 Today's agenda would be hyperbola okay so we are into our last conic which is hyperbola and this is only with respect to your school level syllabus. We will be doing hyperbola in much more detail once your school semester exam is over I have been repeating this since the beginning of the conic. So I'm repeating this yet again that whatever we are going to do in today's class is only meant for your class 11th CBSC or ISC curriculum. It is not going to help you or it is not going to be the complete you can say J main J advance or competitive exam content that content will be definitely taken up once your school semester exam is completed. Okay. So let's get started with hyperbola. So let us look into the locus definition of a hyperbola. I have already explained to you how a hyperbola is obtained from a right circular double cone. If not, I will explain this yet again to you. So let us say this is a this is a right circular double cone. Okay. It's called a right circular double cone because as you can see they are double cones seen in this particular diagram. And it is called right circular because the axis of this cone or axis of this right circular double cone is perpendicular to the base is perpendicular to the base. Now, if this right circular double cone is cut by a plane. Okay, if this is cut by a plane, let me draw a plane here. Okay. Or let me just make it slightly more inclined so that you have an idea about, yeah, let's say I make it like this. Okay. Nevertheless, the cone is not going to the plane is not going to pass through the vertex. Okay. So the cross section area that would be left. I'm just showing it with the blue color. Yeah. This cross section area that you can see on your screen right now will be that of a hyperbola. Okay. Now this angle is what we call as the semi vertical angle. Let me name it in green color. Okay. And this angle is what we call as the, as the angle beta. Okay. So the plane, the plane makes an angle of beta with the axis. So if your beta is between, beta is between zero degree, including zero degree to alpha, then the cross section area obtained would be that of a hyperbola. Okay. So I hope everybody is clear why, why these corner sections are named in that way. So if this plane cuts perpendicular to the axis, we end up getting a circle. If the plane cuts the axis in such a way that the angle it makes that is your beta angle is lesser than 90 degree, but more than alpha, it will be an ellipse. If the angle which the plane makes with the axis is equal to alpha, it's a parabola. If it is lesser than alpha, but greater than, of course, greater than equal to zero, it will be a hyperbola. Okay. So this is our last of the corner circles. Parabola ellipse we have already done in the previous classes. So in this chapter, you realize that you're doing more or less the similar formula, similar points as what you did in case of ellipse. Okay. So there's a lot of, there's a stark resemblance between hyperbola concept and ellipse concept. Okay. So first let us look into the locus definition. What is the locus definition of a hyperbola. So hyperbola is locus of a point or a path traced by a point, traced by a point moving in a plane, moving in a plane in such a way, in such a way that the ratio of its distance, the ratio of its distance. From a fixed point to that from a fixed line, that from a fixed line is a constant where the value of this constant is greater than one, where the value of this, where the value of this constant, constant is greater than one. Okay. By the way, we had more or less same definition for ellipse as well, where I told you that the ratio, okay, the ratio which is a constant is between zero to one. So in case of an ellipse, it was between zero to one. In case of a parabola, it was equal to one. In case of a circle, it will be equal to zero that I've already discussed with you. So, ascenticity for a circle is zero. Ascenticity for a parabola, sorry, ellipse is between zero and one. Ascenticity for a parabola is equal to one and ascent, ascenticity for a hyperbola is greater than one. So these points that we are talking about and this line we're talking about is known as your focus. So from a fixed point, sorry, point I'm forward right. So this fixed point is what we call as the focus. This fixed line is what we call as the directrix. And this constant is what we call as the eccentricity. Okay. So your eccentricity is greater than one for a hyperbola. Is it fine? Any questions? Any questions? So let me draw a figure for the same. Let's say this is our directrix and this is our fixed point S. Okay. And let's say there is a moving point P, which moves in such a way. Let's say this is your P point, which moves in such a way that SP distance is greater than PM. Okay. So SP is more than PM. That means SP by PM is greater than one. Okay. So the path trace by such a point would look like a structure, which is going to appear like this. Okay. I'm just drawing one of the arms here. By the way, with the same eccentricity, you can have the same point showing you the same eccentricity value with another set of point and line. In other words, in other words, the very same point P, the very same point P, the distance from this point, let me call this as S dash. This divided by PM dash, that would be the same E value. That means S dash P by PM dash will be the same eccentric value. In other words, you have a pair of focus and directories for hyperbola as well. Okay. And you can well imagine that there would be one more branch of the hyperbola which will look like this. Okay. I'm just drawing the two branches over here. Okay. I'll show you this structure on the Jujibra tool as well. Okay. So there are two directories, you can say D and D dash. There are two four side corresponding to these directories. Please note, just like in case of an ellipse, this focus and this directories are associated with each other. That means if you're referring to the distance of the point from the focus S, you must take the ratio of assistance from the same point from the directories D. And if you're taking the ratio of assistance from the focus S dash, you must take the ratio with the distance PM dash. You cannot swap. You cannot swap the two four side and the directories. Please note this down. Okay. So I noted this down, a local definition, why hyperbola is called a conic and how the directories and focus situations look like in case of a hyperbola. Let me show this diagram on Jujibra. Let me show this on Jujibra. Let's say I have, I just make a hyperbola for you right now. However, most of you who have not done hyperbola would not know the reason why I chose this equation, but I will formally derive it in a few minutes time. Don't worry about it. Okay, so let's say this is our hyperbola. As you can see on your screen, there's a hyperbola. Now Jujibra has a feature that it can show you the directories. So let me show you the directories of this. So as you can see it has shown you two lines, F and G. Okay, in fact, it has also given you their equations as well. And it can also show you the focus. It can also show you the focus of the conic. So let me show you that. Okay. Now what I'm going to take here is I'm going to take a generic point. Let's say any point on this hyperbola. Let's say I take it here. So what's the distance of this point from B? Let's check. H is 5.48 as you can see on your screen over here. And let's also drop a perpendicular from this point on to this line. Okay. And let's find out CD length, C to D. C to D they're calling as a J. Okay. Now let's take the ratio of H by J. So H divided by J. Oh, I hope I'm writing the right things down. Yeah, H divided by J. As you can see on your screen, it shows it to be 1.2. Okay. Now all of you please pay attention. First of all, I would like to show you that if you take this length also CA. Okay, as you can see, they have called K and CE. They have called it as L. And if you do K divided by L, see what you get again 1.2. So H by J is also 1.2. K by L is also 1.2. Okay. Now what I'm going to do is I'm going to take this point or a trip. Okay. So I'm just going to move it on the hyperbola. When I'm moving it, just observe A and B values. Just observe A and B values. They still remain 1.2. They still remain 1.2. There is no difference in the value of the eccentricity. Even when you're referring to any one of the pair of focus and direct cases. You see that. Any questions, any concerns. Any questions. So even if I bring it on this, it doesn't make any difference. Okay. All right. So now, having seen the locus definition, we will now derive the equations of standard hyperbola. Just like in case of an ellipse, just like in the case of a circle, just like in the case of a parabola, we had standard cases. Let me what was the standard case for circle. Anybody. What used to be the standard form of a circle, circle whose center is at origin. Correct. What used to be the standard cases for the parabola parabola whose vertices used to be at origin, and the axis of the parabola used to be along one of the coordinate axes. Right. What used to be the standard case for an ellipse. When your major and minor axis is where your x and the y axis, in other words, the center of the ellipse in the in that case also used used to be at the origin. Yes, sir. So even for a hyperbola, we will have two standard cases. So let us take those standard cases so standard hyperbole. Okay. Or standard hyperbolas also you can call it both are the same thing. So standard hyperbola are those cases where let me write it down the center of the hyperbola is at origin. Center of the hyperbola is at origin. Okay. And there is something called transfers. Oh, I have not told you about the signal. Let me just write it down first transfers and conjugate access are along are along the coordinate axes. Okay, so first let me give you the nomenclature. The reason why I skip nomenclature probably because it follows the same nomenclature as that offer. As that offer ellipse so let me just do one thing. Let me just take the diagram from here itself. Let me take the diagram from here itself. Better to draw it again because don't save. So I'll be just picking up my diagram from here itself so that we can use it. Yeah. So first of all, we will let you know the terms that we are using. Okay, so first get yourself familiarized with the term. Now forget that there is an XY axis over here just think as if there are two arms. Okay. And let us say these are the two foresight. Okay, so these are the two by the way, I could have as well picked up a diagram. With a foresight. One second. Give me one more second. I just. I'll just put the foresight and direct this is. So sorry and that is. Yeah. Okay, so the terms that we are going to use is more or less the same as what we use in case of a ellipse. So these are the two foresight. So focus number one. This is focused number two. Okay, so these are the two direct this is that you already know. Okay, so let me just write it down on top of it. You can say focus one, focus two. Focus one, focus two. And this also can call it as direct tricks number one, direct tricks number two. Okay. Now. The first thing that you should all know is that. The line joining the two foresight. If I join the two foresight. This line that you see here, this line is called. The transverse axis. Okay, this line. Is called as the transverse axis. Okay. I already use the term over here. And then I realize that I have not given you the nomenclature. So nomenclature or the terms that we normally use along with hyperbola. So this is your transverse access. Transverse access is the line connecting the two foresight. Okay. The midpoint of the two foresight is called the center of the. This is called the center of the hyperbola. Okay. The midpoint. So center is the midpoint of the two foresight. Okay. Please note that. The point where the. Transverse access cuts the hyperbola. They are called. Okay. So these two will be called as the vertices. Okay. Center is also the. Midpoint of the two vertices. Okay. Now. A line passing through the center and perpendicular to the may. Transverse access. This line is called. This line is called. This line is called. The transverse access. This line is called. This line is called. The conjugate access. This line is called the conjugate access. Now there is a story behind why it is called conjugate access. We will talk about that story in some time. Not to worry. Similarly, other things that you have already learned in ellipse. Those will be anyways be valid. If you connect any two points. Okay. It will be called as a chord. Okay. So this is a chord. By the way, many people ask me, sir, can I connect a point from here to here and call this as a chord? Yes. Yes. Definitely you can call both of them as cards. So this is also a chord. This is also a chord. Okay. Please don't think that you have to connect two points on the very same branch. No. That's not required. You may connect a line from one line. Taking a point from one branch to a point on the other branch as well. Okay. Next, double ordinate. Double ordinate is nothing but it's a chord which is perpendicular to the transverse axis. Okay. So if I draw a line like this. Okay. Let me choose a different. Yeah. So this is called a double ordinate. This is called a double ordinate. Double ordinate. Let me write it here. Double ordinate. Okay. And the line or you can say that the double ordinate which is perpendicular. Sorry, which is passing through the focus. Okay. By the way, if a chord is passing through the focus, it is called focal chord. We already know it. Okay. So this is a focal chord. You can draw another focal chord here. Okay. Any number of focal chords can be drawn. So these are called focal chords. Okay. And if a double ordinate is passing through the focus. That means. If a double ordinate. Is a focal chord or a focal chord is a double ordinate. Then that would be called as a. Lattice rectum. So this is called as a lattice rectum. So there are two lateral recta. Just like we have in case of a. Just like we have in case of a. Just like we have in case of a. Just like we have in case of a. So there are two lateral recta. So if I make this thing, this is also a lattice rectum. Is it fine. Any questions. Any questions, any concerns. Okay. So to four side to direct this is corresponding to that foci. You already know what's the vertex. The line connecting the transfer. The line connecting the lines. You already know what's the vertex. The line connecting the transfer, the line connecting the two foci is called the transverse axis, wherever it gets the hyperbola, they become the foci. And the line perpendicular to the transverse axis passing through the center. That line is called the conjugate axis. And a line segment joining any two points on the hyperbola. That's a chord. If the chord passes through the focus that will be called as a focal chord. And the line perpendicular to the, the line perpendicular to the transverse axis. And connecting two points on the hyperbola that is called. That is called a double ordinate. And if the double ordinate passes through the focus that double ordinate would be called as the lattice rectum. So there are two lattice rectums. Normally the word is lateral recta, not lattice rectums, lateral recta. Okay. So now with this information, we are ready to derive the equation of standard form of hyperbola. So if you have copied this, can I go to the next slide? Can I move on to the next slide? So the first standard form that we are going to take up. In fact, I'm going to pull out my diagram. So in the first standard form, what I'm going to do, I'm going to assume, I'm going to assume that this is my X axis. As you can already see in the figure also. Okay. This is your X axis. This is your X axis. Okay. So I'm going to name my transverse axis as my X axis. Okay. And I'm going to take the Y axis as the conjugate axis, as you can see in the diagram as well. Okay. This point that you see. Let me use a green color. Yeah. This point that you see. Vertex. I'm going to call it as a comma zero. Okay. And I'm going to call this as C comma, sorry, minus a comma zero. Okay. Now the focus wherever it is. Okay. Let's say this, this is a focus. Let me show you. Yeah. These focus. I'm going to call them as C comma zero as of now. As of now. Okay. And minus C comma zero as of now. Okay. And the directories. Let me show the directories as well. Okay. Just going to, you know, show you some orbit line. By the way, it's crossing through this. Anyways, let me do one thing. Let me, let me show it at one place only. Okay. But I have shown in one place. So let's show in the other as well. Only half a lot make it full because I don't want the point to be cut. Yeah. I'm going to show you the policies here. And the direction is, the direction is where the direction is. And let's say the directories has passed through D comma zero. And minus T comma zero. Now there are too many variables involved. A is there. D is there. C is there. And there's already, now, also involved that comes from the locates definition itself. That is the distance of any point. Let's add. Any point here as. Okay. The distance of that. okay divided by the distance of that point from the directrix that ratio is already a e okay now what i'm going to do is just like we had done this in case of a ellipse we are going to reduce the number of variables we are only going to use a and e so we are going to do away with c and d okay so what i'm going to do here is what i'm going to do here is i'm going to take this point let's call this as v point okay let's call this point as v now this vertex v is a part of the hyperbola so it must also satisfy the locus definition of the hyperbola yes or no correct so can i say v s okay and let me name this point as n and let me name this point as n dash okay so v s by v n that is the correct yes or no v s that is the distance by v n that is the distance that should be equal to e okay and let's do one more thing let's call this point as s dash and let's call this point as n dash i've already called it as n dash so v s dash by v n dash will also be e this is from the locus definition this is from the locus definition now what is v s what is v s so as for the diagram you will say v s is nothing but c minus a correct and what is v n v n is nothing but a minus d okay so this ratio is e so from here i end up getting c minus a as a e minus d let's call it as one let's call this as one similarly v s dash okay now right on your chat box what do you think is v v s dash s dash is your other focus minus c comma zero so what is v s dash write it down on the chat box v s dash a plus c absolutely or c plus a okay and what is v n dash what are the distance between v and n dash write that down as well absolutely a plus d correct so from here i end up getting c plus a as a e plus d let's call this as second equation now in order to get c just add both of them okay if you add one into you will end up getting two c as two a e that means c is equal to a so i don't need to use a separate variable for c i could use a e for it okay and if you subtract the first result from the second you get two a as two d e that means d is equal to a by e so please note this down so c is equal to a e and d is equal to a by e okay so what we have got is let me make a fresh diagram let me make a fresh diagram over here and since i would be needing this diagram what i'll do is i'll make a proper one so that we can use the time and again okay so this was our this was our transverse axis this was our conjugate axis by the way still i haven't told you why they are called conjugate axis i will talk about it in some time not to worry okay so i'm showing you right now some lines in blue which is your direct axis and these are the two four sides okay you may call them as s one and s two okay now we have already figured out that this is called a comma zero okay this is called minus a comma zero another vertex so two vertices a comma zero minus a comma zero and this just now we figured out it is a comma zero and this is minus a comma zero okay and this line this line that you see is your x is equal to x is equal to a by e i hope i have written it over here x is equal to d which is x equal to a by e okay and this is x equal to minus a by e now for a standard case for a standard case as i already told you let us derive the equation of this hyperbola so what i'm going to do is i'm going to take a generic point like this one yeah so for a standard case what i'm going to do is i'm going to assume a point p moving point p call it as h comma k and the distance of p from s one divided by distance of p from the direct x x equal to a by e that ratio is e so let us use that s one p by p m one is equal to e okay so s one p is h minus a e the whole square k minus zero the whole square under root divided by p m one so what is p m one distance it is nothing but it is nothing but h minus a by e mod right so this ratio is given to you as a e is it fine any question with respect to whatever i have written so far so let's take this to the other side correct and let's square it by the way since e is a positive term you can introduce this e inside so this e can go inside and become e h minus a okay now when you expand it let's see what do we get so i'm quickly expanding it over here this plus k square is equal to e h minus a the whole square so this is x square a square e square minus two a e h and from here i'll end up getting again x square e square minus two a e h plus a square these two will get cancelled off okay now let's do one thing let's take the x square and k square to the other side and let's bring the a square to the left in fact i will write down the right side first s square e square minus h square minus k square is equal to a square e square minus a square in other words you end up getting something like this now divide both sides by divide both sides by a square e square minus one okay so when you do that it leaves you with h square by a square minus k square by a square e square minus one equal to one okay and once we have obtained it let's generalize it so on generalization this gives you x square by a square minus y square by a square e square minus one equal to one now what do we do is here we claim this term to be a b square now let me ask you this question yet again i have already asked this to you in your ellipse chapter what is why do you think why do you think i've called it as a b square means how how am i sure that this quantity is going to be positive okay now the answer to this you already know i'm just reiterating it you already know that e is greater than one so e square minus one will be positive correct so even if you multiply with a square it is not going to become any negative right it's still going to be positive so something which is going to be positive you can easily call it as a b square so this term could easily be called as a b square okay i'm not writing it over here yeah so this term could be easily called a b square so you end up getting the equation of the hyperbola to be this okay please make a note of this very very important now one important point to be noted here is that there is nothing like a greater than b b greater than a the greater than a in this case please don't try to apply your ellipse concept to it many people ask after deriving this sir is a greater than b here or b greater than a nothing like that anything could be greater than anything a could be greater than b also b could be greater than a also a could be equal to b also okay so this is not a case where a is more than b just like you had it in case of a ellipse nothing like that in case of a hyperbola there is no major minor axis please understand it please understand this so let me go back to the diagram see in this case if i change this to four and this to nine you'll see that the orientation will remain the same maybe there would be a change in the curvature size okay so i'm just making it nine over s here curvature size of the branches will change but it's still facing left and right are you getting a point so there's no major minor thing like that unlike in case of an ellipse where that ellipse became vertical that doesn't happen here the only thing that will make this hyperbola vertical is i'll tell you in some time not to worry right now okay okay i'll tell you in the second case don't worry is it fine any questions is it fine we can create a vertical hyperbola we'll come to that that's a different that's a second case of a standard hyperbola we'll talk about that also okay first let us finish this off okay let's finish this off we have some agenda to be discussed over here let's do some analysis okay the first thing that i would like you to observe here very interesting observation so in this equation of course your b square is a square e square minus one in this particular in this in this particular equation let's do a small activity let's put y as zero okay when you put y as zero you end up getting something like this which means x square is equal to a square which means x is equal to plus minus now this basically shows that the the points where the points points of intersection intersection with the x-axis with the x-axis are your vertices okay and what what are your vertices a comma zero minus a comma zero we have already seen it in the diagram also okay now let's put x as zero and see what happens when you put x as zero you end up getting something like this which will actually surprise you a bit okay now here people are like sir how is this possible because y square is supposed to be positive b square is anyways positive we have seen it how can a positive number be negative of the other how can a positive number be negative of the other now here is the catch if you simplify this you get y as plus minus b i the same i that you have in a complex number now this basically tries to signifies that signifies that the hyperbola does not cut the y-axis in real points okay that means it cuts the y-axis in imaginary points okay and these are the two imaginary points plus minus b i so if you look at the diagram let me go back again to the to the diagram in fact i will draw it once again over here you i did not take a snapshot of the diagram go see how i forgot anyways i'll make it quickly if you see it cuts the x-axis at minus a comma zero and a comma zero but it doesn't seem to get the y-axis but actually it cuts the y-axis at imaginary points and because it could set the imaginary point you cannot see it cutting because in order to you know see the point should be real in nature okay so hypothetically we assume that there was a point zero comma b and zero comma minus b but by the way this zero comma b and zero comma minus b they are actually on the imaginary axis okay not the real y-axis so real y-axis it will not cut it will cut the imaginary y-axis and these two points are actually conjugates of each other so as we know that b i and minus b i are conjugates of each other that is why this axis has been named as a conjugate axis so that is why we call this as the conjugate axis okay so please note in reality y-axis is not cut by the hyperbola but what do we say we justify it that the hyperbola is cutting at imaginary points which you cannot see okay and those imaginary points are b i and minus b i okay so b i as a coordinate on the argon plane we normally write it as zero comma b and minus b i we normally write it as zero comma minus b that is why we just show them but it is not the point it is not the points where it is being it is it is cutting the x-axis it is just a imaginary representation of the points they don't exist in reality okay the couple of things here to be noted down number one the distance between the two vertices let me name it as v1 v2 and let me name it as um pq okay so the v1 v2 distance is called the length of the transverse axis this is called the length of the transverse axis transverse axis so in our case this length is two a units okay similarly pq length that is called the length of the conjugate axis this is how it is defined axes axes are infinitely extending lines right so when we define a length we just take that line segment and call it as a length okay so this is your length of the conjugate axis okay now a simple exercise i would like you to do it let's say i would like you to find out the length of the lattice rectum so let's say l r is a extremities of the lattice rectum lattice rectum is a focal chord which passes through the focus please everybody figure out the length of the lattice rectum or length of the rata recta rata recta okay so for that what we are going to do i'm also going to help you out this point let's say l point is a comma k now this point should satisfy a comma k should satisfy our equation so a comma k must satisfy our given equation which is x square by a square minus y square by b square equal to 1 okay so let us put into that so this will become a square e square minus k square b square equal to 1 so these two will get cancelled off so e square minus 1 is k square by b square now all of you please note down if you recall i have also written it over here your e square minus 1 was actually b square by a square correct so can i not use this over here okay so this e square minus 1 i can write b square by a square so when i simplified i get k square as b to the power 4 by a square that means k is plus minus b square by a in other words in other words this is b square by a and down over here this will become minus b square by a okay so what is your length of the ladder spectrum so length of the ladder spectrum let me complete here length of the ladder spectrum will be given by b square by a units now you would say sir this is the same figure as what we got for ellipse yes exactly the same figure so guys you will realize that most of the things that you have learned in ellipse they will be in sync with whatever you have learned whatever you're going to learn in hyperbola like coordinates of the foresight coordinates of the vertices equation of the directices okay even to the extent that normally we say many of the cases if you replace your b square with a minus b square in a in ellipse you will end up getting hyperbola okay so if you realize in the equation itself x square by a square plus y square by b square equal to 1 change your b square with a minus b square you'll end up getting the equation of the hyperbola not only that if you recall in case of an ellipse we had this relationship b square as a square 1 minus e square so if you change your b square with a minus b square you'll end up getting b square as a square e square minus 1 that is the the one which is applicable for a hyperbola so later on also when we do higher versions of the concepts in hyperbola you would realize that if you replace your b square with a minus b square you will end up getting a lot of things for hyperbola as well okay so no new formula to be learned for the length of the later symptom still remains to b square by a okay so uh we have already covered yeah transverse axis length of conjugate axis length of latter symptom okay one final thing that we need to talk about is your eccentricity formula so since your a square e square minus 1 is b square it means i can write my eccentricity as under root of 1 plus b square by a square please note this down please note this down and this is a general formula but i'll give you a universal formula for eccentricity in this case so universal formula for eccentricity is eccentricity is given by under root of 1 plus semi conjugate square by semi transverse square okay this is a formula which is a universal formula for the case of a hyperbola so even when we take the second standard form which is what we are going to take up very soon you will realize that the same formula will be valid for eccentricity even in that case as well is it fine copied everything done any questions okay now as i told you in the beginning of this discussion on standard form of the hyperbola that there are two standard forms so we'll take up now the second form and i think one of you was asking can we not have a vertical hyperbola now we are going to discuss that so let us say i have a hyperbola which is opening or whose arms are opening upwards and downwards all right now all of you we'll discuss that anish okay now all of you in this case please note the critical points here center is still at origin okay now as i told you the the line connecting the two foresight that is called the transverse axis so from the given diagram your y axis becomes your transverse axis okay and this axis which is not cut by the hyperbola in real points at real points that is called the conjugate axis okay so just tell you the name you know nomenclature for whatever you would require over here let's say this foci is 0 comma b e okay this vertex is 0 comma b and this is y equal to b by e similarly here this is 0 comma minus b this line is y equal to b by e and this point is 0 comma minus b e okay now i would request you all i've already given you the directrix and the focus in fact two directruses and a pair of directruses and a focus is already seen on the screen now please derive the equation of this hyperbola and we'll see what is the difference between this hyperbola and the previous one okay one more thing i will tell you let's say hypothetically the imaginary points or the points where it cuts the conjugate axis is a comma zero minus a comma zero now you can use this a in your final figure of the hyperbola as well so all of you first derive it and let me know the answer on the chat box get any shh your answer is right anybody else see not a rocket science again the process is the same okay take a point take a generic point h comma take a specific point h comma k okay and what we have learned is that the distance of p from s1 divided by distance of p from the directrix this ratio is equal to e isn't it so what is s what is s1p under root of h minus 0 b whole square k minus b the whole square by p m1 p m1 will be a mod of k minus b by correct let's take it to the other side let's take the denominator to the other side and square both the sides as well so this will become x square k square b square e square minus 2 k b e this is e square k square b square minus 2 e k or 2 k b e okay 2 k b minus 2 k b will get cancelled off then this case you take this term to the other side this term to the other side and this term to the left side okay let's simplify what happens so this will become k square e square minus 1 minus x square is equal to b square e square minus 1 okay now divide both sides by this term so this will become this will become k square something is happening so divide by that term so it will become k square no this is b square not k b square yeah so it will become k square by b square minus x square by b square e square minus 1 equal to 1 okay now generalize it generalize it by replacing h with x and k with y so when you generalize it this becomes y square by b square minus x square by b square e square minus 1 equal to 1 now as I already told you assume that the point where it is cutting the x axis is a comma 0 so put y as 0 and x as a okay which clearly means that which clearly means that you can replace once again oh have I missed out anything okay so you have to replace it with a i not x is a i yeah so this will become a plus yeah so that means you have written your b square times e square minus 1 as a square so you can replace this whole thing with an a square and your equation will turn out to be something like this more commonly or better we we write it as this minus 1 please note this sound so in the case of a vertical hyperbola the only difference in the equation is not your values of a and b please don't add a and b values don't make the hyperbola become horizontal and vertical don't mix this with ellipse concept in ellipse if a is more than b it's a horizontal if b is more than a it's vertical right but in hyperbola a and b don't make any difference in horizontal and vertical case if your equation is of the nature x square by a square minus y square by b square equal to 1 no matter whether a is more than b or b is more than a it will always be having arms opening left right or you can say horizontal case if your equation is x square by a square minus y square by b square equal to minus 1 no matter whether a is more than b or b is more than a doesn't matter then in that case it will always open up and down that means the arms will be up and down okay you can call that as a vertical hyperbola are you getting my point here everybody any questions so we will do some analysis of this also like we will find out later symptom length we will find out a simplicity expression we'll also see length of the transfers and conjugate axis all those things so let's go to the next page and do the research work on this hyperbola everybody has copied it no questions okay so let's go and do some research work on this hyperbola I'll draw the diagram again just for your convenience now let us find out the length of the latter symptom oh sorry let us first write down the length of the transverse axis so please note please note this is the length of the transverse axis v1 v2 okay so this is your 0 comma b 0 comma minus length of transverse axis so length of transverse axis will be 2 b units length of conjugate axis to a units okay now length of the latter sector let's figure that out see this point is 0 comma b e so let's call this as l and let's call this as r so our point can I say it will only have a different x coordinate y coordinate will still be b only correct so let us use this in our equation of the hyperbola that we have just now derived so we have just now derived that this is the equation of our hyperbola okay so you put x as h and y as b e okay let's cancel it out so h square by a square is equal to e square minus 1 now remember just now when I was deriving the result let me go to the previous slide we wrote a square as b square e square minus 1 correct let me replicate that on the next page also so we wrote a square as b square e square minus 1 so from here can I say e square minus 1 I can write it as a square by b square okay so I can write this as a square by b square so x square is a to the power 4 by b square that means h is plus minus a square by b so this point I can write it as this point I can write it as okay and this point I can write it as this okay so what is the length lr the length lr will become 2 a square by b units okay by the way I would also like to discuss with you a generic formula your universal formula for length of latter sector it is applicable to both the cases so that is why we are calling it as a universal formula so it is always two times remember it is always two times the semi conjugate axis length square by the semi transverse axis okay in case of a ellipse you can write it as two times semi minor axis square by semi major axis so this is a universal formula whether your hyperbola is of standard type one or standard type two doesn't matter this formula will be applicable to both of it is it fine any questions so length of transverse axis is now two b units length of conjugate axis is two a units and latter sector length has now become 2 a square by b okay now from our eccentricity expression here we will can write our eccentricity as under root of 1 plus a square by b square mind you the universal formula is still applicable which was 1 plus semi conjugate whole square by semi semi conjugate whole square by semi transverse whole square okay this formula is still valid is this fine any questions any concerns anything that you would like to note down please do so ASAP now I'll do a comparative study of the two standard cases so that not only you have a clear-cut idea related to their you know positioning of the critical points but you should know when to use what okay so let's do a comparative study of both the standard cases so I'll just make a bifurcation here so the first case of a hyperbola was our this equation okay and the second case was this equation okay the distance that is the difference between the two equation is the presence of a plus one here and a minus one here that is the main distinguishing element right by the way don't expect the question set it to always give you like this only he can like you know Anish was giving the equation he's he stopped the position of y and x terms and made a one there that can also be written by the question set okay so don't expect that he will always write in the very same format in which I'm giving you he may work around with the order I'm right you know in a different way from diagram point of view this is leftward rightward opening I'm just making a rough sketch and this fellow is upward downward opening now before I move any further these two diagrams or these two hyperbola they are called they are called conjugates of each other okay so these two hyperbola are actually called conjugates of each other so this is a conjugate hyperbola of the yellow one yellow one is the conjugate of the white one just like in complex number we have no conjugates of a plus i b is conjugate of a minus i and vice versa isn't it so both are both the hyperbolas are conjugates of each other okay now I'll talk about hyperbola conjugates etc in more detail after the semester exams there are a lot of things that we need to discuss about them okay so this is a rough diagram of the second case now let us write down the parameters on which we will you know distinguish between them center how does the center change uh vertices eccentricity foci equation of the direct cases equation of the direct cases let's write down equation of transverse axis length of transverse axis equation of equation of conjugate axis equation of conjugate axis length of conjugate axis so center for both the cases is zero zero vertices for this is a comma zero and minus a comma zero while for this it is zero comma b and zero comma minus b okay eccentricity in this case is one plus b square by a square while in this case it is one my one plus a square by b square actually one more thing I would like to discuss so as I'm writing you know some important uh you know commonly made errors are coming see in case of an ellipse if you swap your a and b positions let's say by mistake let's say you forgot the formula in the examination room and if you swap the positions of a and b that means your numerator became denominated denominator became numerator then automatically one minus that will become negative and you realize that oh something has gone wrong because I cannot find eccentricity as root of a negative quantity so in in ellipse even if you want to make a mistake ellipse will not allow you to make a mistake because the reason if you swap a and b positions you will automatically get one minus b square by a square or one minus a square by v square whatever is the situation of the ellipse that you're dealing with that will automatically come out to be negative but in hyperbola hyperbola doesn't give you that opportunity or doesn't give you that safeguard so as to say so you can confidently make mistakes here so be careful of course don't don't don't do mistakes but there is no mechanism by which the formula itself will save you from making a mistake so here many people you know make mistakes because there's a plus sign hidden over here so even if you swap b square with a square a square with b square you would not realize that I have made a mistake okay ellipse tells you that oh see you're going wrong okay because you're getting a negative answer okay foci foci is a comma zero and minus a comma zero in this case the foci is zero comma b and zero comma minus b equation of the directories is x equal to a by e and x equal to minus a by e and in this case it is y equal to b by e and y equal to minus b b by e okay so one more thing I forgot ladder symptom length that is also important parameter length of ladder system okay yeah transverse axis equation in this case is your x axis in this case it is your y axis and length of the transverse axis is 2a in this case it is 2b equation of the conjugate axis is y axis in this case and in this case it is your x axis length of conjugate axis is 2b and in this case it is 2a okay length of the ladder system in this case is 2b square by but in this case it is 2a square by b is it fine so this comparative chart everybody please note this down and more importantly it should be noted here in your mind not anywhere else clear any questions clear anybody any doubt please do let me know okay let's see questions can you go a bit down yeah sure of it very good let's do a question I'll first begin with the question where I will test your locus understanding of how to get the equation once you know once you know focus and the corresponding directrix and of course yes the eccentricity so here was the first question find the equation of the hyperbola whose directrix is 2x plus y equal to 1 focus is 1 comma 0 and eccentricity is please solve this if you if you don't want to type your answer just say done that is also fine with me yes done anybody with success so y equal to 1 my y equal to 1 minus 2x it will be a line like this okay focus is at 1 comma 2 somewhere over here this is your focus right now there's a point p which is moving in such a way let's say I call this as a point p h comma k okay and this point p is moving in such a way that the ratio of its distance from s to that from the given directrix that ratio is root 3 so sp by p m is equal to root 3 that means sp square is 3 p m square what is sp square sp square is h minus 1 square k minus 2 square 3 times oh very good we'll see whether that is correct or not is 3 times p m square p m square the distance of p from m will be our 2h plus k minus 1 by root 5 the whole square isn't it so let's try to simplify this a bit so 5 times h minus 1 the whole square plus k minus 2 the whole square is 3 times 2h plus k minus 1 the whole square let's simplify this so when I expand it I already know that I'm going to get a 5 h square from here and I'm going to get a 12 h square from here right so why not be you know simplified and make it as 7 h square okay so h square is taken care of here I'll get 5 k square and here I will get 3 k square so it'll become minus 2 k square and this will become 4 h k and into 3 so that will become 12 h k okay and this is going to give you a minus 4 h minus 12 h and from here I'm going to get a minus 10 h so that is going to give you minus 2h any questions and from here I will end up getting minus 6 k and this is minus 20 k okay so that will go there and become 14 k and constant terms would be constant terms here would be 25 and from here I'll get a 3 so that's minus 22 equal to 0 now I can see yourself that how ugly a equation of a hyperbola can be provided you just make its transfers or conjugate axis inclined at a certain non-zero angle with the you know coordinate axes so just let's generalize it so on generalization you'll end up getting 7x square okay abhay and vavita you got the same result or something else we'll do one thing we will check our answers with jujibra as well so let me just copy this let me just copy this means I'll just take a snapshot on my phone yeah let's copy this on jujibra in these so our equation was 7x square 7x square minus 2y square plus 2xy minus 2x plus 14y minus 22 equal to 0 okay so as you can see it's a slightly inclined version of the hyperbola okay so this is how the hyperbola looks now thankfully I can easily figure out the foci from the tool itself 12xy oh I'm so sorry 12xy thank you thank you for letting me know that okay as you can see a is only 1 comma 2 okay of course there's one more focus which I don't need it which I did not need while I was finding the equation I don't need it right now and let's also figure out the directrix directrix of yeah so two directrises as you can see here okay one directrix is yes as you can see this is 2x plus y equal to 1 please don't go by the you know exactness of the value it's considered to be 0.9 because the tool makes it from their own there is some technical glitch that happens so it is 2x if you divide by 0.45 it is 2x plus y equal to 1 okay other one of course is yeah other one of course is if you divide it by 0.9 it becomes x plus y by 2 equal to minus 1 okay so this is the one which was of you know requirement to us is it fine so whatever we have done is absolutely right any questions any concerns should we go to the next question okay let's move on to the next question so this is a typical school level question find the length of transverse axis conjugate axis eccentricity coordinates of four psi vertices length of lr equation of directrises for the following hyperbolas let's do the first one I would request you all to give me all these answers in a single shot okay don't like you know find one thing and press and enter find another thing press and enter I want it to be at one place so once you've written all of them then only go for a press and enter okay so that I don't get confused who has answered with what first one done correct so this clearly means is one by three b is one okay again don't take a call based on a and b value which is greater than what doesn't make any difference there's a one here so it will be a case of a left and right word opening branches of a hyperbola so this will be your case okay just because b is more than a doesn't become vertical it is not an ellipse situation I'm repeating it time and again so now close your eyes and remember everything that you have learned about the case where the hyperbola opened left and right so in this case our length of transverse axis so the first part of the question length of transverse axis so this length is going to be to a is it fine any questions second was length of conjugate axis which is to be okay so let the conjugate axis be bigger than the transverse axis not an issue at all because there is there they're not called major minor for your kind information major minor is an ellipse in the case of hyperbola any axis can be greater than the other one doesn't matter at all next eccentricity now here eccentricity is under root of one plus b square by a square okay so b square is one a square will be one by nine so nine will go up so that will be a root of 10 units next foci foci is a e comma zero minus a e comma zero so a e and minus a e is it fine any questions vertices vertices a comma zero minus a comma zero okay next length of l r in this case it is two b square by correct so two b square by will be six units next last one equation of the directories so directories in this case will be x equal to a by e so a by e a is this okay and x equal to minus a by e is it fine anybody who got all of these in the same values any questions but got it like never mind the second problem is always there okay so i'll just copy paste it here so i'll take the problem once again the second part on the next sheet let's do this and this time i'll be giving you time okay take time write it down and then let me know the results yes waiting for your response oh well done very good anybody else okay let's discuss it out so first of all we all realize that this equation can also be written like this okay the presence of a minus one shows that it's a vertical case or it's the conjugate of the first case however you know about this kind of a hyperbola just remember that and start answering these questions so in this case the length of the transverse axis is two b units now here remember a is three and b is four so the answer to the first part is eight units which is two b yes or no answer to the conjugate axis is two a six units ascenticity in this case is one plus semi conjugate semi conjugate by semi transverse so that is going to give you a five by four fosai is zero comma b e and zero comma minus b e vertices zero comma b and zero comma minus b latter sector is two a square by b which in this case is nine by two units direct cases is y is equal to b by e b by e's will be 16 by 5 and y equal to minus b by e these is these are your answers to this question let me check momita's answer eight six awesome very good 10 on 10 all right any questions any concerns here anybody all right let's take another question find the ascenticity of the hyperbola whose lattice victim is half of its transverse axis find the ascenticity of the hyperbola whose lattice victim is half of its transverse axis very good nobita okay should you discuss it anybody else who would like to answer this okay lavanya lattice rectum length is half the length of the transverse axis okay so what does it mean it means oh i'm so sorry yeah it means two b square is equal to a square so b square by a square is half and for such cases where you have assumed this to be your lattice rectum length and this to be a transverse axis in those cases ascenticity formula is this right because you may ask me this question why this formula why not one plus a square by b square formula if that is the case then change your lattice rectum formula over here as well change your length of transverse axis here as well okay so if you're going to make a change in the ascenticity formula you must make a change in the lattice rectum formula first of all first of all over here and your transverse axis length over here so ultimately your end result is going to be the same which is going to be under root of three by two is it fine any questions simple version any questions any concerns anybody please do let me know i think root of three by two this will be for answer root of three by two right samath all right let's take this one up the foci of a hyperbola concites with the foci of the ellipse find the equation of the hyperbola if its ascenticity is two if its ascenticity is two done anybody with any response so this is your ellipse of course your a is five b is three and your a is greater than b okay so it's a case of a horizontal ellipse okay so let's figure out first a few things where is what is the you know what are the coordinates of the foci for this case and do we actually need it to be more precise do we actually need it do we need to know the coordinates of the foci in this case many of you are still trying i believe so should i give you more time or should we discuss it what do you think okay lawanya very good so lawanya has responded before anybody else who would like to respond okay now for this ellipse what is the ascenticity it's one minus now see this is what i was trying to say let's say if you misuse the formula you will immediately realize that this is wrong because you'll end up getting a negative answer here okay so that is what i was trying to say even if you want to make a mistake the formula will not allow you to make a mistake okay so ideally sticking this is the ascenticity for this ellipse right so this will become four by five if i'm not mistaken so the distance between the two foci the distance between the foci will be two a e which in this case is eight units and this is supposed to be the distance between the foci of our hyperbola as well let me write it like this so same should be the distance between between the foci of the hyperbola as well so two a e but remember e is two here two a e is eight so a value is two correct yes or no now these are your two foci your hyperbola has to be left word and right word opening right this can be a hyperbola correct so it's a case where your equation is x square by a square minus y square by b square equal to one a value I've already figured out and in this case our b square is a square e square minus one which is nothing but four into e square e square is again two square minus one which is 12 correct so let's write down our final equation so a square is four b square is 12 so our final equation of the hyperbola will become x square by four minus y square by 12 equal to one is it correct any questions absolutely right I think Lavanya got it exactly correct any questions any concerns okay so please do not waste time finding the foci of the ellipse because there will be of no use to you in solving the question okay anyways we'll make a move to the I know next part of the question obtain the equation of a hyperbola with coordinate axis as principal axes principal axes means transfers and conjugate given that the distances of one of its vertex vertices from the foci are nine and one units so one of the vertices is distance from one focus is nine and the other focus is one you may assume any of the two forms that doesn't matter okay I'm fine with you assuming a vertical hyperbola or you are using a horizontal hyperbola so two answers will come as a result of that but you know you can go for any one of the answers okay in this case you can assume this to be one and this to be nine yes anybody with his or her response okay what is given to you is a e minus a is one and a e plus a is nine that's what has been given to you in the question isn't it correct so just add it to a is 10 so a e is five subtract it to a is eight so a is four so e is five by four okay so a is four e is five by four and we already know that b square is a square e square minus one so b square is a square e square minus one that will leave you with a nine right so your answer would be x square by a square minus y square by b square equal to one that's your answer is it fine any questions any questions any concerns let's take if e and e dash if e and e dash are the assenticities of these two hyperbola are the assenticities of these two hyperbola then one by e and one by e dash lies on which of the following curves I'm launching a poll for it okay very good I got one response yes anybody else three minutes about to get over let me stop the poll in the next 15 seconds five four three two one two so four of you have only voted for this question and all of you have voted for option a okay let's check see both of these are actually equations of hyperbola which open left and right so this one okay so e dash is under root of one plus b square by a square which is literally in this case b square by a square which means sorry why e dash e which means e square is a square plus b square by a square let's call it as one the second one will satisfy under root of one plus now in this case it will become a square by b square because a has now become b b has now become a but the formula still remains the same so from here I can say e dash square is b square plus a square by b square okay let's do a simple activity here let us reciprocate both of these first and the second equation and add them so the yeah so this expression will become this which clearly becomes a square plus b square upon a square plus b square which is actually a one right that means if I take this as a point x so one by e is your x and one by e dash is your y it implies that x square plus y square is actually a one okay now here something very interesting I would like to highlight since this question was asked please note this down let us say this is your hyperbola whose eccentricity is e okay and this is the conjugate of that hyperbola whose eccentricity is e dash okay so please note here that one by e square one by e dash square will actually be a one very very important result please note down let me write it properly I'll still think as it's e to the power of 12 yeah so e dash one by e dash square plus one by e square is actually a one is it fine any questions any concerns please highlight so this is the property which is true in general also that means if there is a hyperbola and it's conjugate their eccentricities reciprocal squares will add up to give you a one okay so this is a property kind of a thing all right so we'll take a break right now on the other hand on the side of the break we will take up shifted cases of hyperbola for that I don't need to tell you any theory I've already done ellipse and parabola exactly the same approach is applied for hyperbola as well okay so we'll take that up on the other side of the break let's meet at 6 22 p.m after a short 15 minutes break all right see you on the other side so the next point of agenda for us in hyperbola would be discussing shifted hyperbola okay or what we call as the generalized hyperbola so as we've already seen in case of parabola ellipse and now we are seeing in case of hyperbola that if you shift a standard case so any of the two standard cases whether it's a vertical opening whether it's a horizontal opening so whichever of the two cases you take and you shift that case in such a way that the center no longer remains at origin but yes the principal axis that is your conjugate axis and your transverse axis is still parallel to the coordinate axis those are called the shifted hyperbola case okay so I'll write it down over here that this is a case of shifting off shifting off center to a point other than the origin point other than the origin but keeping the principal axis principal axis or you can say the transverse and the conjugate axis to be more precise I'll write it down but keeping the transverse and conjugate axis parallel to the coordinate axis okay so despite shifting the origin the the transverse and the conjugate axis still remain to be parallel to the coordinate axis okay so a typical such case would look like this so let us say this was your initial position of the hyperbola okay the dotted one now let's say I shift the origin of the hyperbola in such a way that the hyperbola now becomes something like this I'm just making a diagram okay so the center has now come to alpha comma beta okay and this is your new let me make it in blue color this is your new transverse axis and this is your new conjugate axis okay so earlier it was like this earlier it was in this position okay this was your equation also and now this hyperbola has been shifted to this position so now this these are your two arms okay so you note here that the transverse axis still remains parallel to the x axis and this still remains parallel to the y axis okay so in such cases the equation will change to as we have already seen before and we will not be deliberating on it again and again so we have already seen that shifting changes the equation to this okay and not only that we should also be aware that how does the other coordinates like the four side coordinates they change the vertices they change what happens to the equation of the transverse and the conjugate axis what happens to the equation of the directories all of these facts have been already covered under ellipse and parabola and the same is going to apply yet again to hyperbola so there's nothing new to be discussed over here okay so your whole change method is going to work here as well if you have forgotten it maybe we can take it through a simple example just to revisit the the concept again let's say I start with this question find the center find the center vertices four side equation of directories okay equation of directories for for something like this x plus 1 the whole square by 4 minus y minus 2 the whole square by 9 equal to 1 okay so the moment you see this you realize that oh this is the case where your standard hyperbola has been shifted okay so just do a role change here uh compare it with compare it with this hyperbola compare it with this hyperbola so here if you see a roll is being paid by 2 b roll is being paid by 3 capital x roll is being paid by x plus 1 and capital y roll is being paid by y minus 2 okay let's try to understand how do we find these uh coordinates and equations so center where was center for this hyperbola center for this hyperbola was at 0 0 so write that like this x equal to 0 y equal to 0 and just do a roll change so instead of capital x put small x plus 1 instead of capital y put small y minus 2 so x becomes negative 1 y becomes 2 now I hope this helps you to you know recall whatever we had done in our previous two connects right next vertices vertices for this hyperbola used to be at a comma 0 and minus a comma 0 and minus a comma 0 correct so now let's do a roll change capital x is x plus 1 small x plus 1 a is 2 as per the given question correct and y is nothing but small y minus 2 so solving it gives you x as 1 and y as 2 that means 1 comma 2 is one of the vertices and same here also x plus 1 is minus 2 y minus 2 is 0 so minus 3 comma 2 is your another vertex is it clear any questions any questions next foci for foci I need eccentricity so first thing is eccentricity so for such kind of hyperbola eccentricity is 1 plus b square by a square which is under root of 1 plus okay b square b square is 9 a square is 4 so it's under root of 13 by 2 okay so let's now find out the foci foci is at a comma 0 and minus a comma 0 okay so a e sorry capital x is x plus 1 equal to a e I think a value was 2 and e value was root 13 by 2 so this will be root 13 and this is y minus 2 equal to 0 so your coordinate will become root 13 minus 1 comma 2 and this guy will give x plus 1 equal to minus root 13 and y is y minus 2 is equal to 0 so this will give us the coordinate as minus root 13 minus 1 comma 2 is it fine any questions now coming to the equation of the direct cases so direct cases for such cases direct cases for such cases is x equal to a by e and x equal to minus a by e so x equal to a by e a a a a a is 2 so a by e a by e will be 4 by root 13 okay and the next one will be x plus 1 x plus 1 is equal to minus 4 by root 30 now first you can simplify it okay you can put the 4 by root 13 to the left side also and in this case also but that is not my agenda is this fine is the approach clear everybody has now successfully recalled right what we had done in our parabola and ellipse cases as well okay any questions anybody okay so can we take one more set of questions let's take find the center vertices eccentricity foresight direct cases later sector length equation of transverse axis equation of conjugate axis length of transverse axis length of conjugate axis so let's do all of them okay I think we have not left anything for equal to minus 1 equal to minus 1 yes done anybody should you discuss it now see again the minus 1 shows that it's a case of a vertical ellipse sorry vertical hyperbola so center is 00 right so you kept to put x as 0 y as 0 so that will give you minus 2 and y as 2 so center is minus 2 comma 2 okay this is your center now vertices in this case is 0 b and 0 minus b this is your vertices that means x plus 2 is 0 okay and y minus 2 is b now in this case your a is 3 and b is 2 that means minus 2 comma 4 and minus 2 comma 0 that will become your that will become your vertices I'm so sorry that will become your vertices for this question is it fine very good Lavanya Lavanya has given the response eccentricity eccentricity in this case is given by under root of 1 plus a square by b square which is 1 plus 9 by 4 which is root 13 by 2 foci foci is at 0 comma b e and 0 comma minus b so that is x plus 2 is equal to 0 and y minus 2 is equal to b e now b into e will be if I'm not mistaken root 13 similarly here this is 0 and y minus 2 equal to minus root 30 okay so this is minus 2 comma 2 plus root 13 and minus 2 comma 2 minus root 30 is it fine any questions any questions any concerns so far next is director says the equation director says the equation would be y equal to b by e and y equal to minus b by e so y is nothing but y minus 2 correct equal to b by e so b in this case is 2 so b by e will be 4 by root 13 4 by root 30 and y minus 2 is minus 4 by root 30 you can just write it as y equal to 2 plus 4 by root 13 also and y equal to 2 minus 4 by root 13 also is it fine next length of the latter sector since it is a case of a vertical hyperbola length of the latter sector will be 2 a square by b units which is 2 into a square in this case I think a square is going to be 9 so that is nothing but 9 units is it fine any questions next equation of equation of transverse axis transverse axis in this case is your in case of a vertical hyperbola transverse axis is your y axis right which is capital X equal to 0 that means x plus 2 equal to 0 will be your answer okay is it fine and this will be y minus 2 equal to 0 okay now length of the transverse axis length of the transverse axis will be 2 b units in this case which is going to be 2 into 2 which is 4 units no issues and and length of the length of the conjugate axis will be 2 b units which is 6 units sorry 2 a which is 6 units okay please remember this is 2 b and this is 2 a any questions any questions any concerns here okay so I think one question is good enough now we'll take some generic questions on the same but if you have any doubts any questions any concerns please immediately highlight let's take let's take this question find the equation of the hyperbola whose foresight are at these two positions and eccentricity is 3 yes anybody see as per the question these are the two foresight okay so basically it is a it is trying to suggest you that it's a case of a okay it's a case of a shifted hyperbola and that hyperbola was originally opening left and right so what are the things that we need to note while you are writing the equation of course we need to know the center okay the center is very easy to find out center is actually the midpoint I'm so sorry I swapped the position this is minus 2 comma 5 yeah center is actually the midpoint of the given two foresight so the midpoint will be at 4 comma 5 this is the midpoint so this will be a center okay another important thing to be noted down is we need a and b values now we all know that distance between the two foresight is 2 ae so 2 ae is equal to 12 correct so ae is equal to 6 and eccentricity is given to you as 3 so a is 2 correct and we also know that b square is a square e square minus 1 so taking everything into the consideration that it is the case of a shifted hyperbola that means its center is no longer at origin and a value is 2 b value is root 32 in light of all these things the equation will turn out to be x minus 4 the whole square by a square minus y minus b the whole square sorry y minus 5 the whole square by b square equal to 1 is it fine anybody who got this answer anybody excellent moment very good any questions any concerns anybody has do let me know all right so let's move on to the next question okay I have a very plain and simple question over here find the center eccentricity and length of the transfers and conjugate axes for this hyperbola so here the problem is we have been given the equation not in the form that we have been seeing it so far because that makes it very predictable right you're able to just you know do a role change mechanism and find all your required points and equations conveniently from that instead the question setter has basically complicated it by by expanding the entire scenario but nevertheless it is actually testing you on your art of competing the square so let's do that and let's figure out what is the center eccentricity and the length of the axes anybody who's able to do it do let me know how should i wait for some time one more minute i can give you all a very good moment see normally when i solve these kind of the questions i group all the x y's and constants you know in separate camps okay so x camp y camp constant camp okay so here if you take three common you get x square minus 2x here you take minus five common you get y square minus four why okay just do a small activity put a plus one here but when you put a plus one here you have to put a plus three here because you're actually adding a three right put a plus four here but when you're adding a plus four you're actually adding a minus 20 so you have to add a minus 20 here as well okay abhay we'll check we'll check is it fine so this simplifies to three times x minus one the whole square minus five times y minus two the whole square this will be 32 minus 17 which is five correct any questions so x minus one the whole square upon five by three minus y minus two whole square upon one is equal to or is equal to 50 so this will be five and this will be three okay so look and feel says that it's a case of a shifting of a horizontal hyperbola so a is root five b is root three now let's try to answer what do they want they want us to find the center center is where capital x which is your x minus one is zero and capital y which is y minus two is zero so x is one y is two one comma two is the center so this is your answer to the center next eccentricity eccentricity in this case is under root of one plus b square by a square right which in this case is two root two by five by root five okay next is length of the transverse axis which is two root five and two root three so your transverse axis length is two root five units conjugate axis length is two root three units is it fine now let us talk about the parametric equation of the hyperbola just like we had taken parametric equations for circles parabola we take the same for hyperbola as well okay any questions anybody has anybody wants to note down anything now for a standard case of a hyperbola let's take this standard case first this case we normally so let's say this is the Cartesian form for this case we normally use use this as the parametric form where theta is a parameter in this case okay so this is a parametric form or a suggested parametric form as I already told you parametric form is not fixed like nobody can say this is the only parametric form for this no nobody can say like that there can be several parametric forms so this is one of the mostly used or suggested parametric forms it basically comes from your derived Pythagorean identity that secant square theta or secant square any angle minus tan square an angle okay is equal to one so it comes from that but here in this parametric form I would like you to understand certain things what does this theta actually represent over here so let me just show you something very important over here we'll be discussing about it little later on also in our further discussions of hyperbola after your school exams are over okay so let's say this is our x square by a square minus y square by b square equal to one okay now let us say I take a point over here p okay and I call it as ac theta comma b tan theta okay now many people when I ask them what is this theta here can you show me the theta in this diagram many people tell me so theta is this angle okay let me tell you this is wrong this is not theta either way theta is also called the eccentric angle of that point okay I'm not going to go into depth of it because anyways we are going to cover it up little later on okay not required for your school hence I'm not going to the depth of it so theta is actually called the eccentric angle of that point of that point p okay so this point p eccentric angle is theta now let me actually show you what is theta in this case let's make a circle let's make a circle with center at c and and radius as the length of or you can say a as your radius okay this circle later on you will learn the name is called auxiliary circle okay this is called an auxiliary circle so what is an auxiliary circle it's a circle whose center is same as that of the hyperbola and whose radius is nothing but or whose diameter is nothing but the length of the transverse axis okay now from this point p from this point p you drop a perpendicular okay let's say it hits the it hits the x axis or the transverse axis at m from this point m you draw tangent to this circle oh sorry okay let's say the tangent hits the circle at this point q this theta that you are seeing over here is actually the angle between c q and the x axis this is the theta and not this one which many people say no please don't say this is theta it would be wrong okay now I can justify it also it's very easy to justify see this is already a so this will also be a and if this this is theta this length cm that will be ac theta correct and if you put ac theta in this equation of your hyperbola so you'll have a square secant square by a square which is secant square theta minus y square by b square equal to 1 which clearly gives you y square by b square is secant square theta minus 1 which is tan square theta so y square is equal to b square tan square theta so y is equal to plus minus b tan theta by the way plus minus because for the same a you can have a point minus b tan theta also over here so I'm not interested in that but I still get that the y coordinate is white but the y coordinate is b tan theta in this case okay in other words what I wanted to explain here is that when you write down the coordinates of a point using the parametric equation ac theta b tan theta theta is not the angle between c to p don't make that mistake how is theta obtained from p drop a perpendicular on the transverse from there you sketch a tangent to the auxiliary circle point of tangency connected to the center this angle is called the eccentric angle of that point okay is it fine any questions don't worry we'll be speaking about it in details we'll be talking about it in detail also in our higher versions or higher discussions of this topic okay this topic has just actually started okay is it fine any questions now similarly many people ask me sir for the other case of a hyperbola which is the conjugate of this what is the recommended parametric form okay let's write that down also so for this case for this case this is our Cartesian form for this case it is a tan theta and b c theta okay where theta is a parameter okay so this is the parametric form is it fine any questions so with this I'll take a small problem associated with it okay let's take this question prove that this point lies on the hyperbola by the way the right way to read this question is if somebody gives you a parametric form of a curve like this you have to show that this is the parametric form of a hyperbola right that is we need to show that this is a parametric form of a hyperbola in short we need to know what is the Cartesian form so for this hyperbola what is the Cartesian form so please tell me the Cartesian form for this remember while finding the Cartesian form you just have to get rid of the parameter involved that's it done anybody with any success so eliminate the t from these two equations tell me what to what do you see do you get a hyperbola's equation or not get it okay so how do you do this question okay I'll show you a very convenient way to do this question so from the first equation can I say 2x by is this or can I say if I square it up I get something like t square plus 1 by t square plus or 2 similarly from the second equation I get 2y by b as t minus 1 by t that means on squaring it I'll get t square plus 1 by t square minus 2 subtract both the results you'll end up getting a four drop the four and this clearly represents a this clearly represents a hyperbola okay so what do you see what are the moral of the story the thing to be learned here is that I wrote in the previous slide as ac theta and beta and theta right as x and y this is an alternative way as I told you as I keep telling you parametric forms are not fixed so this is yet another parametric form for the same hyperbola okay so this is yet another parameter parametric form for the same hyperbola later on when you learn hyperbolic functions hyperbolic trigonometric functions you will understand that there are you know a few more parametric forms which are very commonly used anyways with this we will now talk about an alternative locus definition of a hyperbola another locus definition so in the very first slide of our session today we discussed that hyperbola is defined as the locus of a point moving in a plane says that the ratio of its distance from a fixed point to that from a fixed line that's a constant whose value is more than one right now there's another locus definition and this locus definition goes like this a hyperbola is the locus of a point or it's a path traced by a point traced by a point moving in a plane moving in a plane in such a way that the difference of its distances difference of its distances from two fixed points fixed points is a constant is a constant is a constant so please note this down i'll repeat this definition once again it's a path traced by a point moving in such a plane that the difference of its distances from two fixed point is a constant now these distances that i'll be talking about these are called the focal radii okay we'll talk about it and these two fixed points that i'm talking about these are your foresight and this constant that i'm talking about is actually the length of the transverse axis okay so now let me show this that why this locus definition actually is fulfilled by a hyperbola okay in a hyperbola why does the difference of the focal radii of any point focal radii if any point means focal distance of that point okay is a fixed value and that fixed value is actually the length of the transverse axis okay so what i'm going to do here is what i'm going to do is i'm going to take any generic point h comma k okay and this is one of the foresight let's say i call it as s1 and this is another foresight let's call it as s2 okay now what is s1p and what is s2p let us try to find it out s1p you don't have to use the distance between two point formula you can use that it is e times p m1 and what is m1 m1 is nothing but the foot of the perpendicular dropped from this point on to the directrix okay i'm just making a small directrix over there so what is this length this length is going to be h minus a by e so if you expand it you get e h minus a this is s1p okay similarly similarly what is s2p s2p will be e times p m2 and that is nothing but e times h plus a by which is e h plus a correct now what do you realize is that if you take the difference of s1p and s2p difference difference means absolute value of the subtraction of these two answers you'll end up getting 2a and this 2a happens to be the length of the transverse axis so this is yet another locus definition of a hyperbola that in a hyperbola any point satisfies the fact that the difference of its distances from the two foresight is a constant again i'll repeat the difference of the two focal distances is a constant and that constant value is the length of the transverse axis always please note this down this is a locus definition of any hyperbola no matter whether it is a standard case or non-standard case this definition will be satisfied okay all right so we'll take some generic questions okay so a few generic questions that is required for your school exams we'll take that up okay because sub we have almost covered all the subject matter required for your school purposes okay so we'll take few questions from here and there let's take this one i'll put the poll on for this what is the eccentricity of this conic now in the present context this question becomes very easy because as you can see since we are doing hyperbola none of these options will you know suit a hyperbola right hyperbola eccentricity cannot be one it cannot be half it can minus one is out of question so only option that seems to be feasible is d okay but anyways i'm assuming that we are doing hyperbola that's why d is the right option okay but let us say if i remove the options from my mind or from my you can say option list let's solve it there now as i told you start computing the square so in such cases eccentricity is one plus a square by v square that's nothing but root two later on we'll come to know that it's that it's a type of a rectangular hyperbola okay so the answer in the present question is option d now i just wanted to see the presence of mind but only three of you responded with d many people did not respond that is not because of the presence of mind because they are not paying attention to the class which they will realize very soon they will realize once the right time comes and they have missed out on a lot of things and that time will become very difficult for them to go back and catch it up time is the best you know teacher it can it can teach you lessons which nobody else can teach okay let's take this one i'll put the poll on for this as well read the question carefully it says find the eccentricity of a hyperbola conjugate to this fellow conjugate to this okay should we discuss it okay anish will discuss it see first of all if you write this equation this you can add a one add a one here so it'll have x minus one the whole square okay divide by nine throughout now here a is three b is root three eccentricity will be under root of one plus b square by a square correct which in this case will be b square by a square okay one by three so it's going to be two by root three uh i'm just stopping the poll in case okay four of you have voted for c option only okay let's check yeah now we have already learned that if e dash or let e dash let e dash be the eccentricity of its conjugate the of the conjugate of this hyperbola then can i say one by e square plus one by e dash square should be equal to one remember this is one of the properties which i gave you early on okay so which means three by four plus one by e dash square is equal to one that means one by e dash square is equal to one fourth which means e dash square is equal to four which means e dash has to be to please note no plus minus business only two will be the answer eccentricity is a positive quantity because it is the ratio of two distances okay so option number option number c becomes the right option right so i think we have done more than what is required for school at least you know first from school point of view we are much ahead okay our next class that we'll be having