 Lecture 2 TA 101 think and analyze, you know I always try to extract some sense from nonsense. So, you know what this word is you guys are students. So, you would be familiar with this word cheating and I am a teacher. So, I am familiar with this word teaching, now what is common between these two words cheating and teaching. You know if you rearrange the alphabets over here you would essentially get this, you know what this called. So, these two are anagrams of each other, you know. So, if you are on the other side of the table you would probably be indulging in this, while if you are on this side of the table like me I would be indulging in this and my job is to have you guys not do this anyhow. Polygons and conics what would a polygon say to a conic. So, you know this is this I am a movie buff and I was watching this movie Jerry McGuire and you know Jerry McGuire was managing one of the football players one of the upcoming football players. And there like two phrases in that movie which I like very much which I guess you all may also be familiar with. Anyway, so what would a polygon say to a conic show me the bunny in the movie it is like show me the bunny show me the bunny here you go. And what word or how would a conic retard draw me draw you help me help you draw me draw you alright. So, thumb proves I am sorry, but anyhow. So, basic construction and conics. So, this is what I am going to be covering in this lecture. So, this is the second lecture and the first week introduction basic constructions alright. So, we will be focusing on how to construct a hexagon you know it is in the family of it is a member of the family of polygons. And there are these two methods that we will be discussing vertex to vertex distance method and distance between flats alright. So, given to vertices this one and this one and let us say we are given this distance D between these two vertices how do I construct a hexagon. So, this is what I would do I would divide this line into two equal parts you know I will draw an arc with this center and radius as D over 2 and another arc with this center and radius as D over 2. So, that these two arcs they happen to meet at the midpoint of the line and also tangent to each other. And with this midpoint as center I will draw a full circle of radius D over 2. Essentially this circle is going to be traversing through the end points of the edge. So, focus on this intersection point this one this one and this one. So, and these two in addition. So, six points they will essentially form the vertices of the hexagon alright. So, if you look at this circle and the hexagon. So, this circle happens to be the circumscribing circle of the hexagon second method distance between flats. So, this is the first flat over here this is second flat over here you know this distance D is given to us how do we draw a hexagon I will mark the midpoint of this line segment draw a circle of radius D over 2 with this midpoint as center I will draw a horizontal line. And then I will draw a line which is tangent to the circle. And this line is such that it is making an angle of 60 degrees with the horizontal I will draw another line which is parallel to this line and tangent to the circle. And then I will draw a horizontal line tangent to the circle at the top I will extend this horizontal line below again tangent to the circle at the bottom. And then I will draw this line which is at 120 degrees to the horizontal and a line parallel to this line alright. So, this point over here this point this point this point this point and the sixth point here they will essentially form the vertices of a hexagon. So, you can actually join these vertices and eventually construct a hexagon. So, the way this construction is done the circle happens to be the inscribing circle inscribed within the hexagon vertex to vertex distance distance between flats in this case the parent circle or the main circle happens to be the circumscribing circle in this case the parent on the main circle happens to be the inscribing circle to the hexagon alright. Now, this is something very interesting how do you draw a pentagon. So, given a line segment whose end points are at distance d. So, the end points are marked as a and c. So, with the midpoint of a c line segment as center we draw a circle with various d over 2. And then what we do is we bisect this line segment b c we draw a vertical passing through b intersecting the circle at d. So, once we bisect this line segment b c we get a point over here this market e. Now, notice that we are going to be working with distance d e let us say this is of magnitude r. Now, with this distance and with e s center we are going to be drawing an arc which is going to be cutting the line segment a c at f. Now, this length over here let us say this is capital R with this radius capital R and with let us say point d as center will cut the main circle at this arc or at this point. So, this length over here from here to here is capital R now with this as center and the same radius will cut the main circle again at this point and will continue doing. So, until we get 5 points on the circumference 1 2 3 4 and 5 these 5 points will constitute or will be the vertices of the pentagon we are seeking. Now, think and analyze how is it that you think that this is pentagon is it difficult for you to appreciate or perhaps not. Let us try to prove that this is really a pentagon let me grab a piece of chalk and let me focus on this region over here. So, this is my point b here is my point d d b is half of the diameter of the circle. So, it is d over 2 now this distance is again d over 2 and b e is half of b c. So, b e would be d over 4 and I look at this distance d e let us first try to figure what this distance is. So, this is R and we will have R square by Pythagoras theorem as d square over 4 plus d square over 16 this is equal to by multiplying divide by 4 here 5 d square over 16 implying that R is d over 4 times root of 5. Now, let us try to figure what this distance is capital R. Now, if I go left forward from b I am looking at point f here now by construction. So, with this point e a center what I had done was I had cut the diameter a c with the arc of radius R. So, this distance here would also be R and therefore, f b will be R minus b e which is d over 4. So, f b is R minus d over 4 and. So, this capital R here d f can also be determined using Pythagoras theorem. So, capital R square equals d square over 4 plus the square of this R minus d over 4 the whole square. Now, let us retain this thing in the square form because we have been needing that. Now, focus on the triangle well let us call this point m m b t. Now, this distance m d is the same as capital R this distance is d over 2 the radius of the circle and this distance is again d over 2 the radius of the circle. So, let me draw this point here somewhere m. So, this distance here is capital R this is d over 2 and m b would again be d over 2 and let me say now that I am interested in trying to find what this angle theta is. So, we can use the cosine rule and the cosine rule is such that cosine of theta equals d square over 4 plus d square over 4 minus this distance over here R square over 2 times d over 2 d over 2 which is equal to this is d square over 2 minus R square over this is again d square over 2 which is 1 minus 2 R square over d square. I hope I am doing this right I can substitute this value over here. So, cosine of theta cosine of theta let me write this thing down here equals 1 minus 2 times R square is d square over 4 plus small r minus d over 4 the whole square over d square which is equal to 1 minus within parenthesis this would be 1 over 2 minus 2 over d square R minus d over 4 the whole square. Let us substitute this value over here and let us see what we get this is equal to well half plus 2 over d square and this is d over 4 root of 5 minus d over 4 the whole square right and. So, this would be equal to half plus 2. So, this d would come out and get cancer with d square here from here we will be getting 16. So, let me write this thing down over here 16 and then this is root of 5 minus 1 the whole square let us see what this is well let me erase this and take this thing up over there. So, this is half plus 1 over 8 and this is 5 plus 1 6 minus 2 root of 5 seems a lock word let us see what takes me half plus 3 over 4 minus root of 5 by 4 this is equal to if I multiply and divide by 2 I have got 5 over 4 minus of root 5 by 4. So, I just realized that I made a mistake here I make a lot of mistakes. So, instead of the positive sign over here I should have really had a negative sign. So, I will go back and correct my mistakes. So, this is a negative sign here because of which this would be negative because of which I guess this would be negative and this would be positive and. So, this actually would be equal to you know 2 minus 3 over 4 plus 1 root of 5 over 4 which is root of 5 minus 1 over 4 and if you go back and try to figure what cosine of 72 degrees is then you would see that this is nothing, but this value here. So, in a sense through geometry construction what we do is we essentially divide the circle into 5 equal to path in such a way ensuring that this angle here theta is equal to 72 degrees this angle. So, once we ensure this it becomes a little easier for us to construct a pentagon again through geometry. So, this was an algebra this was the algebra just to you know support the construction of pentagon, but you would have realized that we did not actually need this algebra and construction of the pentagon. Anyway, so let us move forward. So, these 5 vertices they will essentially be a part of this pentagon. How would you construct an octagon? So, let us say you have a square of equal dimension sides draw two diagonals indicate these vertices from this a center and this length as the radius draw an arc do the same with this a center and the same radius again the same radius with this a center and finally, with this as radius with this a center and the same radius. So, draw these bunch of arcs and wherever these arcs are going to be intersecting the edges. So, here here here here here here here and here you essentially have 8 vertices join them and you will be getting an octagon. So, pretty simple construction I think an analyze you know through algebra or otherwise if this is really an octagon not. How about a regular septagon or any other regular polygon for that matter given a side you know. So, let us say you have given a side you have been given a side all right. So, you have this point over here this distance is known. So, draw a semicircle and divide this semicircle into. So, since we are focusing on septagon. So, divide this semicircle into let us say 1 2 3 4 5 6 and 7 equal parts and name them in such a way that you let go of this first point on the left. So, start naming them as 2 3 4 5 6 and 7 right and then draw about this happens to be the center of this semicircle over here name that center is a and one of the ends of the semicircle is b. Now, draw radial lines in such a way that they happen to be connecting this point a here and all these points on the semicircle. So, once you are done with that take this as your distance your reference distance you know mark that on your compass with point 7 a center mark an arc on this edge with this point a center with the same radius mark a point on this line segment with this a center same radius mark a point over here and keep going forth until you mark a point over here. And then once you have gotten all these points vertex 1 2 3 4 5 6 and 7. So, these are the 7 vertices that will essentially constitute your septic. So, this is a generic method that allows you to draw any regular polygon construction of an ellipse by means of concentric circles. So, you got this larger circle you have got this smaller circle they happen to be concentric same center divide both circles into equal number of parts. So, I believe I would have divided the outer circle as or into 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 16 equal number of parts and likewise the inner circle also into 16 equal parts. Now, this is what the trick is from one of these points draw horizontal from one of these points corresponding points draw vertical or vice versa let us see. So, I draw horizontal from the inner circle and vertical from the outer circle corresponding points over here horizontal vertical horizontal vertical horizontal vertical horizontal from the inner vertical from the outer horizontal vertical horizontal vertical you know horizontal. And these points they essentially will be lying on an ellipse I could do the other way round for example, instead of drawing the horizontal from the inner circle I would be drawing the horizontal from the outer circle. Well this is the construction of the ellipse for the first case as I said I will be drawing the horizontal from the outer circle and vertical from the inner circle inner circle this time. So, horizontal vertical vertical horizontal vertical vertical horizontal vertical horizontal vertical horizontal vertical horizontal and thus I will essentially be getting these points and they will happen to fly on an ellipse which essentially will be you know oriented 90 degrees. So, these are the points on the ellipse and I will join these points and get another ellipse. Now, the question think and analyze for you is this or this an exact ellipse or not let us try to find it out. So, let us join the corresponding points on the inner and outer circle have this horizontal have this line segment over here let this angle be theta. So, let this point p or rather let this point be p and the coordinates of p would be b cosine of theta and b sin of theta where b is the center of b is the radius of the inner circle likewise point over here q will have coordinates a cosine of theta a sin of theta where a is the radius of the outer circle. And if you focus on this point what would be the x coordinate of this point and what would be the y coordinate of this point you guess you have these values over here. So, the x coordinate of this point essentially will be the x coordinate of q and the y coordinate of this point would be e y coordinate of the coordinate of p. So, essentially if this point is represented as x and y then x will be a cosine of theta while b p sin of theta and of course, if you go back to your 11th grade or 12th grade coordinate geometry you would readily confirm that x y is nothing but a point on the ellipse. And therefore, this ellipse is an exact ellipse yeah alright. So, if I want to draw a tangent or a normal at any point on the ellipse what do I do let us say I have this point here I would like to draw a tangent or a normal at this point p. Now, what I will be doing is I will be using one of the properties of the ellipse first I will try to figure the focal or the two fork high. Now, let me go back let me draw the horizontal and let me also draw the vertical I will set at this point let us say and I would also I would rather already know the length of the major axis of the ellipse I will take half of that distance with this as center I will draw two walks over here and here this point being m I will mark a point over here. So, this distance is half the major axis and I will also mark a point on the other side of the major axis. So, these two circles would represent the two fork high of the ellipse and the property I am using is straight forward. So, if I take any point on the ellipse so the distance of this point from this center and this center if I sum these two distances that sum will always be equal to the length of the major axis. So, two fork high f and g and this is the property I just had mentioned. So, m being a point on the ellipse distance f m plus distance m g has to be equal to 2 a where a is the length of the semi major axis. So, having found the two fork high I will join or I will draw a line segment joining f and p I will draw another line segment joining p and g I will take the angular bisector of the angle f p g you know how to draw the angular bisector. And this line over here essentially will be the normal at point p all I need to do is you know using a compass or using a ruler I have to bisect this angle this angle is 180 degrees I have to bisect it and I will do exactly the same thing you know bisect the angle. Once I do that this line segment here will be the tangent to the ellipse normal to the ellipse n and tangent to the ellipse. Now think and analyze T a y is the n you know this line segment normal to the ellipse construction of ellipse by intersecting arcs again I will use the same property I had discussed before. So, if I have a point m on the ellipse and if I have these two fork high of the ellipse well this is not these are not the fork high, but they happen to be the points at the extreme ends of the semi or of the major axis. So, if the length of the major axis is 2 a if the length of the minor axis is 2 b should be this way I guess never mind. So, the property is any point well let me see if f and g are 2 fork high and if t is any point on the ellipse then f t plus t g has to be equal to 2 a alright. So, what I will do is I will divide 2 into equal number of parts let us say the entire length this length over here which is given to me into equal number of parts. Let me mark the extreme points on the ellipse as p and q there I go well and then what I would do is if I choose let us say this length and draw an arc with center f and this length then I will have to choose this length for the radius of the other arc that I am going to be drawing with g as center that would be my point one of my points t on the ellipse and I will do that you know for as many points as possible once again. So, if I choose you know any other point over here let us say this one if I choose this length with this length as radius and this a center I will draw another arc and with this as the radius and center as g I will draw the second arc the intersection these 2 arcs will give me another point. So, I will keep getting these points and that would essentially help me draw the ellipse straight forward think and analyze what is the maximum radius of the arcs possible from point f or g why you think always a good idea to think let us do a little bit of algebra choose any point t over here let us mark f t as distance d 1 and let us mark g t as distance d 2 let the coordinates of t b x and y and it. So, happens that point t is a result of this property that t 1 plus t 2 has to be equal to the length of the major axis of the ellipse 2 l and essentially d 1 plus d 2 is a constant because 2 l is constant for a given ellipse. Now you know for a simple case if I choose g such that it has coordinates a 0 then f would be such that it would have coordinates minus a 0 with this point as center and if I do a little bit of algebra. So, d 1 essentially would be you know x plus a the whole squared plus y square and d 2 will essentially be x minus a the whole squared plus y square and that is equal to constant and with a little bit of algebra I will get this equation x squared over c squared over 4 plus y squared over c squared minus 4 a squared the entire thing over 4 is equal to 1 and it. So, happens that if you choose you know if you represent this as you know some a 1 if you represent this as some b 1 it is like x squared over a 1 squared plus y squared over b 1 squared is equal to 1 which is the equation of an ellipse. So, apparently this happens to be an exact ellipse and precisely. So, if this constant c is greater than or equal to 2 times a I think yeah alright. So, you can use the same property let me go back you can use the same property have a pin here at one of the 4 pi have another pin here at the second focus and you know you can have a loop of string. So, this distance and this distance making sure that the sum of these two distances is a constant you can draw an ellipse. of our committees another way of constructing an ellipse alright. So, let us say you have two line segments m and n well. So, you have this length equals 2 a you know have two points red and green and have this entire ruler place on the vertical with the same two points red and green and start moving this ruler in such a way that the green is always on the horizontal and the red is always on the vertical. And let me go back see what is happening focus in this point green always on the horizontal red always on the vertical you know this point is actually leaving some traces out over here and these traces are nothing but points which are going to be lying on the ellipse. So, you can do the algebra and figure that this also kind of gives an exact ellipse let me skip this algebra well let me go back and let me try to work out the algebra for you alright. So, for any position of this line you will have this as distance y you will have this as distance x join point r with the vertical over here. So, this triangle and this larger triangle they are essentially similar triangles let me call this y 1 let me call this x 1 and let me have this length as l 1 and this length as some factor f of l 1 because the triangles are similar it should be possible for you to work out the ratio x 1 over x is the same as 1 over 1 plus f y 1 over y is the same as f and x square plus y square is equal to l 1 square do a bit of algebra and you will get x 1 over 1 plus f the whole square plus y 1 over f the whole square equals l 1 square precise ellipse I make a lot of mistakes. So, do not trust me work this algebra out by yourself you will realize that I may have made a mistake construction of ellipse the parallel gram method given a parallel gram how do you construct an ellipse how do you construct an ellipse well divide this parallel gram into a bunch of parts at the top I guess 8 parts here you know like. So, and also well, but before that mark these points 1 2 3 4 5 6 7 8 and also divide this side of the parallel gram into those many equal number parts now watch out for the way I am numbering them and numbering these points from here 0 1 2 3 4 5 6 7 8 and over here is going to be slightly different it is going to be from here to here. So, once out that then I will have 1 reference point over here and the other reference point over there and watch out for the line segments I am constructing well. In fact, I can have the same points on this side watch out for the line segments I am constructing. So, I would imagine that this point in this point essentially will be lying on the ellipse with this as reference I will have my line segment pass through point 1 with this as reference I will have my line segment pass through point 1 this is a point on the ellipse intersection of this line segment and this line segment from here I will have a line segment pass through this point 2 and from here I will have my line segment pass through this point 2 here. So, the intersection of this line segment and this line segment over here will give me another point on the ellipse. So, it is always a good idea for you to you know have some reference associated with the points or the geometric entities that you are using for construction. For example, from here I draw a line segment passing through 3 from here I am going to be having a line segment passing through this point 3 here intersection of these 2 line segments will be a point on the ellipse and I will continue like. So, I can do the same thing and on this side of the diagram once a word just follow the construction and what I can do is I could mirror these sets of points over there mirror this set of points over here and once I have these points essentially all these points they happen to be lying on the ellipse. Now, you try to figure if this is an exact or an approximate ellipse also is it possible for you to identify the conjugate diameters try to figure it out construction of ellipse another method by means of conjugate diameters. So, given to conjugate diameters let me name them as a b and c d the insect at point o a b and c d are conjugate diameters let us say o is the center of the ellipse. Now, with o as the center and a b as the diameter I am going to be drawing this outer circle in red and let me draw o q such that o q is perpendicular to a b. Let me join point d and q let me divide this outer circle into say equal number of parts let me also join point c with the n point over here. Now, you would see that I am using line segments with different colors just to indicate that they are parallel. So, this line segment is parallel with this line segment now watch out what I am going to be doing with these points be careful. Now, I am going to be drawing a line which is parallel to this line here o q and likewise over here again I am depicting these lines using color blue just to emphasize that these lines are parallel. Now, from here I am going to be drawing a line parallel to c d and with one of the points over here I am going to be drawing a line segment parallel to d q and likewise from here as well I am going to be lying rather I am going to be drawing a line segment I am not lying I am going to be drawing a line segment parallel to d q from here. So, of course, this point and this point theoretically will be a part of the ellipse likewise I will draw a line segment parallel to c d such that it passes through the intersection between this blue line and the line a b from here I will draw a line parallel to o d from here I will draw a line parallel to again o d and of course, this red line this red line over here they would be intersecting with the purple line. So, these two points again will be belonging to the ellipse that I am seeking. So, once again point here here intersection between the purples and the reds will essentially give me points on the ellipse intersection between the purples and reds. So, this is how I draw an ellipse using conjugate diameters is this an exact ellipse go figure all right one of the final methods of constructing an ellipse the four center method. So, given a rohambus once again given a rohambus these four lengths are the same just that the angles are not 90 degrees otherwise the problem is square identify the centers of each of the line segments. Now, the first thing you would want to do is identify and draw the longest diagonal. So, of course, this one is not the longest diagonal, but the other one which is the longest diagonal. Now, look at this point go to the opposite vertex join this point with this vertex and likewise look at this point look at the opposite vertex join this point with this vertex. So, you will essentially have this as one of the centers this as one of the centers and two more centers. So, let me call the center as C 1 with C 1 as center and this is radius to an arc of course, this arc is going to be having these two points lying on it with this point as center and this as radius draw another arc that will be having these two midpoints of the respective edges of the rohambus on it the third center or rather the fourth center would be here the third center would be here. So, with this as center this as radius draw the third arc and with C 3 as center and this is radius draw the fourth arc looks like an ellipse, but is this an exact ellipse I will give you a clue. Well, if you think about this this is an actually this is actually an arc think about this this also is an arc rather or likewise these two are also arcs. So, of course, I mean this cannot be an exact ellipse it has to be approximate again. So, to be able to construct an ellipse using the four center method step one identify the longest diagonal, but before that make sure you look at the midpoints of the four sides of the rohambus. Now, there would be a vertex on this side of the longest diagonal and on this side of the longest diagonal. So, one vertex over here one vertex over here identify those vertices and join the corresponding vertices with the corresponding midpoints of the edges of the rohambus. So, from here to here and then from here to here. So, you will essentially have four centers center C 1 center C 2 center C 3 and center C 4 and follow the procedure as I said this is or this has to be an approximate ellipse because all these four geometric entities are arcs they cannot be a part of the ellipse and of course, this as I said the clue you have to identify the longest diagonal first, but this method works only with rohambus it does not work with pylograms for example. So, for example, if this length and this length you know they have to be the same and if they are different from these two lengths then this method will not work. Incidentally this arc has to be tangent to these two line segments this part it has to be tangent to these two line segments and that would not be the case for any generic pylogram construction of a parabola all right. So, given one line segment divide that into equal number of paths let us say 1 2 3 4 5 6 7 another line segment divide them also into equal number of paths just make sure that you follow the numbering. So, for this line segment I am now numbering from this point over here which is the intersection between this line segment and this line segment for the other line segment I am numbering the other way around. So, I am starting from one of the end points of this lines and join 1 2 1 2 2 3 2 3 keep going and try to figure an arc that is tangent to all these line segments once you get that. So, it is this green curve which is the parabola that you are seeking keep thinking and analyzing until next time.