 Hello, welcome all to the YouTube live session on Koenig section. So guys, those who have joined the session, please type in your name. Please type in your name in the chat box so that I know who are attending the session. So this session would be on parabola continued. So from where we left off, we'll be starting from there itself. So today's lecture, I'll be starting with co-normal points. So guys, what is a co-normal point? Let me define this for you. So let's say we have a point h comma k over here, okay? And from this point, let's say I draw normals to the parabola. Let's say I draw normals to the parabola. So all these lines, they are normal to the parabola. Meeting the parabola. First of all, my question is, how many normals can I draw from a external point on to the parabola? How many normals can be drawn from a point external to the parabola on to the parabola? Please type in in the chat box. How many normals do you think can be drawn from a point to a parabola? Is it one? Is it two? Is it three? Is it four? Any guesses? No guesses? Okay. So guys, the answer is hidden within the equation of a normal to the parabola. So we all know that the equation of the normal, equation of the normal to the parabola y square is equal to four a x is given by y is equal to mx minus two a m minus a m cube, okay? So basically, if the slope of the normal is m, m is the slope of the normal, m is the slope of the normal, you would realize that this equation is actually a cubic in m, this is a cubic equation in m. That means there can be three possible values of m, okay? So there can be three possible values of m, okay? There can be three possible values of m for this line to be normal to the same parabola, okay? In other words, if I say that this point h comma k lies on this normal, then I can say y can be replaced with k, x can be replaced with h. And now we have a cubic equation with us that is a m cube plus m to a minus h plus k equal to zero. And this equation will have three roots. This equation will have three roots. Therefore, you can only draw three normals from an external point on to the parabola. Let me tell you these three may not be all real, right? You may have all the three real or you may have one real and two imaginary also, okay? Now what are co-normal points? Co-normal points are nothing but they are these points a, b and c, right? So a, b and c, which are nothing but the feet of the perpendicular or feet of the normal drawn form the external point, let's say p on to the parabola are called the co-normal points. a, b, c are called the co-normal points. Is that clear, guys? Okay, any question with respect to what's a co-normal point? So those who have joined in the session, please type in your names in the chat box. So now there are certain properties which we'll be looking at with respect to the co-normal points. So let's say these are my co-normal points a, b and c, okay? Let's say the slope of these lines is m1, m2 and m3. Clearly m1, m2 and m3 are the roots of, are the roots of, let's say this point is h comma k. They are the roots of the equation am cubed plus m times 2a minus h plus k equal to zero, okay? Now if you could see here, there is no m2 term. This expression does not contain m2 term, right? What does it imply? It implies that the sum of the roots which normally is given by minus b by a, right? And since there is no b term over here because m2 term is absent would actually be zero, okay? So this is the first thing that we all need to remember that the algebraic sum, the algebraic sum of the slopes of the slopes of three concurrent, three concurrent normals, okay? Is zero, whereas the algebraic sum means along with the sign, okay? Not just the magnitude but along with the sign. This is the first thing which we need to keep in mind. Now, second is a question which I'll ask to you. What can you comment about the algebraic sum? What can you comment about the algebraic sum of ordinates of the feet? Ordinates of the feet of these co-normal points. That means if let's say this point is at1 square comma to at1. This point is at2 square comma to at2. And this point is at3 square comma to at3. Then what can I comment upon? What can I comment upon to at1 plus to at2 plus to at3? What will be the value of this? Any response, guys? Any idea what would be the sum of the ordinates of the feet of these three co-normal points? Now, this answer is related to what we had learned in the first property. How, let me show you. First of all, m1, how is m1 connected to t1, right? It's very simple. If you see, m1 is nothing but minus dx by dy. That is minus dx by dt divided by dy by dt, correct? Minus dx by dt is minus 2a t1, and dy by dt is 2a, that means it's minus of t1, correct? So m1 is minus of t1, correct? So if I write this expression as t1 plus t2 plus t3, the same could be written as minus m1 minus m2 minus m3, correct? Taking a minus common, I get 2a times m1 plus m2 plus m3, okay? So can I say it will become 2a into zero because some algebraic sum of the slope is zero, correct? That means always remember, if you add the ordinates of the feet of the conormal points, your answer will always be zero. Your answer will always be zero. So let me ask you another question. If three normals, if three normals drawn to y square is equal to 4ax from a point h comma k, be real and distinct, be real and distinct, then prove that 27ak square will be less than four times h minus 2a whole cube, is the question clear? That means if you are going to draw three normals if you are going to draw three normals whose slopes are real and distinct, that means the normals are real and distinct, then it can only happen when 27ak square is less than four times h minus 2a the whole cube. Can you prove this? See, we are talking about this equation, right? We are talking about a function in m which is am cube plus m times 2a minus h plus k, okay? We are talking about this function. Now let me call it as a function of m, okay? Now just try to recall that we had discussed in the case of cubic equations, right? We had already discussed the nature of the roots of cubic equation in the application of derivatives chapter in the increasing-decreasing function. When does a cubic equation has real and distinct roots? Try to recall that situation. So first of all, we used to find the derivative of this function. Let's find the derivative of f of m. The derivative of this function will be a 3am square, okay? Plus 2a minus h, correct? Now, if you see the roots of this derivative, the roots of this derivative are, when you put this to zero, you get the value of the roots. Let's say the roots are alpha and beta. The roots are alpha and beta. Then alpha is going to be h minus 2a by 3a under root and beta is going to be negative of that, correct? So both the roots will be exactly opposite in sign, same in magnitude, okay? Now just try to recall the situation that your cubic equation has three distinct roots, okay? Let's say this is your m1, this is your m2, this is your m3, okay? And this value that you just figured out is actually your alpha and beta, alpha and beta, okay? Now if you see here, alpha and beta both are opposite in sign. So actually I should have drawn this figure in this way. Since they are equal and opposite in sign, I should have drawn this figure in this way. So I'm just redrawing it again. So it's something like this, okay? So this is your alpha, this is your beta, right? Or you can call this as your beta and this is your alpha because they are having signs which are negative and positive respectively. So this is your beta and this is your alpha. This point is your m1, this point is your m2 and this point is your m3. So clearly here you can see that the roots are distinct. For that to happen, if you see this figure clearly, the value of the function at beta, that is f of beta, that is positive and the value of the function at alpha, that is f of alpha, that is negative, correct? So this value is actually negative. If you combine these two situations, you can say f of alpha into f of beta should be negative, correct? Because they're opposite in sign. And here you can see f of beta can same, will be same as f of minus alpha, right? Because beta is actually negative of alpha. Okay? So what I'll use is I'll use this situation, f of alpha into f of minus alpha less than zero to reach out to this required condition, okay? So f of alpha, when I say that means a alpha cube alpha to a minus h plus k times, f of minus alpha would be minus a alpha cube minus alpha to a minus h plus k less than zero, correct? Now this is actually of the form a minus b a plus b, right? So if you take this as a and b, this will be minus a and b. So that would become b square minus a square, b square minus a square. So I can write it as a alpha cube alpha to a minus h the whole square, okay? This is less than zero. So let us try to simplify it. So k square minus, you can take alpha square common, a alpha square to a minus h, the whole square is less than zero. And let's put the value of alpha square as, we know that alpha square is going to be h minus two a by three. So I can write k square minus h minus two a by three a, sorry, by three a by three a, a times h minus two a by three a plus two a minus h, the whole square will be less than zero. So these two terms get canceled. And if I simplify the term within the under root sign, I get something like this. It becomes four a minus two h, four a minus two h by three whole square, less than zero, right? Combining this, combining these two expressions, I get k square minus four times h minus two a whole cube by 27 a less than zero. Multiply throughout with 27 a, multiply with 27 a, remember a is a positive quantity. So you will start getting this expression, which means that 27 a k square can be less than four times h minus two a cube. And I think that is what we wanted to prove. That is what we wanted to prove. Yes, that is what we wanted to prove over here. Okay? So guys, a pretty long problem, but this could be a potential GE advance question for you. Is this fine, guys? Any question? Please type CLR on the chat box if it is clear to you. Okay?