 Hello all welcome to the YouTube live session on conic sections. So in this class we will be continuing with the ellipse which we had left the previous class. So I will start today's concept with the concept of pole and polar in case of an ellipse. So let me first load in a graph for an ellipse over here okay. So what is a pole and a polar concept? So let me let P be any point inside or outside an ellipse okay. It can be even on the ellipse. Let's say there is a point P inside the ellipse and through this point P you are drawing many cords okay. So let's say I draw this cord through this and again I draw another cord like this okay. So I can keep on drawing so many cords and now through the extremities of these cords and now through the extremities of these cords let me draw tangents okay. So let's say these are the two tangents which I draw at the extremities of these cords yeah. So these tangents they will meet on a line which we call as the polar. So as you can see they meet on this line and this line is called the polar of this pole. So this point is called the pole okay. And also remember this pole could be inside the ellipse as well. The pole could be sorry pole could be outside the ellipse as well okay. Now the equation of the pole equation of the polar with respect to a given pole x1 y1. So equation of the polar with respect to an ellipse for pole x1 y1 for pole x1 y1 is again given by t equal to 0 okay. That means if x square by a square plus y square by v square equal to 1 is the ellipse then the equation of the polar would be this. So this would become the equation of the polar. So as I told you earlier t equal to 0 can behave as many things it behaves as a tangent. When the x1 y1 point is on the ellipse it behaves as polar. So I'll write it down over here it behaves as tangent when x1 y1 is on the ellipse okay. It's not that it is not polar in this case it can also be called as a polar in this case. So tangent is the polar of the point of contact. So t equal to 0 can also behave as chord of contact. Now this is also a type of a polar. When the pole is outside and when t equal to 0 purely behaves as a polar purely behaves as a polar when the point is inside when x1 y1 is inside right. So let us take some questions based on pole and polar. First question is find the polar of the focus or you can say find the polar of a focus of an ellipse. Very simple question you should all be able to answer this. So I am asking you if pole is the focus if pole is your focus then what is the polar and please feel free to type in any response in the chat box. How was your practical guys today? Alright so pretty simple. So again let's say this is the equation of the ellipse and we are trying to find out the polar of the focus. So focus is at let's say a comma 0. So t equal to 0. So first of all this is the expression for t and here we will be putting our x1 as a and y as 0. So it becomes x equal to a by e right and this is nothing but the direct tricks. This is nothing but the direct tricks of the ellipse. So just remember this as a theorem that polar of the focus is nothing but the direct tricks. Polar of the focus in case of an ellipse is nothing but the direct tricks of the ellipse. So let's move on to another question. Next question is the perpendicular from the center of an ellipse x square by a square plus y square by b square equal to 1 on the polar of a point is constant. Prove that the locus of the point is x square by a to the power 4 plus y square by b to the power 4 equal to 1 by c square where c is the given constant over here. Where c is the given constant. So this constant over here is your c. If you are done guys you can also send me the snapshot on my whatsapp id. Alright let's assume that the point is h comma k. So the equation of the polar will be t equal to 0 which is nothing but xh by a square plus yk by b square equal to 1. Now the distance of the origin that's the center of the ellipse from this polar is a constant c. So I can clearly write it as first of all I can just rephrase it as this equal to 0 and put 0 0 in place of x and y so it will be mod of minus 1 by under root of h by a square whole square k by b square whole square equal to c. That means 1 by c is under root of a square by a to the power 4 k square by b to the power 4 so which becomes 1 by c square h square by e to the power 4 k square by b to the power 4 and you can always generalize this as x square by e to the power 4 plus y square by b to the power 4 equal to 1 by c square and hence proved. Again be very very careful with locus type questions because they are very very important. Let me take one more cd is a variable chord of the ellipse x square by a square plus y square by b square equal to 1 always subtending 90 degrees at the center always subtending 90 degrees at the center show that show that the locus of the pole of cd show that the locus of the pole of cd with respect to that ellipse is x square by a to the power 4 plus y square by b to the power 4 equal to 1 by a square plus 1 by b square. So basically the question says there is an ellipse and there is a variable chord that subtends a right angle at the center. Okay that subtends a right angle at the center. All right now we have to find out the locus of the pole of cd. Okay so let us say the pole of cd was pole of cd was let's say h comma k point. Okay now the equation of cd now will be equation of cd now will be xh by a square plus yk by b square equal to 1. Okay right now if I have to find out the combined equation of oc and od that means if I have to find out the equation of the pair of lines oc and od right how will I find that out I can find it out by homogenizing I can find it out by homogenizing the equation of cd with the ellipse equation. So how do we homogenize remember the concept when a line and a conic section gets intersected the pair of straight line joining origin to the point of intersection is obtained by homogenizing the equation of the line with the conic section. So when I say homogenizing do not disturb the second degree terms of the conic but this one term this one over here you can actually write it as xh by a square plus yk by b square whole square. Okay because one is this and you can write that one as one square so from this equation you can write x square by a square plus y square by b square equal to xh by a square plus yk by b square whole square. Now you know it's 90 degree that means it represents a pair of straight lines which subtends a 90 degree at the origin that can only happen if the sum of the coefficients let me call this as one sum of the coefficients of x square and y square in one add to zero. Okay so what are the coefficients of x square coefficient of x square will be this what are the coefficients of y square the coefficient of y square will be this and if this adds to zero that would be the condition for the two lines represented by OC and OD to be at 90 degrees to each other. So if you write it in a resolve form you will find that it actually comes out as the required locus so if you generalize this so if you generalize this it becomes x square by a to the power 4 plus y square by b to the power 4 equal to 1 by a square plus 1 by b square and I think this is what we are supposed to prove as you can see yes this is what we are supposed to prove over here right is this clear guys any questions please feel free to ask so the next concept that we are going to talk about is the concept of diameter. Now first of all how is the diameter defined diameter is basically focus of the midpoints focus of the midpoints of system of parallel cords focus of the midpoints of a system of parallel cords is called the diameter okay so basically let's say if this is your ellipse and you draw a system of parallel cords like this okay and start connecting their midpoints and start connecting their midpoints then this line that you see will be called the white line that you see will be called as the diameter so this line will be called as the diameter now let me ask this as a direct question to you find the equation of the diameter that by six a system of parallel cords with a slope of m so let's say we have our standard case of an ellipse and all these cords they have the slope of mm each okay then find the equation of the diameter then find the equation of the diameter so that's the question again feel free to type in your response in the chat box or send me a pic on the whatsapp okay it's not important to know the concept of diameter per say but how do you actually solve it as a locus question is what is important again i'm repeating the concept of locus is what i'm more interested in testing probably it will never ask you a question on diameter but this is like a building block for you to understand locus very well so the answer should be in terms of of course the variables x and y and a b and m only we don't have to assume that that will actually be evident from the equation only all right so let's see this case again let's say h comma k whose locus we are finding okay so h comma k is the midpoint of a chord so what is the equation of that chord if h comma k is the midpoint of a chord what is the equation of that chord simple question the equation of that chord is t equal to s1 right so t equal to s1 correct so this is the equation of the chord now this chord is having a slope of m this chord is having a slope of m right so let us write down the expression for the slope of this slope of this will be minus h by a square divided by k by b square so this will be your slope right and that's it that's what we wanted we wanted a relationship between h and k right so now we generalize this now we generalize this and we write m as minus b square x by a square y in other words y is equal to minus b square x by a square y sorry a square m this becomes the equation of the diameter and as you can see atmesh it passes through origin so yes it passes through the center of the ellipse it passes through the center of the ellipse okay plain and simple