 So I am going to begin with the basics of hyperbola in case you have forgotten it, just a quick recap will do, okay. So just the basic concepts will recap of hyperbola, again hyperbola is basically defined as the locus of a point which moves in a plane in such a way that its distance from a fixed point which is the focus, okay. So I will just quickly write it down. So it is the locus of a point which moves in a plane, which moves in a plane such that its distance from a fixed point which we normally call as the focus is E times its distance from a fixed line that is called the directrix. This E value is greater than 1, right. So just to quickly sketch hyperbola, in fact I will try sketching it on the GeoGebra tool so that you have an exact idea of the shape of a hyperbola. So I will just sketch x square by let's say 36 minus y square by 25 equal to 1, okay. So earlier I have taken a bigger value, let's say I take 9 over here, 9 and this is 4, alright. So this is basically the diagram of a hyperbola. So focus of this conic C, as you can see it is at point A and B and directrix of this conic C as you can see is this line F and G on your screen, right. So what I'll do now, I'll just take a snapshot of this so that I don't have to sit and draw this every time, right there, yeah. So let's go back, so I don't know this line anymore, so I'll just dump it, yeah, yeah. So this is your hyperbola. Now the equation of this hyperbola, I will not derive it, we have already learned in our class 11, so x square by y square, x square by a square minus y square by b square equal to 1 is a standard form of a hyperbola, where the 4 psi, where these 4 psi, okay, so let me name the 4 psi as A and B, they are called the 4 psi. The 4 psi are taken as a comma 0 and minus a comma 0, okay, so these are the 4 psi of the ellipse, sorry, hyperbola, okay, I'll actually write them down so that you can just read it from the graph, this is a e comma 0 minus a e comma 0, okay, and these 2 are called the vertices, okay, the vertices is at a comma 0 and this is at minus a comma 0 and this is x equal to a by e and this is x equal to minus a by e line, alright, now quick things about the hyperbola, so normally what you saw here was a hyperbola like this, sorry, a hyperbola like this, you could also have a hyperbola where the equation actually becomes this with a minus 1 over here and this is called the conjugate of the hyperbola, okay, so this is called the conjugate of the previous hyperbola, so just a comparison chart between them, so this pink border, graphically speaking these 2 graphs would be like this and this would be like this, so the orientation is different, here your x axis is basically containing the transverse axis, okay, so this is your transverse axis and this is called your conjugate axis and here this is your conjugate axis and your y axis becomes the transverse axis, y axis becomes the transverse axis, okay, in both these situations this will be the center, okay, so I will just keep writing them down, so center for both of them would be 0, 0, that is something which you all know, FOSI, before FOSI we will talk about vertices, vertices would be a, 0, minus a, 0, here it would be 0, b and 0, minus b, eccentricity, eccentricity, 1 plus b square by a square under root and in this case it will be 1 plus a square by b square under root, in both of these cases it's actually the length of, eccentricity is given by under root of 1 plus the semi conjugate by semi transverse, so this is a universal formula for eccentricity, so please keep this in mind, yeah, so FOSI, FOSI is at a e comma 0 and minus a e comma 0, in this case it is 0 comma b and 0 comma minus b, the equation of the directives is x equal to a by e and x equal to minus a by e, here it is y is equal to b by e and y is equal to minus b by e, length of transverse axis, length of transverse axis is 2a units, here it is 2b units and length of conjugate axis here it is 2b units and here it is 2a units, okay, length of the lattice rectum, length of the lattice rectum again 2b square by a units and in this case it will be 2a square by b units, so just a quick recap, I mean I knew you already knew this but just in case it has slipped out of your mind, okay, now we will talk a bit about the formation of a equation of a hyperbola given you know its directrix and focus and eccentricity, so we will start with questions on the basic definition of a hyperbola, so let us begin with this question, find the equation of the hyperbola, find the equation of the hyperbola, of the hyperbola, whose directrix, whose directrix is 2x plus y equal to 1, whose focus is 1, 2 and whose eccentricity is root 3, whose eccentricity is root 3, if you are done please type in your response in the chat box or you can message me on my whatsapp, again we will use the basic definition, the distance from the focus should be e times the distance from the directrix, so let us say the point is x comma y, directly we can assume x comma y instead of basic time taking h and k, so the distance from 1 comma 2 is e times the distance from the directrix, okay, so let us take the root 5 on the other side and square both the sides, so we will get 5 times x minus 1 square y minus 2 square plus 3 times 2x plus y minus 1 the whole square and if you open the brackets you will be getting 5x square plus y square minus 2x minus 4y plus 5 is equal to 3 times 4x square plus y square plus 1 plus 4xy minus 2y minus 4x okay and simplifying it will give you 7x square minus 2y square plus 12xy and then we will have plus 14y and we will have a minus of 12 and 10 which is minus 2x and 25 minus 3 is minus 22 is equal to 3, so this is the desired equation of the hyperbola, is that fine guys, now we will quickly take up some question on the general form of a hyperbola, when I say general form of a hyperbola means its origin is no longer at sorry its center is no longer at origin so it is of this nature, so these cases are called this or this, so these cases are called the general form where the center of the hyperbola is not at the origin okay, so what I will do is I will start up with a question okay directly asking you some of the critical points okay, so in this case the center is at plus minus alpha comma plus minus beta okay, however the transverse axis and the conjugate axis are still parallel to the x and the y axis is respectively, so let's take a question on this find number one center number two vertices number three eccentricity number four let's say four psi number five let's say equation of the direct races, so find these parameters for 7y square minus 9x square plus 54x minus 28y minus 116 equal to 0, so guys please start solving it and this time I want you to give your responses because I can see no response coming from any one of you, all of you are like mute spectators alright guys so finding the center there are a lot of methods, one of the method is the partial derivative method, so let's say I call this term as s okay, you can find out the center of any aconic section by solving these two equations simultaneously, solving these two equations simultaneously dou s by dou x equal to 0 and dou s by dou y equal to 0 if you solve these two simultaneously you may get the you will get the equate get the coordinates of the center okay, so I'll try this method as of now, so partial derivative of this with respect to x if you do we get minus 18x plus 54 equal to 0 that means x is equal to 3, if you do a partial derivative with y you get 14y minus 28 equal to 0 which means y is equal to 2, so you can say 3 comma 2 is the center of this particular conic right, so the use of this partial derivative can be to find out the center only not for any other thing, so if I try to solve it by our regular method the method that we have learned in our class 11th right, so we need to complete the squares, so first we need to group up terms having y's, so let's say I group up these two terms then I group up terms having x okay and now we are trying to complete the square over here, so it becomes 7y square minus 4y x square minus 6x minus 116 equal to 0, so let's complete the perfect square this will be y minus 2 the whole square this will be 9x minus 3 the whole square and you have to compensate by adding 28 and minus 81 on the other side, so you get 7y minus 2 the whole square minus 9x minus 3 the whole square is equal to 63 that gives you y minus this by 9 minus x minus 3 square by 7 equal to 1, in other words you can write down the equation of this hyperbola as x minus 3 square by 7 minus y minus 2 square by 9 equal to minus of 1, now if you want to get the center please note that the center of the hyperbola let's try to do a role change over here, let's try to do a role change over here let's compare this with let's compare this equation with let's compare this with x square by a square minus y square by b square equal to minus 1, so it's obvious that x is now the role of x capital x is now being paid by small x minus 3 role of capital y is now being paid by small y minus 2 your a square is 7 and your b square is 9 that means a is root 7 and b is 3 okay, now we are trying to find this center for center we know x is 0 y is 0 that means x minus 3 is 0 y minus 2 is 0 which means x is 3 y is 2 so 3 comma 2 is your center the next question was vertices the next question was the vertices, so vertices in this case is 0 comma b and 0 comma minus b but instead of writing it like 0 comma b and 0 comma minus b I would suggest write it like x is 0 y is b that means x minus 3 is 0 y minus 2 is b b is going to be 3 right so 3 comma 5 is one of the vertices the other vertex the other vertex is at x equal to 0 and y equal to minus b that means again x will be 3 and y minus 2 is minus 3 which means y is minus 1 so 3 comma minus 1 is the other vertex next is the focus sorry the eccentricity eccentricity in this case will be under root of 1 plus a square by b square so which is 1 plus 7 by 9 that's going to be 4 by 3 so eccentricity is going to be 4 by 3 right next is focus we know that for focus of 4 side in this case x is equal to 0 and y is equal to plus minus b e so I'll take these two cases x equal to 0 y equal to b so x minus 3 is 0 and y minus let me write plus minus directly y minus 2 is equal to plus minus b e so that gives you 3 as x and y could be y minus y could be a 2 plus minus 4 so y could be either 6 or y could be minus 2 so 3 comma 6 and 3 comma minus 2 oh sorry 3 comma 6 and uh oh yeah that's fine 3 comma 6 and 3 comma minus 2 okay now direct to cis so the equation of the direct to cis would be y is equal to plus minus b by e okay so y in this case is y minus 2 is equal to plus minus b by e so that will be this so y is equal to 2 plus minus 9 by 4 so your y could be 17 by 4 or your y could be minus of 1 by 4 okay so this basically helps you to understand the initial concepts that you had learned in your class 11