 Next question is prove that the co-normal points, let me write it like this, the co-normal points with respect to the ellipse x square by a square plus y square by b square equal to drawn from the point h comma k satisfy satisfy a square minus b square x y plus b square k x minus a square h y equal to 0 equation which is also called the aplonion is also called the aplonion rectangular hyperbola which is also called the aplonion rectangular hyperbola this is just for your x sign information nobody will ask you what is an aplonion rectangular hyperbola so prove that the co-normal points with respect to this ellipse drawn from h comma k satisfies this equation please write very very simple alright guys so it's very simple should not take you more time see we all know that a normal drawn to x1 y1 a normal drawn to x1 y1 equation is this right so if i'm drawing normal at x1 y1 its equation will be given by this expression correct and this is going to satisfy h comma k so h comma k is going to satisfy this given equation right i can also say that the normal drawn from x2 y2 just keep noting it what i'm going to write x2 y2 normal drawn from x2 y2 will be a square x by x2 minus b square y by y2 is equal to a square minus b square correct and just like this was h comma k satisfied this so h comma k will also satisfy this so guys in a similar way in a similar way similarly i can say that a square h by x3 minus b square k by y3 equal to this equation will also be true and a square by x4 minus b square k by y4 equal to a square minus b square will also be true do you see that if i have to represent a curve on which x1 y1 x2 y2 x3 y3 and x4 y4 lies all i need to do is generalize this all i need to generalize this means wherever you have your x1 y1 x2 y2 x3 y3 x4 y4 written i just have to replace it with x and y i just have to replace it with x and y now get the feel of this that is what j wants whether you know the art of generalizing whether you know the art of generalizing it so if you simplify this if you simplify this dear students you will get a square h y minus b square kx is equal to a square minus b square x y which you can actually write it as a square minus b square x y plus b square kx minus a square h y equal to 0 and this is called the aplonion aplonion aplonion aplonion rectangular hyperbola okay is that clear or not please type God level clear if it is clear so this is one of the most difficult concept that you may come across you may come across i'm saying you it's not necessary that will always come so now we'll talk about the concept of pair of tangents just like the same way we did for a circle please remember pair of tangents equation is t square is ss one okay and if i talk about the standard form of the ellipse you know your t is going to be xx one by a square y y one by b square minus one s is going to be s is going to be this expression and s one is going to be this expression okay moving on call of contact call of contact let me just up draw the diagram over here so that all of you can make sense so these are the pair of tangents let's say you're drawing okay so these are called the pair of tangents and this line connecting the point of contact this is called the chord of contact this is called the chord of contact again same old stuff nothing new here the chord of contact equation is again t equal to zero again t equal to zero let's take a question on this before we move on to some other concept question is the chord of contact the chord of contact of tangents of tangents drawn from the point h comma k to your standard form of the ellipse will will subtend a right angle will subtend a right angle at the center at the center if h square by a to the power four plus k square by b to the power four is equal to one by a square plus one by b square okay and hence also find the locus of h comma k so whosoever is able to do it will get a samosa from me next class paka i don't i don't promise pizza and all because i can't afford it so samosa is well within my budget guys anybody any progress so guys when you draw the diagram you'll come to know about it so the answer is hidden in the diagram itself so there is a chord of contact let's say this is your chord of contact okay and this chord of contact subtends a right angle at the center so this is your right angle okay so does it remind you of the concept of does it remind you of the concept of homogenization homogenization so homogenization is basically a concept which says that the equation of the pair of straight lines from the origin to the point which has been formed by intersection of a line with a conic is obtained by homogenizing this line and this conic equation what are the meaning of homogenization means making the degree of the entire equation uniformly as two so chord of contact what is the equation of the chord of contact we know chord of contact equation in this case is going to be this correct yes or i have to homogenize this equation with the equation of the ellipse so i have to homogenize these two equations homogenize means by using these two equations i have to uniformly create a equation of degree two so what i'm going to do is in place of one over here i'm going to write the square of this term so please see carefully what i'm going to do i'm going to write one as one square and this one i'm going to replace it with x h by a square plus y k by b square whole square now you realize that it's a homogeneous it's a homogeneous second degree equation it's a homogeneous second degree equation okay which actually represents the pair of straight lines op and oq right it represents the pair of straight lines op and oq together so this concept i have repeated so many times in past also and i'm yet repeating it now guys okay it's fine it represents a second degree equation and it represents the pair of straight lines op and oq but how does this help us to solve this present question okay now we all know that in any pair of straight lines please know this concept very very well in any homogeneous equation like this which represents a pair of straight lines if the lines are perpendicular if the lines are perpendicular a plus b has to be zero that means the sum of the coefficient of x square and y square has to be zero so what i'm going to do next is i'm going to extract the coefficient of x square and y square from this huge equation and i'm going to add them and i'm going to make it zero right so i'll be you know hand picking these terms let's say coefficient of x square will be 1 by a square from here and minus h square by a to the power 4 from here plus coefficient of y square is 1 by b square from here and minus k square by b to the power 4 from here this should be equal to zero which clearly brings me in one shot to my required result that is h square by a to the power 4 plus k square by b to the power 4 is 1 by a square plus b square right hence proved and when you talk about locus when you talk about locus locus of h comma k okay you can just replace your h generalize it replace your h with x and k with y that gives you the locus of h comma k as well that gives you the locus of h comma k as well so the locus of all those points from where the cord of contact drawn to this ellipse will subtend the right angle at the center is given by this equation okay do you realize that it itself is an ellipse with the same center as the center of the given ellipse is that clear guys please type clear on the chat box if it is clear so we'll move on to the next concept that is the cord bisected at a given point the next concept is the cord bisected at a given point so guys again you'll you'll start realizing that concepts are quite repetitive the whatever you have learned in circles is getting repeated of course with some minor changes in terms of their representation but overall the structure is the same so if you have a cord like this okay whose midpoint is x1 y1 whose midpoint is x1 y1 then we say the equation of the cord equation of the cord whose midpoint is x1 y1 is given by t equal to s1 that is xx1 by a square yy1 by b square is equal to x1 square a square plus y1 square by b square okay