter no matter wherever you shift it will end up cutting the hyperbola Are you getting this point? So you can never get a real tangent You can never get a real tangent if your m happens to be let me write it. No real tangent No real tangent If your m happens to lie between minus b by n b by So it has to be either less than minus b by Or greater than b by it to get a real tangent To get a real tangent and another thing which I want to tell over here is that which many people tend to get confused there cannot be There cannot be any Tangent which actually is tangent to both the arms simultaneously apart from these two asymptotes So asymptotes are only two such tangents which are Simultaneously touching both the arms So these are like tangents to both The branches at infinity So asymptotes can only touch both the arms simultaneously and this point of touching is at infinity rest note no tangent Can touch both the arms simultaneously? For example, let's say I make a tangent like this Okay, now this will never this tangent can never touch the other branch if it is touching over here It will not touch the other branch Okay so no Tangent can touch both the arms of the hyperbola simultaneously except for the asymptotes and for the asymptotes they touch the arms simultaneously at infinity is that clear guys So what we'll do is we'll take some quick questions and wrap up this session because I know many of you are having exams also So let's take this question show that X cos alpha Plus Y sin alpha equal to P touches the hyperbola Touches the hyperbola x square by a square minus y square by b square equal to 1 if if a square cos square alpha Minus b square sin square alpha is equal to p square done So Lolita is done. So very simple Use the condition c square is equal to a square m square minus b square So remember C over here will be p by sin alpha Okay, a square m will be minus cos alpha by sin alpha whole square Minus b square multiply throughout with sin square alpha you'll get a square cos square alpha Minus b square sin square alpha. Okay, hence proved Next question for what values of value of lambda Does the line? Y equal to 2x plus lambda touches 16 x square minus 9 y square is 144. Please type in in your response in the chat box For no, that's not correct. How about one? Rowan any response? It's a similar nature question the root 20 That's correct. Okay. Let's let's apply this first of all. We write it in a standard form. So 16 x square by 144 minus 9 y square by 144 equal to 1 which means x square by 9 minus y square by 16 is equal to 1 Okay Now a square is 9 b square is 16 And for our case our C is equal to lambda and m is equal to 2 Now it will be tangent only when now since this is a standard case It will be tangent only when C square is equal to a square m square minus b square Which means lambda square is equal to a square 9 m square which is 2 square Minus b square is minus 16 that means lambda square is equal to 36 minus 16 which is 20 Which means lambda could be plus minus root 20. That means lambda could be plus minus 2 root 5 Okay So all right guys, so we'll call it a day now However, we have many more concepts left to be covered in case of a hyperbola So we'll be talking about other forms of the equation of the tangents like the point form parametric form, etc Then we'll talk about the other equation of normal again in Cartesian form and parametric form Sorry a point form and parametric form again repetitive concepts like chord of contact pair of tangents Cod whose midpoint is given to us concept of pole and polar Equation of the diameter Concept of conjugate diameter Concept of co-normal points So all these concepts and apart from this there is a very small part of a hyperbola, which is called rectangular hyperbola and Asymptotes that we need to cover up So I think the last next class that we'll be having will be covering up all these concepts as quickly as possible So that you are you know well prepared with your chronic section part And it's good that we are doing it towards the end because these these concepts will be fresh in your mind Because they're done normally in class 11 then people tend to forget them Okay So thank you so much for coming online. So over and out from Centrum Academy. Thank you and best of luck for your exams