 If you have understood this, then I can follow it up with another question, which is of very very similar nature. Let's see whether you are able to crack that question is show that the locus of points the locus of points such that of the three normals drawn from them to the parabola y square is equal to 4 a x coincide is 27 a y square is equal to 4 times x minus 2 a cube. Is the question clear? So you have to find the locus of points, let's say that point h comma k such that two of the three normals drawn from them coincide that means two of the roots are two of the roots from M1, M2, M3 are the same very very similar to the approach that we had taken you guys this time it is something like this it touches and goes back. So let's say this is your M1 and here M2 and M3 would be the same okay it could be this way as well I mean I am not denying that it could not be this way it could be this way as well it could come like this touch and go back that means this is your the two roots M1 equal to M2 and this is your M3. If done let me know so guys here the things will not change we are looking at the same scenario right so h comma k will satisfy the equation of the normal like this right okay and again let's say I call this term as f of M this is your f of M f dash M again if you put it to zero you get two roots alpha and beta and you know both are negative of each other okay. So in this case what will happen this is your alpha and this is your beta or it can happen this is your alpha and this is your beta in both the situations you would realize that f of alpha into f of beta would be equal to zero isn't it because either the value of the function at beta is zero or the value of the function at alpha is zero correct. So if you remember the previous slide in the previous slide where we talked about less than zero so if you recall here where we talked about less than zero now we are talking equal to zero and if you're talking equal to zero it's but obvious that will end up the same expression just with a equality sign so we'll end up 27 a k square equal to four times h minus 2 a cube correct isn't it yes or no yes or no guys and then we generalize this it becomes 27 a y square is equal to four times x minus 2 a cube is it clear any question so far so please ask me if there's any question so far atmash Rohan Lalita Snigda okay next we are going to talk about circle through co-normal points circle through co-normal points so let me ask this as a question to you if three normals are drawn from an external point alpha comma beta to the parabola y square is equal to 4x then prove that the equation of the circle passing through passing through the co-normal points passing through the co-normal points is given by x square plus y square minus 2 a plus alpha times x minus beta by 2 y equal to 0 excellent question so what do we need to prove over here is that so I'm an external point alpha comma beta this is your alpha comma beta when you're drawing normals to this parabola y square is equal to 4x which meets the normal at a b and c then if you pass a circle through it one thing that you would have observed that this circle is passing through origin this circle passes through origin so that's why I have drawn a diagram in such a way that I'm made to pass it through the origin also so prove that the equation of the circle is this first prove that it will pass through origin that is most important step before generalizing once more okay see Rohan in the previous problem in the previous slide we had talked about f alpha f beta less than 0 correct so that there are three distinct real values of m correct now in this case what is happening alpha and beta are what alpha and beta are the points of maxima and minima of the function because they are the roots of because they are the roots of the derivative of the function equated to 0 right correct so alpha and beta should be such that in this case it would be such that the value of the function at beta would be 0 you can call beta to be a repeated root okay so you can see in this situation number one and in both situation number two the value of the function at beta is becoming 0 f of beta is becoming 0 here here f of alpha is becoming 0 because to have two real and equal roots it must touch the x-axis it must touch the x-axis so that is what is happening over here so when it is touching it is creating repeated roots over there right so whatever is the root of f dash m the same would be the root of f of m okay so f of alpha into f of beta one of them would be 0 I don't know which one of them but one of them would be 0 that's why their product would be 0 so I just use the previous result so instead of inequality sign I put a equal to sign and therefore I got the locus is that clear Rohan so here guys let me help you with this let's say a is basically corresponding to that slope which is m1 or let's say this point a a m square comma minus 2 a m1 remember t is related to m as this relation so t and m are negative of each other okay so let's say point b is a m2 square minus 2 a m2 point c is a m3 square minus 2 a m3 okay and let's say point b which I don't know what it is right now I'm just calling it as a m4 square minus 2 a m4 okay so what I'm saying here is that the circle is cutting the parabola not only at those points where it is where the co-normal points are getting formed but also at some fourth point now how did I know that first first of all how do I know that is cutting at the fourth point it's very obvious it's very obvious that if you generalize if you write the equation of any circle we usually write it like this correct and if you try to solve the parametric equation of this parabola if you try to simultaneously solve these two okay and you replace your x with a m square then this is what we get right and this is clearly a fourth degree expression this is clearly a fourth degree expression in m so this is a fourth degree or you can say a bi quadratic a bi quadratic equation in m bi quadratic expression in m that means four roots are possible for this that means a circle will cut the parabola definitely at four points and let's say the roots of this equation are m1 m2 m3 m4 okay these are basically your co-normal points so these first three that is m1 m2 m3 they are nothing but they are associated with the slope of the normal at the co-normal points now what is m4 that is what we have to find out okay now we have already seen in our previous concept that the sum of the algebraic sum of the normals drawn at these co-normal points will always be 0 is it so from an external point if I draw three normals to the parabola the sum of the slope of these three normals would always be 0 this we have seen as a property correct so here also if you see the sum of the roots that is m1 m2 m3 plus m4 what will be that that will be minus of b by a right and b is the coefficient of m cube so if you see there is no m cube term that means 0 m cube term is there so it will become 0 by a square which is again 0 correct which implies this term was already 0 this term was already 0 and adding m4 will also giving you 0 that means m4 is also 0 correct if m4 is 0 then you can see that this point here would be a into 0 square minus 2 a into 0 that means my circle will definitely pass through the origin that means it is cutting the parabola not only at the co-normal points a b c but also at a fourth point let's say d this d which is your which happens to be your origin so far clear guys no problem in understanding this correct now if your circle is passing through origin that means if the circle is passing through origin that means c term in the circle would be 0 that means your circle would actually be of the form x square plus y square plus 2 gx plus 2 fy equal to 0 the c term would be 0 in the circle correct so if I remove this c term from here this c term if it goes off then you can see that you will be getting something like this let me write it in some other color let's say blue then you get something like a square m to the power 4 plus 4 a square 2 a g m square minus 4 f m equal to 0 okay take an m common and you get a m cube plus 4 a g m minus 4 a f equal to 0 now we clearly note that this m equal to 0 that is coming from here is corresponding to the origin it is corresponding to this fellow correct that means this equation is the one which is corresponding to m1 m2 m3 okay if you want you can drop a factor of m also from here sorry drop a factor of a also from here so that means a m cube 4 a plus 2 g m minus 4 f equal to 0 this will be responsible for giving you the roots m1 m2 m3 m1 m2 m3 okay this was responsible for the root m4 are you getting this point slightly heavy concept but if you understand this things will be fine for you okay now what I can claim here is that the original equation whose roots was m1 m2 m3 which was actually a m cube plus 2 a minus alpha m plus beta equal to 0 this also has roots m1 m2 m3 correct that means these two equations are identical these two equations are identical correct and if they are identical I can compare the coefficients of m cube m and constants on both the sides okay so if you compare the coefficient of m on both the sides we can say we can say 4 a plus 2 g is equal to 2 a minus alpha that means 2 g is equal to minus 2 a minus alpha okay so g we obtained similarly minus 4 f is equal to beta so 2 f is equal to minus beta by 2 so 2 f also be obtained so the required equation of the circle which is x square plus y square plus 2 g x so plus 2 g x plus 2 f y plus 2 f y plus c c is already 0 it would be this correct and if you simplify this it becomes the required equation that we need this is going to be our answer okay so I think it matches with this equation yeah it matches with this equation so all of you can see over here I just repeat the process once again so that everybody understand this clearly this is quite a difficult concept actually so first of all I had to find out the equation of a circle which passes through the co-normal points correct so what I did is I took the equation of any general circle I took the equation of any general circle like this okay and try to solve it with the equation of the parabola okay and I got a bi-quadratic equation and I got a bi-quadratic equation okay now bi-quadratic equation will have four roots m1 m2 m3 m4 correct out of that three roots are the ones which are belonging to the co-normal points so that will be 0 that means the fourth root that is m4 that will also be 0 right once I got the fourth root as 0 that means the fourth point through which the circle passes is definitely origin correct that makes my C also 0 that makes my C also 0 that means the circle will be of this nature that means my circle will be of this nature correct and if C is 0 the new equation becomes this if I take m common it becomes this so either m could be 0 or this expression could be 0 correct m equal to 0 corresponds to m4 so forget about it the other cubic equation which is left that corresponds to m1 m2 m3 as the roots so I am talking about this fellow correct so this should be same as the original equation of the of the conic section or of the parabola from where m1 m2 m3 roots where whose roots where m1 m2 m3 so these two should be identical so the identical I just compared their coefficients here I got the value of 2g I got the value of 2f and I then substituted it in the equation and I got my