 So first of all, parabola, how is it a conic? How is it obtained from the interaction of a plane with a right circular double cone? In fact, I had already told you this in the first class of conic, but still I'll be repeating it. If a right circular double cone, okay, so this is a right circular double cone. If one of the naps, if one of the naps, so these are the two naps, and we all know that this is called the semi-vertical angle. If one of the naps is cut by a plane at an angle which is equal to the semi-vertical angle, so something like this, okay? So as you can see, this is a plane which is cutting this nap at the same angle as alpha. That means this angle, this angle, this angle is also alpha, okay? So this angle is anyway alpha and this is also alpha. So that means this plane is parallel to the generator over here. So this and this are parallel. So because of this, there is a cross-section area that you would be seeing over here, and that cross-section area will be that of a parabola, okay? So parabola is nothing but a conic which is obtained by interaction of a plane with a right circular double cone where the plane cuts one of the naps at an angle equal to the semi-vertical angle, okay? Is this fine? Now, as a locus, how do we define a parabola? As a locus, how do we define a parabola? So what is the locus definition of a parabola? By the way, the definitions which I'm going to use for a parabola, the same would be carried forward, in fact, with a minor change in case of an ellipse and hyperbola as well, okay? So it's very important that you listen to this definition, you know? Clearly, because the same definition I'll be using in case of ellipse and hyperbola as well. By the way, in how many of your school has parabola been already completed? Has it been completed in Kaur Mangala, HSR? NPS is, okay, Arun Dati, completed in HSR, okay. What about Kaur Mangala? Completed in Kaur Mangala, okay, good. So I think you will be all making sense of whatever I'm going to say. What about DPS and all? Okay, so parabola is basically a path traced by a point, moving in a plane, by a point, sorry, I forgot to write point, point, moving in a plane, moving in a plane in such a way that the ratio of its distance, from a fixed point to that form of fixed line, that form of fixed line, is a constant, is a constant where the constant value is equal to 1, where the constant value, where the constant value is equal to 1, okay? Now let me revisit this definition once again. So it's a path traced by a point moving in a plane, yes, parabola is a two-dimensional figure, okay, so you draw it on a plane, okay, in such a way that the ratio of its distance from a fixed point, which is called the focus, to that form of fixed line, which we call as the directrix, is a constant, where this constant is called the eccentricity, where this constant is called the eccentricity, that means your eccentricity value is 1 for a parabola. Now many people ask me, sir, you told that circle is also a conic, because it is obtained by interaction of a plane with the right circular double cone, where the plane cuts one of the naps at an angle 90 degree to the axis. So what is the eccentricity of a circle? I don't know whether I've told you this or not, but for a circle, the eccentricity value is equal to zero, okay, so please note this down, this will be helpful for your comparative level exam as well. So for a circle, we say the eccentricity value is zero, okay? For a parabola, which we are doing today, eccentricity value is equal to 1. For ellipse, which we'll be doing in our next week's session, for ellipse, the eccentricity value is a fraction lying between zero to one. For a hyperbola, it's a value which is greater than one. And for pair of straight lines, which is another degenerate conic, nevertheless, it is a conic. So for pair of straight lines, E value is very, very large, okay, almost tending to infinity. Is it fine? Eccentricity is the ratio of the distance of the point lying on that conic from the focus to the distance from the directrix. That ratio is defined as eccentricity. Got it, Kartik? Okay, I'll show you here. I'll show you. So let's say this is our fixed line and this is our fixed point, okay? And there is a moving point, let's say p, okay? This point is a moving point. Let's call it a moving point here. This is a fixed point, okay? This is a fixed point and this is a fixed line. This is a fixed line, okay? Now, at this point, p moves in such a way that wherever it goes, wherever it goes, I'm just drawing few situations here, wherever it goes, it's distance from s divided by its distance from e, sorry, d, okay, the fixed line. So sp by pm, this ratio is always 1. That means sp is equal to pm and this ratio is what we call as eccentricity, okay? So this ratio is what we call as eccentricity, okay? So eccentricity is 1 for a parabola. It is 0 for a circle. It is some value which is between 0 and 1 for an ellipse. It is some value greater than 2 for a hyperbola and if it becomes very, very large, it becomes infinity almost, ending to infinity. Then for a pair of straight lines, the eccentricity value is very, very large, okay? We'll be talking about all those things later on. So if you see the path trace by this point, the path trace by this point would look like a structure of this nature, okay? Of course, this is a familiar structure to you. You have already done quadratic polynomials and a graph of quadratic polynomials are also parabola. Remember, parabola is actually the word has come from the path taken by the motion of a body thrown under gravity, okay? Cutting ellipse, sir, how is that both this definition and cutting ellipse is same? I didn't get your question. See, when it comes to the cone part of it, that requires your understanding of 3D geometry, okay? First 2D geometry also, you have not done the concept of, you know, many things. In 3D geometry, later on, you will understand the concept of a cone and in class cells, you will understand the concept of a plane. So when you study those concepts, you will be able to relate these two definitions. That is why we don't actually go into the depth of the interaction of a plane with a cone because the knowledge of 3D geometry or the knowledge of three-dimensional bodies, so as to say, is very limited for us. You will explore more about it in your higher courses of mathematics, okay? Because right now you will not make sense about the equation of a cone, equation of a plane and how they interact to give you a parabola, okay? So we'll be talking about those. In fact, your engineering professors will be talking about those little later on. It is just to tell you in the class 11th and 12th that why it is called a conic. That's the only thing you need to know, okay? So the actual thing that you are supposed to know as a 11th grade student is how a parabola is obtained on a two-dimensional plane and that is basically explained through the locus concept. This is the locus concept or this is the locus definition, okay? So there are a lot of things that you need to explore and do a research work on it. Could you scroll up? Which part do you want me to scroll till? Locus definition or below it or to the right side? Now anybody who wants me to scroll up or down, please give the direction. Up and left, up and right, down and left, down and right, so that I am also clear where to go. Okay, Manu, you are done with your copying of the definition? Okay, so now up and right, Seethu wants Firli to copy this part. Okay, meanwhile he is copying where all you have come across parabola other than your quadratic polynomials. Have you seen parabola in your daily life? Of course, I already gave an example. When you throw a ball at somebody, the ball takes a parabolic path before reaching the hands of that person, right? Anyway, else you have seen a parabola? Okay, Firli would have observed the head lamps of your car. They are actually parabolic projectors or they are parabolic lamps. One of the very important features of a parabola is in a parabola there is something called axis. So we'll talk about that also. Okay, so if there is a ray of light which is coming parallel to the axis after hitting the parabolic reflector, it always will pass through the focus. So it's always going to pass through the focus. Okay, and the same is true for the reverse direction also, you know, light reversibility of direction of light is always true. So if a ray of light starts on the focus after hitting the parabolic reflector, it becomes parallel to the axis. Now there is a proof for this, but for that proof, you need to know the concept of normal and tangents also. We'll talk about it when we're doing this chapter in more depth. Right now, we are only limiting ourselves to school curricula. So because of this, because of this, if you have a parabolic reflector in your car, when you when you light the bulb, which is at the focus, the bulb light after hitting the glass of the parabolic reflector or the mirrored surface of the parabolic reflector will become parallel to the axis. That means if the driver of the car wants to see far on the road, okay, he will switch on this light. Okay, if he switches on the other light, which is not at the focus, then basically after hitting it, it will become it will converge, it'll come back to the, you know, it'll just let's say if there is a bulb over here, let's say I'm just putting a bulb over here, then this bulb after hitting the car will maybe converge like this. Okay, it'll not become parallel. Okay, so there are, you know, multiple bulbs in the headlight of the car, as you can see, the driver will always, you know, use a dipper and all those things. So basically he changes the position of the, basically he lights up a different bulb kept at different positions. Okay, that is how he can see far off on the road. Earlier there were used to be spherical mirrors, but there were, there were problems with the spherical mirrors like spherical aberrations. That is why the parabolic mirrors have replaced those spherical mirrors in the modern cars. Another application is in treating the stones, kidney stones. So what does, what do doctors do? They will keep the kidney stone at the focus of the parabolic reflector and they will pass a laser beam. So the laser beam on hitting the parabolic reflector will pass through the, yeah, will pass through the, that will pass through the focus, thereby damaging that stone, kidney stone, and thereby curing that problem. Is it fine? Any questions? All right. So any point that you take, let's say I take this point, so this divided by this, okay, that will also be, you know, the ratio will be one. So any, any point you take on this parabolic surface, okay. So this length will always be equal to this length. Okay. So let me show you this demonstration on GeoGebra also, that if I take a parabola and choose any point on the parabola, the ratio of its distance from the focus to that from the directrix will always be a one. So let me show that to you. Oh, one second. I'll take a simple parabola y square is equal to 4x. Maybe most of you who have already done it would know that this equation is of a standard parabola. We will talk about it, don't worry. So those who are wondering how did they write this parabola equation. This is one of the standard cases of a parabola, which I'll be taking up very soon with you. Now, this tool has a facility that it can actually tell you the focus. Okay. But please don't misuse it for your came on test. Okay. So it can tell you the focus. See, it has told you the focus, focus is at one comma zero. And this tool can also tell you the directrix. Wonderful tool. Okay. So see, it has shown you the directrix. Now if you take any point on this parabola, let's say I take this point. Okay. And from this point, I drop a perpendicular, I drop a perpendicular on to the directrix. Okay. And let's, let's see the lens. So length from B to the focus is H and H value is right now 4.64. As you can see on the left side of the, the software. Okay. And from B to C, the length is I, which is again 4.64. Okay. So this ratio of H by I, that is what we call a, in this case, they have called it as a, that is what we call as the eccentricity. Okay. So that ratio is called eccentricity. In case of a parabola, it is one. Okay. Now see, I will make this point B dance on this parabola. Just have a close eye on a value. You would realize that a is not changing. A is always a one. Or in other words, H and I are always taking same values. See 1.43, 1.43, 1.9, 1.9, 2.91, 2.91. So wherever you go, that ratio will always be equal to one. Is it fine? Any questions? Any concerns here? No, we don't define eccentricity for x cube and all. Okay. So eccentricity is only defined for a conic sections, which happens to be a second degree general equation curves. Okay. This is a third degree. So conic, eccentricity and all will not be defined for such terms. Okay. All right. So with this, we are now going to learn certain terms associated with a parabola. These will be the new terms for you vis-a-vis your circles because in circles you had hardly center and radius and maybe a chord or something or x intercept, y intercept is what you learned. But in case of a parabola, you will be learning plethora of terms. So what are those terms? Let's talk about it. And these terms would also be used in your other conic as well, like ellipse and hyperbola. So terms used in parabola. So first let me draw a parabola. Normally for making a parabola, I, I do some weird things. I just make an ellipse. I just chop off the half part. But let me not, you know, let me, let me tell you that this doesn't mean that half of an ellipse is a parabola. Please don't take that into consideration. It is only to facilitate drawing of the graph. That is why I'm using it. Okay. So don't start saying, oh ellipse is half, parabola is half of an ellipse, nothing like that. Okay. Now all of you, please pay attention. Let me also draw the directrix. Okay. Please ensure that you are roughly following the fact that directrix, okay. Directrix, any point on the ellipse is equidistant from the focus as well as the directrix. Okay. Now all of you, please pay attention. This you already know the name focus. No issues with that. Okay. This is called the directrix, directrix. Don't start calling it as a dielectric. There was some teacher I was interviewing two years back. And I still could not forget that day, not two years, maybe three years back just before COVID began. So he was calling directrix as dielectric. The one which you'll study in physics, maybe when you do our capacitors and all. So he was constantly calling it as dielectric, dielectric, dielectric. How was like five minutes? I could not understand them. What is this dielectric doing in mathematics? So then I will, oh, he's talking about directrix. Okay. So please do not mispronounce it. Thankfully, this year you also may not have a KVPYSA vaiva. But if it does, and you're using a wrong pronunciation, that really turns off the professors who are very old and experienced. They don't like misuse of words, especially when it comes to your mathematical terms. Anyways, now there's a line which connects the focus and perpendicular to the directrix. That line is called the axis of the parabola. So this line that you see, this line is called the axis of the parabola. So axis of a parabola is a line perpendicular to the directrix passing through the focus. Axis divides the parabola into symmetrical halves. It divides the parabola into symmetrical halves. Okay. The point where the axis cuts the parabola, that point is called the vertex of the parabola. Okay. So this is axis. This is vertex. This is focus. This is directrix. A line segment joining any two points on the parabola is called a chord. So if I connect any two points on it. Okay. So this is a chord. Okay. We'll talk about chord also a little later on. Okay. If the chord passes through the focus, I'll use some different different colors for everything. Okay. If the chord passes through the focus, then that chord will be called as a focal chord. So this is a focal chord. Okay. A chord which is perpendicular to the axis. Let's say I draw a chord like this. Okay. This is perpendicular to the axis. This is called as a double ordinate. D, U, B, L, E, double ordinate. R, D, I think I'll be double ordinate. So double ordinate is a chord which is perpendicular to the axis of the chord. Okay. Next. The most important term that you'll be in a hearing from your school teachers or in the exam also is asked, latus rectum. Okay. So what is the latus rectum? Latus rectum is a focal chord which is also a double ordinate or vice versa. It's a double ordinate which is also a focal chord. That means this angle is a 90 degree and it is passing through the focus also. This word has come. This is a Latin word. Latus means side. Rectum means line. It's a side line. Okay. So some names were given to these geometrical figures way back very early when maths was evolving and those same names have been carried forward even till date. Okay. Just to honor those initial mathematicians who named them. Okay. So this is called the latus rectum means side line. Okay. Later on, you'll come to know something about latus rectum. That latus rectum is the shortest focal chord in a parabola. That means there is no focal chord shorter than the latus rectum. Okay. So latus rectum is the shortest focal chord in the parabola. So a few terms or a few things that I would like to make you note down so that you are clear about it. Let's say this is endpoint. Please note down number one. The vertex, vertex is the midpoint of S and N is the midpoint of S and N. See, you've seen the diagram. This is your S. This is your N. Vertex is always a midpoint. Okay. Second thing, axis is always perpendicular. Axis is always perpendicular to the directrix and focus always lies on the directrix. Sorry. Focus always lies on the axis. Okay. So remember, axis will always contain the focus. Okay. Vertex and it will be perpendicular to the directrix. Latus rectum is the focal chord, which is also a double ordinate and vice versa. Also a double ordinate. And later on, we'll prove this in today's session. It is the shortest focal chord. So latus rectum is the shortest focal chord of the parabola. Okay. Any questions? Any concerns? Now having known these terms, please note that the same terms will be carried forward to our other conics as well, like ellipse and hyperbola. So if you have any doubt with respect to any term, why it is called double ordinate? Many people ask this question. It is because the endpoints of the double ordinate have the same y coordinate just differing in sign. That is why it is called double ordinate. Because the endpoints of the double ordinate are symmetrically positioned about the axis. Okay. That's the reason why it is called double ordinate. Okay. So now with this, we are going to introduce the elements of coordinate geometry to our chapter. So we are ultimately studying coordinate geometry. So equations, et cetera, are bound to come. Okay. So we have to study about equations, coordinates, et cetera. So I'll be starting my today's discussion with standard parabola. Equations of standard parabola. By the way, parabola or parabolas, both are acceptable word for plural of parabola. Okay. You can use both of them. So there are four standard cases that we are going to discuss today. But first of all, why they are called standard cases or which cases are called standard cases? So first of all, standard cases are those cases which will be used most frequently in deriving certain property. As I already told you, standard cases are like frogs in the laboratory. So we do so much of experiment on these animals, on frog, lizard, et cetera, in the bio lab, right? Because they're easy to handle. They can easily be found out. Okay. And they help us understand the human anatomy as well. So standard parabola is the most simplest form of a parabola. And many a times any property or something related to a generic parabola, we derive on a standard case because easy to manage. As you can see from their equations when we derive it, they look very simple and hassle free. Okay. So standard parabolas are those parabolas whose vertices is kept at origin. Okay. And axis is or axis is either of the coordinate axes. That means those parabola whose vertex is at origin and axis is either the x-axis or the y-axis. Those cases are basically called standard cases. Remember when we did circles, I had also told you the standard case for a circle. So circle whose center is at origin that is called standard case. In our parabola, the standard cases are those cases where the vertex is at origin and your axis of the parabola is either along your x-axis or along y-axis. So as a result of satisfying these two conditions, four standard cases actually stand out or four standard cases get formed. So let us take our first standard case. This first standard case, I call it as a right word opening parabola. Now this is a word which I use. You might not see this word in your in your textbooks. Okay. So I call it as a right word opening parabola. So how does it look like? So let me again draw a parabola for you. Okay. So I just chop this arm off. Okay. So what do we do in this case? We keep the vertex at the origin. In other words, our axis, our axis, we will call it as let's say the x-axis. Okay. So let's say this is our axis of the parabola. And I'll take it as my x-axis. And this is origin. Okay. So if this is origin, you're bound to have your y-axis going like this. Okay. So this is your y-axis. Okay. And you can also show your direct x also somewhere over here. So direct x is also okay. Now in order to write the equations, we need to use our locus condition. What is the locus condition? That any point on this particular parabola, its distance from the focus should be equal to its distance from the direct x. So for that, at least have to fix where is the focus. Now since focus lies on the x-axis, I will consider some point a comma zero to be focused. Okay. Now please understand that this a term that we are using is representing a distance over here, the distance between the vertex and the focus. Okay. So this represents the distance between the vertex and the focus. And that is why we always take our a value as positive. So this is very, very important. Right. You rarely find me writing VVIP, IMP. Okay. So I normally write important or not. But this is very, very important to note that a represents the distance between the focus and the direct x. And hence a is taken to be positive always. Never make a mistake about it. Okay. So many people will put a point on the negative side also when we talk about leftward opening parabola and they start calling it also as a comma zero. No, that point should be called minus a comma zero because a is always taken to be positive because it represents the distance between the focus. Sorry, what did I write? Focus and the vertex. Focus and the vertex. Sorry. Okay. Is it fine? Now, since origin is the midpoint of n and s, it's automatically understood that n will become minus a comma zero. And since your directrix is perpendicular to the axis and axis is your x-axis, that means this line becomes parallel to y-axis. So this line equation will become x equal to minus a or you can say x plus a equal to zero. Yes or no? Yes or no? Ashleisha, blue line is the directrix. White line I've already written over here y-axis. If you want, I can write directrix on top of it. Got it? All right. Now, we will use our locus definition to get our equation of the ellipse. Sorry. Why did I say ellipse? Oh, yesterday I had a class on ellipse. That's why. Sorry. Okay. So let's take a generic point h comma k. This distance and this distance should be equal. Okay. So as for the definition, sp should be equal to p m. Okay. That means this ratio should be equal to 1. Now, what is sp? Sp, as you can easily use your distance of two points formula, x minus h minus a the whole square k minus zero the whole square under root equal to p m. Now, what are the distance of a point from a line? We already know that in our straight line chapter, we had done it. So we put the coordinates of the point in the left side here, x plus a, we take a mod of that and divide by the squares of the coefficient of x and y, which will give you mod h plus a by under root of one square. Isn't it? Anybody has any doubt with the right side expression to let me know? Anybody? Okay. So once we have got this, let us square both the sides. By the way, root one square is as good as a one. So you can skip that part and just write h plus a the whole square. Remember, mod h plus a whole square and h plus a whole square means the same thing. Okay. Now just expand it. You'll have x square a square minus two h. This will be x square a square plus two h. You will end up losing, I think, x square a square. Yeah. And you'll end up getting k square is equal to four a h. So once you have got it, generalize this. Generalize this by writing h as x and k as a y, which leads to our equation of the first standard case, or you can say a rightward opening parabola as y square is equal to four a x. As you can see, this equation seems to be so simple. And that's the reason why it is basically used for finding or proving many properties in parabola. So they're called standard cases. Is this fine? Any questions? Any questions? Any concerns? Okay. So once you have found it out, let us also find out the length of the lattice rectum. Okay, let us also find out the length of the lattice rectum. Maybe I'll take a separate diagram here. Let's find out the length of the lattice rectum. Okay, I'm already showing you the x axis and the y axis. Okay, directorates I need not show, it's not required. And let's say this is our lattice rectum. So lattice rectum will pass through the focus and perpendicular to the axis. So let me call it as L r. This point is your a comma zero. So how do you find L r length? Very simple. Let's say L coordinate is a comma something. Okay, please note, since the lattice rectum is parallel to the directrix or y axis, you can say on the lattice rectum, your x coordinate will always be a. Okay, so I can easily take the L coordinates to be a comma k. Now a comma k will satisfy your equation that we have just now derived. So this satisfies y square is equal to four ax. So in place of in place of y put a k in place of x you put a so you end up getting k square as four a square, which means k is plus minus two a. Okay, now, which is plus two a and which is minus two a, which is plus two a and which is minus two a? Who will tell me? Is L coordinate to a or is r's coordinate to a? Who will tell me? And why? Why not this is minus two a? Why not L is having an ordinate of minus two a? Right, because as I already told you, and I wrote VVIP, IMP also for that, your a is always positive. So since you're above the x axis, the above one is two a, the below one is minus two. So in light of this, you can easily figure out what is the length of the lattice sector. So the length of the lattice sector will be four a units. Four a units. It's not like first quadrant is bashed now again, same mistake you're doing. Whether first, second, third, fourth, seventh, eighth, hundredth quadrant, sorry for that hyperbola, but a will always be positive because a represents the distance between the vertex and the focus. It doesn't depend on the quadrant. Let it be clear here. Okay, the reason I wrote VVIP for that is because despite telling people make mistakes and that cost them the problem. And that costs them the problem. Okay, so please make no mistakes about it. Is this fine? Any questions, any concerns with respect to the first standard case, which is your right word, opening parable. Okay, now this image that you have, this should be implanted in your mind. That means you should clearly know when you close your eyes, where is the focus? What is the equation of the characteristics? What is the length of the lattice sector? Where is the vertex? Which is your axis, etc. Everything should be there in your mind. The moment you close your eyes and imagine this, everything should be clear. Okay. So, with this, we can now move on. Yes, yes. Sure. So, with this, we now move on to the second standard case. Yes, any questions? Lattice sector, first of all, it is lattice sector times the equation of the parabola is a coincidence. I didn't get that question. You're saying the coefficient of x here is the lattice sector. That's what you're trying to say. Yes, actually, it's not a coincidence. There's an alternate definition. We will talk about that little later on. See, equation of a parabola is actually like this. But I don't want to discuss about it because it is beyond your school level right now. Equation of a parabola is nothing but it's any point on the parabola. Okay. It's distance from the axis, distance from the axis square is lattice rectum times distance, distance from the tangent at the vertex, distance from tangent at the vertex. Later on, we'll discuss about it. See, when you wrote y square is equal to 4x. So, y is what? y is actually, let's say this point is x, y. So, y is the distance on the axis. So, you're squaring that up. See, distance on the axis square is equal to 4a. 4a is actually the length of the lattice rectum that you were telling Sharduli. And what is x? x is nothing but it's the distance from the tangent at the vertex. Okay. So, that x that you wrote is the distance of the tangent, sorry, distance of the point from the tangent at the vertex. Okay. Point from the tangent at the vertex. But right now, we are not going to discuss it because it's anyways a pattern parcel of our regular chapter when we do it for competitive level exam point of view. But I think you identified it. So, that's why I am discussing it. So, this y is actually distance from axis. Okay. So, many times this definition is used when somebody provides you with a, you know, let's say some orbit axis equation and let's say it provides you with the equation of a tangent at the vertex. So, this equation can also be utilized. We will discuss about it later in more detail. Now, we are going to talk about the second standard case. So, second standard case would be a case where the vertex is at origin. Okay. The vertex of the parabola is at origin. The axis is still the x-axis. But this time, the focus, let me make a white line. Yeah. This time, the focus of the parabola is on the negative x-axis. So, as you can see here from the diagram, this is your positive x-axis, positive y-axis. So, this is a leftward opening parabola. And the focus here is at minus a comma zero. Now, see, I could have easily wrote a comma zero also where I would say a is a negative quantity. But as I already told you, we always take a to be a positive quantity. And hence, I have to put this external negative sign to show that the focus lies on the negative x-axis. Okay. So, the vertex here is still at origin. And the directrix will be, I'm just going to make them very quickly. The directrix is going to be x equal to a line. Okay. So, for such cases, you don't have to derive the equation once again. However, if you want to, you can. But why to waste time when we have already done transformation of graphs in our bridge course. So, you tell me, how is this equation of this parabola related to the equation of the previous parabola? Of course, after observing that one is the mirror image of the other about y-axis, how would you get this equation without deriving it from scratch? What will you do? Write it down on the chat box. Replace x with a minus x. Absolutely, Sethu. So, in the previous parabola, if you replace your x with a minus x, you end up getting the equation of this parabola. Because whenever a graph is reflected about the y-axis, we change the sign of x to a minus x. Is it fine? Any questions? Any concerns? So, please note this down. Critical things are equation of the directrix, which is x equal to a focus knob becomes minus a comma zero, vertex is zero zero still. Equation becomes y square is equal to minus four x and there would be no change in the latter-sector length. Ladder-sector length will be still four a units only. So, just taking a mirror image about y-axis doesn't change the dimension. So, many people also want to know that is ladder-sector always four times the distance between the vertex and the focus. Yes. So, this can be noted down. Note that in fact, I would write it in the very first slide where we talked about ladder-sector for the first time. I think we had taken here. So, ladder-sector is always four times the distance between distance between focus and the vertex. Is that fine? Any questions? Any concerns? All right. So, in the same page, I will take the other two cases as well. So, altogether there are four cases. Okay, two of them we have already taken care. Can you go back to the slide to the LR node you just gave four times, so four times the distance between I'm sorry, I think it is on slide number four. Four times the distance between the focus and the vertex. Copy it. So, two cases we have already done. Now we are going to talk about the case where the parabola is going to open upwards. Okay. This is the most commonly seen parabola by us. We have also seen this in our polynomial function chapter. Okay. So, this is an upward opening parabola. The vertex is still at origin. So, let me just showcase your coordinate axis here. This is your y-axis. This is your x-axis. Okay. y-axis, x-axis. This is your... So, in this case, the directrix equation will become... Let me just write down the critical points. Focus is 0 comma i. Directrix is y equal to minus a. Y equal to minus a. Okay. Now, in this case, if I ask you this question, that using our first standard form, can you get the equation of this standard form? What will you do for that? How is this figure and the first standard parabola related to each other? Right. Absolutely. Please note that this figure is nothing but the first standard case turned a 90 degree. Now, turning a 90 degree is equivalent to saying taking reflection about y equal to x-line. Okay. Now, see here, I'll show you in a very small diagram here. This is your y square is equal to 4x. Right? Let's say I make a line like this, which is y equal to x-line. And I start taking the mirror image of this white parabola about this blue line. Assume that this blue line is lowered on both sides. So, this bulge, as you can see, I'm moving my cursor on it or pen on it. This bulge will look like this. Okay. This will look like this. And this fellow will look like this. So, ultimately, you'll end up getting an upward opening parabola. Okay. So, you don't have to derive the equation from scratch. So, in y square is equal to 4ax, just swap the positions of x and y. That is how we normally reflect any curve. So, if any curve is reflected about y equal to x-line, in that for the reflected curve, we just swap the position of x and y. Okay. So, this will become x square is equal to 4a y. Please note this down. And later-sector length will still remain 4a units. No impact on the later-sector length because of this reflection. Is it fine? Any questions? Any concerns? Okay. The fourth case is where your parabola is opening downwards. That also I will show it here itself. So, vertex is still at the origin. Okay. This is your vertex. The focus now becomes 0 comma minus a. Okay. And this becomes your directrix. Directrix is y equal to a. y equal to a. So, for such case, as you have already seen, change your y with a minus y because you're reflecting something about the x-axis. So, when there's a reflection about the x-axis, we normally change the sign of y. And again, in this case, later-sector length still remains to be 4a units. Okay. So, all together, we have four standard cases of the parabola, y square is equal to 4ax, y square is equal to minus 4ax, x square is equal to 4a y, and x square is equal to minus 4a y. And let me tell you, out of these four cases, the very first one is the more common one. The y square is equal to 4ax. Okay. Many experiments which we perform in parabola, we perform it on y square is equal to 4ax. Okay. Charlotte has a question. If anything is reflected about, say, y equal to 3x sign or, okay, will y be reflected replaced with 3x? Okay. Now, see, if anything is reflected about y equal to 3x sign, we have to use our concept of a rotation by a matrix. Okay. Normally, this matrix we'll be talking about in class 12. Any y equal to tan x theta, there is a rotation matrix which is associated. Okay. And that rotation matrix has got 2 theta into it. Are you getting my point? So what do we do? Let me just tell you very straightforward what happens actually. When you're reflecting something or when you're rotating something, okay, by an angle of theta, you are actually multiplying it with something like this. A matrix comes into picture. Okay. So this matrix, okay, this matrix, in fact, so this matrix is responsible for giving you the new coordinate. Okay. So this matrix is what we call as the, you know, rotation matrix, rotation matrix. Okay. By the way, I've written it for a point, for a curve, it will be minus will come on this side. But this is something which you will not understand right now, because it requires your understanding of linear algebra, a bit of linear algebra, which means some matrices information is required. So if you're reflecting a curve about y equal to 3x, 3 becomes 3 becomes your tan theta. Okay. And with that theta, you have a rotation matrix state which gets formed. Let me tell you, it will have two theta into its angles. So what you are saying that will not exactly hold valid. Okay. That is going to be going to give you a completely different result. Okay. That's not going to be as simple as replacing y equal to 3x. So a matrix gets formed rotation matrix, which has got twice of the angle by which you want to rotate it that involved there. Okay. I'll give you the result in some time. Maybe after the break, I'll give you that result. What does result give you? Okay. Yes. See what happens? Yes. In our result, theta was 45 degrees because you're rotating it about y equal to x line. So cos theta minus sin theta, sin theta cos theta, when it gets multiplied, it basically sends that point to a, you know, sends that point. You know, it rotates it. So that two theta will come into picture. So two theta will be 90 degrees. So cos 90 degrees, sin 90 degrees, minus sin 90 degrees, cos 90 degrees will come into picture. So if you know a bit of matrix multiplication, you can realize that some point, let's say a comma b will become b comma a in that case. But in general, we will not like to waste everybody's time over here. We will talk about it. In fact, if you want to see that, you can watch one of my videos on matrices, especially the last session of the matrices where I've talked about, how do these points, how do the curves get transformed when you're reflecting it about certain line? But it is definitely not the way you're thinking it to be. Okay. So the next thing that we're going to talk about is questions. So first, first type of question, which I'll be asking you is your understanding of the locus definition. Let's take a question which is based on the locus understanding of your parabola. Here's the question. Find the equation of the parabola whose focus is at minus one comma minus two and the directrix is a straight line. x minus two y plus three equal to zero. Decide this question and give me a response on the chat box. Yes, anybody any success? I can give you a response on the chat box. Okay, okay. Anybody else? This is just based on your locus definition. That's it. Nothing else you have to do. Very good. Anybody else? All right, let's discuss this. This is very simple. See, if I just draw the scenario on the coordinate axes, so there is a directrix x minus two y plus three equal to zero. If you just guess this up, it's going to be a line like this, okay, having a slope of one by two and cutting the y axis at three by two. Okay. And there's a point minus one comma minus two, maybe somewhere over here. Okay. So your directrix is in this fashion and your focus is at minus one comma minus two. And let's say this is your, this is your parabola. I'm just drawing a rough sketch of a parabola. Now, every point that you take on the parabola would be satisfying the fact or will be satisfying the locus definition that the distance of P from the focus should be same as the distance of P from the directrix. Okay. Let me draw this S slightly closer to the point. It's looking somewhere in between. Yeah, this is your S. Okay. So if you make up, if you make this locus condition get satisfied by this point P, let's say H comma K, you can write SP. SP would be under root of H minus minus one, which is H plus one K minus minus two, which is K plus two equal to PM. Now PM then would be nothing but distance of H comma K from this given line, which will be H minus two K plus three mod by under root of one square minus two square. Okay. Noel, we'll check. So take this term to the left side and square it by the way, this is root five. And if you square it, you will end up seeing something like this. When you expand it, by the way, you already know that I don't write everything. I expand it, you know, just by finding the coefficients of each type of term. So let us first focus on X square. So five X square will come from here. X square will come from here. So you'll be left with four X square. You'll have five K square and you'll have four K square. So you'll be left with the K square. And here you'll end up getting minus four HK, which when comes to the right side, left side will give you four HK. Then from here, you'll have 10 H and from here, you'll have six H. So that will leave you with a four H on the left. On this side, you will have 20 K. And this side, you will have minus 12 K. So 20 and minus 12 will add, so it will become 32 K. And constant terms will be 25, 25 minus nine, which is 16. Okay, so now you can generalize it by replacing your H with an X and K with a Y. That leaves you with the final answer as four X square plus Y square plus four X Y plus four X plus 32 Y plus 16 equal to zero. Now this question was given as an eye-opener question to many of you who actually just remember the standard form not knowing from where it has come. So it has come from a locus condition and locus condition is something that helps us to find answers to any general form of a parabola as well. Another eye-opener here was, oh, the parabola equation can get so ugly also. I always thought it was Y square is equal to four AX or Y square is equal to minus four AX or X square is equal to four AY. Please understand, those are standard forms. Those are the simplest form of a parabola. Okay, so your parabola as any other conic can have all types of terms, X square, Y square, X, Y, X, constant, Y, etc. So don't make a false image in your mind that whenever somebody says parabola, you always think of Y square is equal to four AX. So that's a very, very special case. Is it fine? Well done. I think most of you got it right, but let me see who was the first person to get this right. Nikhil was the first person to get this right, followed by Arav Sena. Well done, guys. Setu, I don't know what you have written also. Setu, you have done the same thing which I told many people do. All parabolas are not Y square is equal to four AX. So it's a simple case out of a lot of cases. Okay, are you getting my point? So let's take a few other types of questions. Many examples that are present over here, they are basically related to, okay, so I'll take another case here. Find the position of the vertex, focus, equation of directrix, length of lateral sector for the following cases. Number one, Y square is equal to X. Number two, Y square is equal to minus AX. Number three, X square is equal to four Y. And number four, X square is equal to minus 16 Y. Okay, so for each of these cases, please note that I have given you purposely the simplest one, the standard cases. So for each of these cases, we need to write down the four things here. Please do it and give you a response in one go. Yes, I'll talk about those necessary conditions. Nikhil, in our pair of straight lines chapter, in a pair of straight lines, when we have completed all our conic, I will talk about the generic condition for our second-degree equation to represent a, for a second-degree equation to represent a parabola. Okay, there is something called asymmetric determinant. You will read it in some time. Okay, so that all conditions we'll be talking about in that chapter. Yeah, sorry, I think there was a power cut because of that. You lost the touch. Can you all hear me properly? Okay. Okay, excellent. Let's discuss it one by one. Let's discuss it one by one. Alright, so for the first case, if you just compare it with Y square is equal to four AX. It means your four A is one. So your A is one fourth. Okay. And now where's the vertex? So for the first one, vertex will be at origin because it is a standard case. Correct. Focus will be at A comma zero. Let me write that down. Focus will be at A comma zero. Okay, next is your equation of the directrix. Equation of the directrix will be X equal to minus A. Equation of directrix. X equal to minus A, which is minus one fourth. And finally, you have latter-sector length, which is four A units. Four A unit means one unit. Clear? Any questions with the first one? Okay, so let me write it like Roman number one. So for the second one, are you all ready with your answers? Should we discuss the second one or you want some more time? Done? Should we discuss it? Okay. Yeah, so let's discuss it. So for this case, you have to compare it with Y square is equal to four AX again. Sorry, minus four AX again. So who will tell me what is the A value in this case? For the second case, what is the A value? Ayo, Noel. I'm sure you did that to frustrate me, right, Noel? I keep on saying this. I'm repeating it again. A is positive always. Always positive. No minus two business. Please note that. Okay, so A here is two. So you have to compare it with minus four AX. So A is two. So the moment you know that it is Y square is equal to minus four AX. You know, it's a left-world opening parabola. So vertex is still at origin. Okay, vertex will be still at origin. Focus will be, I'm so sorry, second one. Yeah, focus will be minus A comma zero. Okay. Equation of the direct X will be X equal to A. Okay. And your length of the ladder spectrum will be four A units, which in this case will be equal to eight units. Is it fine? Any questions? Any questions, any concerns? Next. X square is equal to four Y. Are you all done with the third one? Let me write it here. X square is equal to four Y. Yes. They're all standard cases. So you realize the vertex will always be zero, zero. Okay, by the way, here A is one. So focus here will be now close your eyes, everybody. Everybody close your eyes and imagine that there's an upward opening parabola. Where should be the focus? Zero comma A. So it'll be zero comma one. Correct? Equation of direct tricks will be Y equal to minus A. Or you can say Y plus one equal to zero. And the length of the ladder spectrum will be four A units, which is equal to four units only. How do we know A is one? Compare this with four A Y. So four is equal to four A. A is one. Last case. X square is equal to minus 16 Y. So before you start, I would like to ask you what is A value here? Write it down on the chat box. Thanks for writing that extra plus chart, will you? A is, Noel? Yeah, Noel, don't make that mistake again. So A is four. Good. Wow, minus, minus four. Put two more. No, minus, minus, minus, minus. Okay, now let's close your eyes. Everybody close your eyes and imagine a downward opening parabola. Okay, so everything related to that should be appearing in your mind. Vertex was zero, zero. No issues with vertex. Focus was zero comma minus A. Correct? Equation of the direct tricks, equation of the direct tricks will be Y equal to A. And the length of the ladder sector will be four A units, which in this case will be 16. So this number itself is the length of the ladder sector. Okay. So these are all simpler cases, which probably your school exams or unit tests you will be asked. Okay. So we'll quickly make our move towards the shifted form of a parabola. Okay, we'll discuss about shifted form of a parabola. And we will also do a question or certain set of questions related to shifted parabola. That will be more challenging than this. Is it fine, any thing that you would like to ask, copy, get clarification on, do let me know. So shifted parabola. Now, let me guess your understanding of your coordinate geometry or review of coordinate geometry a bit here. Let's say I had a standard form of a parabola. Y square is equal to four AX. Let me draw it. This was your y square is equal to four AX. Now, the same parabola, the same parabola now shifts to this position. Okay, the same x axis, y axis. Now the same parabola shifts to this position. That means it becomes like this. Let me change the color. I have already used green for the axis. Maybe I'll use a blue one. Now, my question to all of you here is that if the vertex is alpha comma beta, if the vertex here is alpha comma beta, where did you shift your origin such that this parabola, the white parabola became the blue parabola? Think carefully and answer. I'll repeat my question once again. If you want to make your white parabola as the blue parabola, where did you shift your origin? Wrong, Sharguli. So how would you say that, Arav? You're on the right tracks. So where has your origin gone? Right, Nikhil. Your origin has actually gone to minus alpha minus beta. Right, please note the curve doesn't shift. What shifts? The origin shifts. So if you want to make that curve go up and right, let us say, I'm taking a sample here, then your origin should go exactly the same units down and left. So as a result, see your origin was at vertex over here. Now your vertex became alpha beta. That means origin would have gone to minus alpha minus beta. So because of this, what will be the effect on the equation of the parabola? So how would this get transformed? What would be the new equation? Who will tell me? If origin shifts to minus alpha minus beta, what will happen to the equation y square is equal to 4ax? What would be the new equation? Tell me. Write it down on the chat box. What are you going to change your x with? What are you going to change your y with? Wrong, Nikhil. Wrong, Sharguli. Guys, I need to revisit now, shift shifting of origin concept. I think you have all forgotten that and with no fault of yours because in school and all, we don't practice any questions based on this. See, everybody, please pay attention. If your origin shifts to a point, and I'm just doing a recall for you all, review your coordinate system. If your origin shifts to h comma k, how does your x comma y get changed? Let's say x comma y becomes capital x capital y. So what is your capital x? Capital x is small x minus h. Or in other words, x is small x is capital x plus h. And how does your y get changed? Capital y is small y minus k. That is to say small y is capital y plus k. That is how the transformation happens to a point. How does a curve get changed? So let's say there's a curve whose equation is this. So this is the old equation. What would be the new equation here? What would be the new equation here? So you will say the same thing. Change your small x with capital x plus h and change your y with capital y plus k. This will become the new equation. Right? Now here, mind you, you shifted your origin to minus alpha minus beta. So minus alpha is your h. Minus beta is your k. Are you getting my point? So your small x, let me write the equation again. So your small x will be changed with x plus h, which is x minus alpha in this case. And your y will be changed with capital y. Let me write it in capitals only just for the timing. And your y will be changed with y plus k, which in this case will be capital y minus beta. But as I told you in that coordinate geometry class also, it is not a good practice to write equations in capital alphabets. What do we do? We write it in small alphabets, small alphabets or small letters. So it becomes y minus beta the whole square for a x minus alpha. This becomes your new equation. Or you can also apply your graphical transformation themselves that if a curve moves, let's say, assuming that it is moving right and up. So right means x will get changed with x minus that quantity by which it is moving right or that many units by which it is moving right. And if it goes up, y will be changed by y minus that many units by which it is moving up. Is this fine? Any questions? Prashim also, that is not correct. Is it fine? Any questions? Any concerns? So you have to be careful while doing this concept because right now it was a simple case. You could have managed here. But when it comes to your complicated problems, especially when it comes to the problems related to homogenization, which you will study in pair of straight lines, any such mistake will lead to a waste of lot of time effort. And of course, you'll get a wrong answer. So be very, very clear with respect to your shifting of origin. How does it impact an equation? How does it impact the coordinate of a point? Is it fine? All right. Now, many types of questions can be framed on this. They may give you a shifted version of a parabola and they may, they might ask you for vertex coordinate, focus coordinate, equation of directories. They may also ask you for the equation of the axis, lattice system, et cetera, et cetera. Okay. So let us take an example based on the same and try to understand from that example that if a question comes where my parabola is a shifted parabola, then how do I figure out its vertex, focus, equation of the directories, equation of the axis, length of the lattice system, and so on and so forth. Okay. So let us learn through an example. So can we move on? By the way, the same concept can be applied to any standard form. So I'm not separately talking about y square is equal to minus 4x or x square is equal to 4a y or x square is equal to minus 4a y. In all those, you have to apply a similar strategy. Okay. Is it fine? Okay. By the way, just to ensure everybody has understood this concept, I would, I will take a problem here. Before I go to the next page, I'll take a problem. Okay. Let's say there was a parabola which was opening upwards. I'm making it as a dotted one because I don't need it actually. So its equation originally was x square is equal to 8y. I shifted this parabola now to this position. Okay. Same parabola. Earlier it was at this position. Now it has come to this position. Okay. And the vertex now becomes 2 comma 5 comma 2. Okay. Tell me what would be the new equation? What would be the equation of this fellow? Write it down on the chat box. Sir, what sir? We have done this in our graph and its transformations. But no, I mean, some of you have forgotten it and that's, you know, painful for me to learn that. So I want to reconform myself whether you are all aware of this. Okay. So please once again, for my sake, please do this question and tell me the answer. Thank you. Very correct. Yeah. So it becomes x minus 5. The whole square is equal to 8y minus 2. Excellent. Okay. Try one more, just one more. So let us say there was a parabola which was initially a downward opening parabola. Okay. And let's say its equation was x square is equal to minus 16y. Now this parabola has been shifted to such a place that its vertex has now become minus 3 comma 4. Okay. So what is the equation of this fellow? No need to simplify and tell all that, Situ. It's fine. In raw form, you give me. Don't worry. Sharduli, a small mistake you have done. Correct question. So it will be x plus 3 the whole square is equal to minus 16y minus 4. Okay. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Some of you have messed up with the science. So please be careful. Okay. Okay. Anyways, so my agenda was to discuss that if somebody gives me a standard form shifted, that means it shifts a standard form and ask me for the vertex, focus, direct x equation, etc. How to do those type of questions. So let's take a sample question here. Find number one, vertex, focus, equation of direct x, length of ladder symptom, or before that equation of axis, length of ladder symptom for, for, I'll just take a case here. Let's say 8x plus 2. Okay. Now, there are two ways to solve this question. One is a graphical route. But trust me, from my experience, I figured out that the graphical route is going to be a lengthy way to solve it. Okay. So many people, what they do, they make a graph. Okay. So let me do that way also so that you can get an idea that how lengthy or how cumbersome that approach can be. So the first route is a graph route. So you basically look at this equation and you figure out that, oh, this was actually y square is equal to 4x case shifted somewhere. Of course, here a value is going to be two, right? Yes. So it was originally y square is equal to 8x. So I just make that graph also for you. Dot it because that is not our actual graph. This was the old graph or old equation. And now this equation got shifted. And that is why this equation became this equation. And that is why this equation became this equation, right? So now tell me where did it get shifted? Now, looking at it, you will be able to figure out very easily that it got shifted to minus 2 comma 3. So minus 2 comma 3 is somewhere over here. Let's say, okay. So your original equation graph would be appearing like this. Okay. I'm just making a dummy diagram here. Don't be judging. Don't judge it by its accuracy. Okay. Just a dummy figure. So this vertex has now come to minus 2 comma 3. Okay. So first part answer is done. It's minus 2 comma 3. Second part, where is the focus? Now you already know that focus will be located on the very same line and at a distance of a units from the vertex. Okay. So this is minus 2 comma 0. Where do you think should be the focus? Where do you think should be the focus? So two units from here, if you go, this becomes your focus. So focus will become 0 comma 3. Am I right? Correct me if I'm wrong. Correct? Yes or no? So second part of the question also that next equation of the directrix. So please note directrix would be, again, let me just sorry for cutting through this point. I'll put it on top, by the way. Yeah. So directrix would be a line, which is going to be a line parallel to the y axis and passing through a point, which is again two units from here. So this is minus 2 comma 3. So this point will be minus 4 comma 3, isn't it? So can I say this line would be x equal to minus 4? So this answers your third part. Directrix equation is x equal to minus 4. Yes or no? Yes or no? Now, axis equation will be y equal to 3 because it has to pass through the vertex, the focus, whose y coordinate is always 3. So y equal to 3. So that answers the answers the fourth part of the question also. LR length, you already know it's 4A units, so which is 8 units. Now, many people find this method slightly cumbersome because they have to first make the graph and you have to account for the units here and there. So people find it slightly cumbersome and difficult to handle. But don't worry, I have an easier approach where you would not require any graph at all. So you don't have to scale the graph at all and still you can find all your answers exactly to the figures which we are working. So let us see the second approach. So this is our first approach through the use of a graph. Second approach is where I will not use graph at all. So what I will do here is I will do a role change activity. I call that activity as a role change, role place. I am sure most of you when you were small, you were all trying to copy something, right? I was like very infatuated by a police officer. So I was always like, take a dummy gun and do discount, discount everywhere. So I will be like playing a role of a cop. That's how many kids also play, isn't it? So if you move around your apartment, you will see small, small girls, they will play some kind of a role. Someone will become mother, someone will become daughter, auntie, uncle like that and all. So the same strategy I am now going to adopt to solve this question. So I am going to say that this guy y minus 3 whole square is like this. So this capital Y is basically playing the role of y minus 3 or you can say y minus 3 is playing the role of capital Y and x plus 2 is playing the role of capital X and of course, 4A is 8, so A is 2. Okay, now what I will do, I will recall everything that I know related to this parabola. So just close your eyes. Imagine y square is equal to 4Ax. So imagine where was the vertex? Imagine where was the focus? Imagine what was the equation of the directrix? Imagine what was the equation of the axis? Imagine what was the lattice vector? Once you've imagined, let us try to solve this. So vertex, if you close your eyes, vertex was at 0, 0. So what I will do here, I will instead of writing 0, 0, I will write x is 0, capital X is 0 and capital Y is 0. And then I will make a role change. Capital X is actually small x plus 2 and capital Y is small y minus 3. I am going to equate it to 0 and solve for my x and solve for y. So minus 2 comma 3 becomes your vertex position. Did we get the same answer here? Yes, minus 2 comma 3. Okay, so this is all you need to do. There's nothing extra that is required when you are using this method. Second thing, focus. Again, close your eyes. Focus was at a comma 0. So instead of a comma 0, just write one extra step x equal to a and y equal to 0. And now change your capital X with x small x plus 2, change your a with 2, change your capital Y with small y minus 3 and 0 is 0. So solve for x, solve for y. So 0 comma 3 becomes your focus. 0 comma 3 becomes your focus. Yes, 0 comma 3. I got it again. Next, try this on your own. Directrix. You know directrix for this curve, capital Y square is equal to 4ax is capital X equal to a, right? Sorry, capital X equal to minus a. So just make a little change now. This is x plus 2 and this is minus 2. So it becomes x equal to minus 4. This becomes your directrix. Did we get the same answer from graphical root as well? Let's check. Yes, same answer x equal to minus 4. Okay. Now equation of the axis, equation of the axis for y square, capital Y square is equal to 4ax is your x axis. x axis is capital Y equal to 0. Now capital Y role is being played by a small y minus 3. So this answer is your equation of the axis. Did we get the same answer with the graph approach? Yes, see y equal to 3. Okay. Next, length of the ladder sector that is pretty straightforward. That will be 4a units. That is nothing to do with this thing. So 8 units. So the, this method is very simple to apply. You don't have to make any kind of a graph. You don't have to use your imagination. I would call this method to be, you can say very, very kind of a straightforward method. Okay. You don't have to sketch any graph. You don't have to imagine where to move, by how much to get to which point, etc. Nothing to be done here. Just a role change mechanism. Now I'm sure you would like to try out your hand at few more questions. Okay. So I will, I will take up few more questions, not to worry. Any questions, any concerns with this? So all you need to do is just imagine which parabola was shifted. So imagine your vertex, focus, equation of directors, whatever for that parabola and just do a role change. So just write a capital X equal to zero. Let's say zero zero was the vertex. So say capital X equal to zero, capital Y equal to zero. Just change your capital X in terms of small, capital Y in terms of small and get your small x and small y that will become your vertex. Same process is to be followed for other coordinates and equations as well. Okay. Let's take the same set of questions. So I'll just take a snapshot of this so that I don't have to write it again and again. Yeah. Find the following five for, yeah. And please put your responses one, two, three, four, five. Okay. Okay, Noel. Okay. You can't write for a, you have to give the value. Yeah. Anybody else other than Noel? Okay. Should we discuss it? So as I told you, compare it with the standard case which you have seen. Okay. It is just a shifted version of the standard case, correct? Where your capital X role is being played by X plus four, capital Y role is being played by Y plus one and A is nothing but four. Okay. Now close your eyes and remember everything with respect to this left word opening parabola. So first of all, vertex zero, zero, right? So zero, zero, you just have to write capital X and capital Y as zero, zero each and just equate capital with small X plus four and capital Y with small Y plus one. So X is minus four, Y is minus one. So minus four, minus one is your vertex. Okay. So if you've got this, absolutely right. Now the focus for this parabola used to be at minus A comma zero. So I will write capital X is minus A and Y is zero because the focus was minus A comma zero. Okay. So capital X is X plus four. This is minus four. Okay. Capital Y is Y plus one equal to zero. Correct? So X becomes minus eight, Y becomes minus one. So minus eight comma minus one will become your focus. Okay. Some of you have got the focus wrong. Some of you have got the focus wrong. Please check everybody. Those who have got it wrong, why it is wrong? Get it clarified. Okay. Next equation of the directrix here is capital X equal to eight. So X plus four is equal to four. That means X equal to zero is your directrix. X equal to zero is your directrix. Again, many people have got it wrong. Many people have got it wrong. Yeah. Next equation of the axis. Equation of the axis in this case for a left open parabola is Y equal to zero. So that will give you Y plus one equal to zero. Okay. And finally, the length of the ladder sector is four A, which is 16 units. Okay. Let me check. Noel, some mistake was there in your answer. Prashim was absolutely right. I think. Yeah. Karthik, some mistake in your answer. I think Setu also some mistake. Is it fine? Any questions? Any concerns? See, you need practice. You need a bit of practice. Don't worry. We'll take one more set of questions. Not too worry. Okay. Okay. Very good, Karthik. Let's take another one. Same data I want. X minus three the whole square is minus Y plus one. Please find the following for this parabola. This time be very, very careful. Okay. Don't make the mistakes which you made for the previous one. Yes. Any success anybody? Anybody who is willing to respond? Take your time. One minute. More I can give. Yeah. Okay, Setu. Okay, Prashim. Yes. Shall we not discuss it? Anybody else who would like to respond? Once there are extra few seconds to write 30 seconds. Okay. Fine. Okay. Chalo finally we'll discuss it up. So if you compare this, you have to compare this with X square is equal to minus four AY because your four A has to be one. That means A has to be one by four. So capital X whole is being paid by small x minus three and capital Y whole is being paid by small y plus one. Okay. Now close your eyes and start recalling where was the vertex? Vertex was at origin. So write capital X and capital Y as zero, zero each to signify origin. So this was zero and this was zero, which means X is three and Y is minus one, which means three comma minus one is your vertex. Okay. Next focus for a downward opening parabola focus used to be at zero comma minus a zero comma minus a. So capital X is small x plus three, sorry, small x minus three and capital Y is small y plus one minus a is minus one by four. Okay. So X is equal to three Y is equal to minus five by four. So three comma minus five by four will be your focus. Three comma minus five by four will be focus. Okay. Gone. Chargillie wrong. Pasham correct. Seethu is also correct. Araf Sinha also correct. Okay. Next. Third, directrix. Directrix for such case used to be Y equal to A. So Y plus one is equal to one fourth. So you can write it as four Y plus three equal to zero or Y is equal to minus three by four. Both are fine. This time I think most of you are correct in this. Okay. Good. Next. Axis. Axis equation used to be capital X equal to zero. So in this case, X equal to three will be your answer. Okay. This also most of you have got it right. Next. Length of the ladder is four A which in this case is four into one by four which is one unit. Is it fine? Any questions? This time I will give you one question which might you know, mislead you. Let's say the same set, the same five things, vertex, focus, directrix equation, axis equation, length of ladder for this parabola. Please try this out. Yes. Anybody? Okay. Seethu. Very good. Anybody else? So tomorrow we have a class again. Okay. In tomorrow's session, we are going to talk about, we'll be completing our mathematical reasoning chapter where very small part of it is left and we'll also talk about trigonometric equations. Okay. So that is another chapter which we had to cover under trigonometry. It has been removed from your CBC curriculum. I agree to that. But still, nevertheless, it is the important part for us for our cooperative exams. So we'll take that up. All right. So let's discuss it. I could see different, different responses coming in now. Okay. No, not to worry. See, actually, if you try to, you know, write this as a shifted case of a parabola, it must appear something of this nature. Okay. This is how a shifted parabola should actually appear. Let me write it like this. Plus minus 4A, y minus beta, something like this. But right now, what I see is that these, basically they're coefficient of x and y that you see here should be directly be one. I mean, indirectly it is something else, but directly it should be actually one. Okay. So I have to convert, first of all, my equation to something of this nature before I start working on it. So first of all, I would try to get this 2 away from x. So for that, I need to take 2 out. So but that 2 will come out as a 4 and that will leave me with something like this. Similarly, I want y to come over here. So minus 4 has to be taken common out. In fact, this 4 and minus 4, you can drop it out. You can do like this. Now, this is the form which I desire. This is the form which I had basically started learning this concept. So now this is where we'll start solving the problem by role change mechanism. So your cap, this is like capital X square is equal to minus 4 minus 4A y. So A here will be 1 by 4, capital X will be small x minus 3 by 2 and capital Y will be small y minus 3. Okay. Now close your eyes, recall everything that you knew about this parabola. So it's a downward opening parabola. So let's talk about vertex. Let's talk about vertex. So vertex was at 0, 0. So x minus 3 by 2 is 0, y minus 3 is 0. So 3 by 2 comma 3 is the vertex. I believe most of you have got it right. Next, focus. Focus for such case was x equal to 0 and y equal to minus A because it's a downward opening parabola. So this is equal to 0 and y means y minus 3 equal to minus A, minus A will be minus 1 by 4. Okay. So if I'm not mistaken, x value will be 3 by 2 and y value will be 3 minus 1 by 4 which is actually 11 by 4. So 3 by 2 comma 11 by 4 is your vertex. 3 by 2 comma 11 by 4, 3 by 2 comma 11 by 4. No prism, that is not right. Noel, that's correct. Karthik, not right. Setu, correct. Next, equation of the directrix. Equation of the directrix was for this case, y equal to A. So y minus 3 is equal to 1 fourth. So y is equal to 3 plus 1 fourth, which is going to be 13 by 4. So you can say 4 by minus 13 equal to 0. 4 by correct Noel, prism again wrong. Karthik, no. Setu, correct. Next, axis equation was, axis equation for this case was your capital x equal to 0. So that will make x equal to 3 by 2 or you can say 2x minus 3 equal to 0 also. Okay, latter symptom length, latter symptom length will be 4a units which is 4 into 1 by 4 which is 1 unit. Is it fine? Any questions? Any concerns? Why are we not taking minus y minus 3? What do you mean by that? I want to take in minus y minus 3. So 4 minus is this minus. 4a is 1, so a is 1 by 4 and capital Y is 4 by minus 3. In calculating the third part, for a downward opening parabola, directrix equation is y equal to A only. In this case, for our case, it will be capital Y equal to A because that's what I've taken up. What mistake you have done? Early, you have taken your capital Y as minus y minus times y minus 3, which is wrong. We can't take y like that. That minus is just signifying that it's a downward opening case. If this is not correct, don't expect your answer to be right. All the things will go for a toss other than the vertex probably because there's a 0-0 that will save you. Okay, so this type of, okay, now another type of question which is asked with respect to the same thing is that the questions it might not give you like this, like how I used to give in the previous few questions. So I was actually kind enough to give you the question like this. Don't expect the examiners to be kind enough to give you like that. They will actually give you questions like this. Let me show you. Okay, so they'll give you something like this. They will say, find the vertex focus ladder spectrum axis, directrix of this parabola. So it is your headache to complete the square and do the needful. Okay, so let's do this problem. So for this parabola, find the following vertex, focus, ladder-sector length, axis of the parabola, directrix equation. Parabola's general form is something which contains everything. X square will be there, Y square will be there, X Y will be there. Okay, so unlike in circle where there was no X Y term and the coefficient of X square and Y square were equal and kept as one, in case of a parabola, we don't have that kind of a restriction. Of course, there is a condition that will be told to you when I do pair of straight lines. Okay, but as such, there is no general form to answer your question. It is as good as a general form of any conic. Yes, any questions? See, first of all, I would like to ask you this question before you solve it. Is this a standard form of a parabola or a shifted form of a standard parabola or any parabola in general? That means it could be oblique also. Which case does it fall under and why? Is it a standard parabola? First of all, answer that question with yes or no. Is this a standard parabola? Yeah, we will come back to that. Is this a standard parabola? First of all, this answer yes or no for that. Say something, 23 of you are sitting, nobody knows the answer to this. It is not a standard form because standard form either will look like Y square equal to something X or X square equal to something Y. But it has other terms also. For example, X square, X also, Y also, constant also. So it is not a standard case. Is it an oblique case? Could it be a parabola like this, inclined like something like this? Yes or no? Could it be like this? No, it cannot be Nikhil. If it is an inclined parabola, it will definitely contain an XY term because in that case, your director will also be inclined. And remember, when you will write that locus condition, you will get that in that distance expression, you will end up getting HK term also. So is there any XY term in this parabola? No. If there is no XY term, it is not an oblique case also. So what is left? So if it is not a standard case, it is not a Teha case. It is a shifted case in that case. Exactly. Are you getting on one? So the point which I wanted to convey to this through this small question that I asked was that if you see an XY term in the parabola, then only it is a case where the parabola is slightly tilted one like that. If there is only X square, Y or Y square, X kind of a term, then it is a standard case, something like this. But in this case, it is neither of the two, which means it is a shifted version. So for shifted version, we need to complete our squares. So for that normally, this is the mechanism that I suggest. So look at these two terms, keep these two terms separately. So in order to complete a square, what would you require here? You would require a 16. So just add a 16 on both the sides of the equation. So this becomes X plus 4 the whole square and send these two guys to the other side. Okay. And as I told you in the previous question, you should always have the direct coefficient of X and Y to be 1, 1 each. So I have to take a minus 12 common. Getting the point here. So looking at this, now you realize, oh, it is now this parabola which was shifted. So your A is 3, capital X is X plus 4, capital Y is Y minus 1. Correct? Yes or no? So now close your eyes and recall everything related to this. So vertex was 0, 0. Vertex was 0, 0. So write capital X as 0, capital Y as 0. So X plus 4 is 0, Y minus 1 equal to 0. So 4 comma minus 4 comma 1 is your vertex. Let's be check. Prisham, we are doing again same mistakes. Sharduli also same mistake. Anybody who got this right? I think nobody got it right. You should actually know why you're making that mistake. Okay. Focus. Focus for this case is at 0 comma minus A. So X plus 4 will be 0 and Y minus 1 equal to minus A. Now minus A here is minus 3. So X is minus 4, Y is minus 2. So minus 4 comma minus 2 is your focus. Many people got this wrong. In fact, all of you almost got it wrong. Anyways, we'll take more questions. Since you have got it wrong, I'll take more questions. Directrix equation. In such case, directrix equation is Y equal to A. Okay. So our Y is Y minus 1. A is equal to 3. So Y equal to 4 is our directrix, or Y minus 4 equal to 0 is our directrix. Next, what they have asked, latter-sictum length. Okay, latter-sictum length will be 4A units, which is 12. They also asked us the equation of the axis, I believe. Axis in this case is capital X equal to 0, which is this. Is this fine? Any questions? Any concerns here now? Pause this exercise. I don't think so. There should be any issue. So completing a perfect square along with the fact that your direct coefficient, I'm not talking about indirect coefficient. Indirect coefficient is like when you expand it. Your direct coefficient should be made 1-1 for them and that part should be taken as your capital X, capital Y, et cetera. And then you have to work with this standard form. Is it fine? Let's take one more. Sorry, I know your break is getting delayed, but till you get this right, sir, we are feeling hungry. Okay, let's take this one. Just one question here, vertex. You don't have to do all the parts. And I'll put the poll also on for the same. Find the verdicts of this parabola. Again, first ask yourself, is it a standard form? No. Is it an oblique form? Slightly tilted parabola? No, because there's no XY term. So it has to be a shifted form. Shifting of what standard is? Some standard form was shifted. Then you got this equation. Okay, very good. I've got one of the options getting some response. So pose this question. We'll take a break. On the other side of the break, I will introduce you to the parametric form of the standard cases. And we will talk a little bit related to questions related to parametric form also. Okay, two minutes have already gone. I can give 30 more seconds, not more than that. One. Okay, just seven of you could muster the courage to respond. Bye. You're all scared. He'll get it wrong. Don't be scared. Okay, so out of seven of you, five say A and two say B. Okay, so A and B, there is some split off votes. See again, the presence of a Y square term says that there has to be a perfect square created in Y, which is clearly obtained when you put a one here. So that becomes Y minus one in the whole square. This is in fact, this will become minus 12 minus 6x. Correct me if I'm wrong. Note that I need to ensure my direct coefficient of direct coefficient of X and Y both are one each. Okay, so this is how I need to write it first. Okay, so now compare it with Y square is equal to minus 4ax, which means your capital X role is being played by a small x plus 2. Capital Y role is being played by y minus 1 and A role is being played by 3 by 2. 3 by 2. Correct. Now, anyways, we don't need to go into a value also. We just needed the vertex. So vertex is where x is 0, y is 0. That means x plus 2 is 0, y minus 1 equal to 0. So x is minus 2, y is 1. So minus 2 comma 1 is your vertex. And that's clearly option number A up to you. So I'll leave this question on your screen only. Please find out everything else. Meanwhile, those who want to take a break, you can take a break. 625 right now as per my laptop. And okay, we'll meet exactly at 640. On the other side, we'll talk about parametric forms. See you after the break. Our next in line is our parametric form of the standard parabola. In fact, we'll just name it as a parametric form. Okay. As I already discussed with you during the circle chapter, parametric form is just an alternate way of expressing the same equation where instead of directly relating the variables, you relate the variables to a parameter. And if you eliminate that parameter, the relationship between the variable becomes the same as the equation. We are done in case of a circle x square plus y square is equal to r square. We had learned that the parametric form was x equal to r cos theta and y equal to r sin theta. In the same way, we'll have parametric forms for our parabola as well. So let us begin with first our standard cases. So if this is the Cartesian form of our standard form of a parabola, what could be a suggested parametric form? Again, I'm using the word suggested because parametric forms are not fixed. Okay. Unlike the equation in Cartesian form, which is fixed, that means there is no alternate equation to the same thing. Okay. Of course, you can change the way you write the same expression. But parametric forms can vary as per person to person. So that is why we don't use the parametric form. We use a parametric form or suggested parametric form. So for such cases, the one, the parametric form that we normally use or which is conventionally used is this x equal to a t square y equal to 2 a t, where t is a parameter. And you can see for yourself that if you eliminate the t from both the equations, you will end up getting the same equation back. So if you eliminate your, if you eliminate your t, you end up getting the same equation back. Okay. Right. Should we check it out? So from the last equation, I can say t is equal to y by 2 a and put this in the first equation. So x equal to a and instead of t, I will put y by 2 a the whole square. So that will give you x equal to a times y square by 4 a square a will get cancelled. So this gives you y square is equal to 4 a x. Okay. So the user parametric form, as you already know, is to represent a point, any point, which is not actually known to us. So let's say if I want to write, choose a point on this parabola y square is equal to 4 a x. I would actually choose it like something a t square comma 2 a t. Okay, I will not choose x 1 y 1 or I will not choose alpha beta, etc. Because unnecessarily two variables will get involved. So here if you see the benefit of parametric form is that you can do away by choosing a point with only one variable involved, which is t in this case. Okay. And not only that, if you see this automatically speaks for itself, it says that, Hey, I have a point on the parabola. Okay. So you don't have to give an extra information to the viewer that x 1 y 1 are such points which will satisfy y 1 square is equal to 4 a x 1. So it has double advantage. One, the number of variable use is only one. And secondly, the point speaks for itself. That means the point automatically says that it lies on that given. Is it fine? Any questions, any concerns here? Okay. Now I would request you to tell me or suggest me a parametric form for the second standard case y square is equal to minus 4 x. Let me see what is your suggestion in this case. By the way, can you all see me? I'm in darkness actually. It's because the power is gone. It's on the backup right now. Yes, very good to kill. So as you can all, as you can all observe here that the sign of x became minus x. So you can write x as minus 80 square y s to 80. Very good. Okay. So this is a parametric form for y square is equal to minus 4 x. Is it fine? Any questions? Any questions? Okay. Try the third one. Tell me for x square is equal to 4 a y. Suggest me a parametric form for x square is equal to 4 a y. Very good. Excellent. So x equal to 280 and y is equal to 80 square. Very good. So basically you have used the fact that this equation is obtained by replacing x and y, swapping x and y positions. So even in the parametric form, whatever was your x earlier, that will become the y here. So earlier, my x was 80 square. Here, my y will become 80 square. And my y was 280. So here, x will become 280. Excellent. Where t is a parameter. Tell me the parametric form for x square is equal to minus 4 a y. Now this is not a challenge for you because you just say change the sign of y. Absolutely. So it will be y equal to minus 80 square. So this is a suggested parametric form. Parametric forms are going to be of lot of use when we do this conic section chapter to more depth. In fact, any conic section, the parametric form has a huge role to play. Don't take it lightly. I know it is not there in your CVSC curricula. But this concept is going to be very, very useful when it comes to solving locus based questions. Later on, we'll talk about locus based questions on conic sections. So now I'll just give you a small question to solve here. Suggest a parametric form for x minus 2 the whole square minus 4 y plus 1. Yes, anybody any success. See all you need to do is just compare this with y square is equal to minus 4 a sorry x square is equal to minus 4 a y. So for such case, if you recall, we had just done it in the last page, your x is equal to 280 and y is equal to minus 80 square. Am I right? Okay, now all you need to do is change your capital X with x minus 2 small x minus 2. And here remember a is one. So this will become 2t. Correct. Remember a is one. So it will become 2t and your y plus 1 is equal to minus t square. Correct. So you can write the suggested parametric form for this as x as 2 plus 2t and y as minus t square minus 1. Okay, where t is a parameter. Please state that also. Okay. Is it fine? Any questions? Any concerns here? Do let me know. Okay. So now I have a small set of questions based on the parametric form for you all. Please copy this down. And if you have any questions, any concerns, do let me know because I'm going to switch to the next slide. Clear? No doubt. Okay. Okay. So here is a question for all of you. This is our standard parabola y square is equal to 4ax. Okay. Y square is equal to 4ax. Okay. And this is a focal chord. Okay. Of course, focal chord is a chord passing through the focus. Let's say this extreme end is p, this extreme end is q. Okay. This point, the parameter is t1. In other words, the coordinate of this point is at1 square comma to at1. And this point, the parameter is t2. In other words, the e coordinates of that point is at2 square comma ta t2. Okay. Question is find the value of t1 t2. Find the value of the product t1 t2. Is the question clear to everybody? Question is this is our standard form of the parabola y square is equal to 4ax. There is a focal chord. One end of the focal chord has a parameter t1. See, when I say that there's a point with parameter t1 indirectly, I'm trying to say that the point coordinate is at1 square comma to at1 because in light of the parametric form that we discussed for this, the suggested parametric form is y equal, sorry, x equal to at square, y equal to 280. In that t is t1. That is what is the meaning of the point having a parameter of t1. Now, many people ask this question, sir. How do I know that the question center has used that parametric form? So this is my convention. If let's say the question says parameter is t1, conventionally the same parametric form that I discussed that will be used. So do this problem. Tell me what is the value of product of t1 t2? Very simple question. Very, very simple question. Give me a response on the chat box. Very good, Nikhil. Excellent. That's correct answer. Okay, so Nikhil has already got it. Anybody else? Okay. This is not a rocket science. You can clearly see that PSQ are collinear, right? Okay. Because they are on the same focal chord, right? What does it mean? It means slope of PS or slope of SP should be same as slope of SQ. That's it. This will help you to get your answer. So what is the slope of SP? y2 minus y1 by x2 minus x1. Similarly, what is the slope of SQ? y2 minus y1 by x2 minus x1. Okay. Drop your two a's and a's on the both the sides. So two a, two a, I'll cancel. In fact, your a, a, a, a will also get cancelled. So that will leave you with t1 by t1 square minus 1 equal to t2 equal to t2 by t2 square minus 1. Just take a cross multiplication. Okay, just expand it. So that'll give you t1 t2 square minus t1 equal to again t1 square t2 minus t2. Bring it this to the other side and bring this to the other side. So t1 t2 square minus t1 square t2 is equal to t1 minus t2. Very good, Shalvili. That's right. Take t1 t2 common and here you take a minus sign. These two will get cancelled giving you t1 t2 as a minus one. Okay. Now remember here t1 is not equal to t2 because p and q are different points. Remember t1 t2 are like the Aadhaar card numbers for those points. So assuming that the point is a resident and the curve is a country, every resident has a different Aadhaar card number. That is what I told you when I was talking about the parametric form in your circles chapter. So this point has a different parameter than this point. Two parameters cannot be the same unless and till the points are same. Please get this right. Two points cannot have the same parameter. It's like two people having the same Aadhaar card, not possible unless and till somebody has committed a fraud. Okay. So please understand p and q are different points. So t1 and t2 cannot be equal. If they're equal, they have to be same points. That is why you can easily cancel out t1 t2 from both the sides. And that's what I was expecting you to give me minus one is your answer. Okay. So this is actually a very important property that you would have to remember in a long run because a lot of questions have been based around this fact that the extremities of a focal chord, the extremities of the focal chord of a parabola, the parameters are related by the relation t1 t2 equal to minus one. Please note this down. Very, very important. That means if the parameter of one end is t1, the parameter of the other end will be minus one by t1. Okay. So let me show you if the parameter of one end is t1. Let's say this is a focal chord. Okay. So if the parameter of this is t1, that means if this is a t1 square comma to a t1, the parameter and the other end will be minus one by t1, which is nothing but a by t1 square comma minus two a by t1. So automatically the same parameter could now be used to get the extreme, the other extreme end of the focal chord. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Okay. Now, once you have known this, let's talk about, let's talk about the focal radius. Focal radius means the distance of the focus from that given point. We also call sometimes this as focal distance. So for this purpose, again, I will take up my standard form. As you can see, standard form is the frog in the laboratory. We keep on doing a lot of findings using the standard form. Okay. So let's say if I have a point here. Okay. Again, I'll take a parametric form 80 square comma to 80. What is the distance of this point from the focus a comma zero? What is SP distance? Please tell me everybody. Tell me this answer in five seconds. Let's see if anybody is able to get this in five seconds. What sir? Five seconds is too less, but it is actually worth a five second. That is why I am giving you only five seconds. Give me the distance SP where P is your point 80 square comma to 80. I think it's almost a minute now. SP, SP, I want SP length. S is this, P is this. What is the length SP? Right. How did you find it out, Sethu? You didn't have to use the distance formula. Use the fact that SP is equal to PM. And this is already X equal to minus A. That means this length is A. This length is already 80 square. Right. So answer is A plus 80 square. Okay. Now that is the reason why I said use only five seconds to solve this. Anyways, since you've found it, there are some things which I will, you know, prove by using this particular form. All of you, please now follow the proof of the fact that lattice symptom, so proof of the fact that lattice symptom is the shortest focal chord. So I'll be using this focal radius or your focal distance concept to prove that the lattice symptom is the shortest focal chord. Okay. All of you, please pay attention. That's why I don't draw a free hand because it doesn't come out to be very aesthetic. Anyways, I think I've done a good job here. Okay. Let's say this is your focal chord. Any focal chord where this point is, let's say P, this point is Q. Now, all recall that if this point is 80 square comma two 80, what will be the coordinates of Q? What will be the coordinates of Q? Write down on the chat box. If P is one extreme end of the focal chord, what is the coordinate of the other? Look at this P and tell me what is the coordinate of Q? If PQ is the focal chord, ah, Nikhil, you're missing something. You're not behind where? They're behind. Or in the previous slide on the mirror, oh, you could see a mirror in my house. Is this slide? Okay. Now, all of you, please pay attention. SP length just now we told A plus 80 square. So can I say, if I want SQ length, it will be A plus A by T square. Am I right? Chaudhary, minus sign, you have put it the wrong place. Are you all convinced with SP and SQ? Everybody's convinced? So just like SP was A plus 80 square, that means add A to the X coordinate of P in the same way SQ will be A plus the X coordinate of Q, which is A by T square. So now the focal chord length will be what? A plus 80 square, A plus A by T square, correct? In short, it is going to become 2A plus 80 square plus A by T square. Okay. Now everybody pays their attention. Do you recall that in one of the chapters, I had done the fact that for positive quantities, arithmetic mean is always greater than or equal to geometric mean. Yes or no? Or positive quantities? Do you remember that? Which chapter we have done this? Let me see whether you remember the chapter name. This is sequence series. Yes. Very good. Now I'm going to apply that to this guy. So let's say 80 square is one number and A by T square is another number. Do you all agree that these are positive numbers? So can I say AM? AM will be greater than equal to GM, correct? Now T square, T square goes, this will be A. That means can I say 80 square plus A by T square will be greater than equal to 2A? Yes or no? So can I say this guy will be greater than equal to 2A plus 2A because 2A is already there and this fellow is just now we figured out that this is greater than 2A. So using this over here, can I say this fellow PQ will always be greater than equal to 2A plus 2A? Yes or no? Right? Yes, if you're convinced. Right? No, if you're not. Yes, you have full freedom to write no also. I will explain once again there. Right? Yes, if you're convinced, right? No, if you're not. Very good. So yes, people are saying yes. Okay, good. That means PQ is always greater than 4A. Correct? Correct? That means the least value of PQ will be 4A. Correct? And 4A is nothing but the length of lattice rectum. Correct? That means the least value of the length of any focal chord is your lattice rectum length. In other words, lattice rectum is the shortest focal chord. Convinced? That means lattice rectum is the shortest focal chord because any focal chord is actually greater than equal to 4A. So the least value of any focal chord is 4A and that is nothing but the length of the lattice rectum. Convinced? Any question? Any concerns? Do let me know. Do let me know. Copied? Noted down? Any questions also if you have? Please ask. The least value. Great. So let's have a few more questions over here. Let's take this question. Prove that the semi-latticectum of the parabola y square is equal to 4Ax. Is the harmonic mean between the segments of any focal chord of the parabola? Read this question. Read it twice, twice till you understand it. And somebody please explain me what does the question actually say? Let's see whether you're able to understand the meaning of the question first of all. Yes. Has anybody understood the question? Okay. All I could know is the semi-latticectum length here and there is some focal chord. Okay. Let's call it as TQS. Now tell me what does the question want you to prove? Anyone who has understood the question, please explain it to me. What does the question want you to prove? Nobody has understood the question. Half of the length of the lattice rectum is harmonic mean of any two focal chords. Sorry, Nikhil. The question doesn't mean that. But good try. I always appreciate trying. Okay. Whether you're right or wrong, it is immaterial to me. I want you to participate. I want you to make an attempt to solve the question. Anybody else? Nikhil, well done. But that is not the meaning of the question. Anybody else? See, the reason why I'm asking you this because this has been written in a very vague way. Okay. That is what I wanted to say. See, the question actually tells you that you need to prove that let's say this is L1, this is L2. Okay. You need to show that L1, 2A, 2A is your semi-latticectum and L2, R in HP. Okay. This is what the question wants you to prove. Okay. So prove that L1, 2A and L2 are in harmonic progression. By the way, this is a very important property also. Later on, we will see while we are doing parabola in much more depth. It's an important property also. Now do this. I'm waiting for you to do this. Now that you know your concepts, you can do it now. Try this out. Just say done when you are done. Okay. It's a proof that question anyhow. There's nothing to find. By the way, let me tell you, we have crossed our school requirement. Okay. I have gone, you know, more than what the school requires for you to know in parabola. In fact, for all the chapters that we have done so far. But let me tell you, this is not the end of this chapter. We have 80% concept more left. Okay. 80%. That means four times whatever we have done is still left to be covered. So big is this chapter. In fact, so big is every conic section chapter. Okay. So once we have some more time left after we are done with that, you know, finishing off our syllabus and doing some practice, we will come back slowly and slowly to these topics in more detail. Done so very good. Say to anybody else. Chalo, we'll discuss this out. See if you have to prove this, that means you're supposed to prove that one by L1, one by two way, and one by L2 are in AP. That means twice of the middle term is some of this. That is, you need to prove this. In other words, you have to prove one by A is one by L1 plus one by L2. Okay. Let's try to prove it. Let us start with my right hand side. One by L1. Now, first of all, what is L1? I've only discussed L1, that is your SP length is A plus, let's say, let's say I call this point to be AT square comma 280. So by the fact that they are ends of a focal chord, this becomes A by T square minus 2 by 2A by T. So L1 is nothing but SP and SP is nothing but PM and PM is A plus the x coordinate of that point. So one by L1 is one by A plus AT square. Okay. Similarly, L2 is what? L2 will be A plus A by T square. In other words, it is A plus AT square by T square. So one by L2 is what? T square. Where did I write a square here? Okay. So you have written it in subscript. Yeah. This is your one by L2. Now add it. See what happens. Add one by L1, one by L2. A plus AT square common. Okay. Take A common in the denominator. These two will get cancelled, giving you one by which is nothing but your left hand side. Is it fine? Any questions? Any questions? Okay. So we are now going to take some questions in general. We have already done more than enough, more than what is required for your school. So we'll take some questions. Okay. So for the next 10, 12 minutes of our class, we will solve few questions. So keep this in mind. This is actually a very important property. You would be requiring it. I have seen this question coming in several competitive exams, including CET, Manipal, VIT, of course, JMAIN, etc. These questions, this question has actually come as it is. Okay. So let's take few more questions and then we'll close the topic. Let's take this question. Find the equation of the parabola whose focus is 4 comma minus 3 and vertex is 4 comma minus 1. Okay. If you make a diagram, things will be clear here. See, they have said the focus is at 4 comma minus 3 and the vertex is at 4 comma minus 1. So 4 comma minus 1 would be somewhere over here. Okay. And 4 comma minus 3 will be somewhere down here. Okay. So this is 4 comma minus 1. This is 4 comma minus 3. Okay. So this is your parabola. Yes or no? Now, okay, I'll not give you any further hint. I'll just wait for you to complete this. So we all realize that it's a shifted version of a downward opening parabola. So that should answer everything now. Very good, Nikhil. Anybody else? I thought it was an easy question. Many of you would be giving the answer well within a minute. Okay Prisham, Setu, Vallan. Prisham's answer seems to be slightly different than Setu's answer and I think Nikhil's answer. Should we discuss it? See, yeah, yeah, you can change your answer. See, guys, here we realize that this is a shifted version of your downward opening parabola. Correct. So its equation should be inspired by x square is equal to minus 4 a y. Correct. Now, since your vertex has come to 4 comma 1, your capital X will now be small x minus 4 and capital Y will be y plus 1. But what is my A? You know that I have already told you A represents the distance between the focus and the vertex. So our A is 2 here. Correct. So all I need to do is write this. Okay. So that is nothing but x minus 4, the whole square is minus 8 y plus 1. That's it. So if you expand it and write it in a proper way, you are most welcome to do that. But I was basically interested in just knowing the format here. So it's x square. I'm so sorry. It's x square minus 8 x plus 8 y plus 24 equal to 0. Well done, Nikhil. Nikhil got it absolutely right. Others who have made a mistake, please know why you have made a mistake that is more important. Come on. We are closing this chapter and such mistakes are happening. Let's take this question. Find the equation of the parabola whose axis is parallel to the y-axis and which passes through these three points and determine the length of its lattice rectum. Determine the length of its lattice rectum. Equation of the parabola with axis parallel to y-axis and passes through these three points and you have to also find the length of the lattice rectum. Yes, anybody? Okay. Now everybody please pay attention. See this type of a parabola which is parallel to the y-axis whose axis is parallel to the y-axis is nothing but a quadratic polynomial. See in quadratic polynomial, we always used to make a parabola like this or like this. So all these have their axis parallel to y-axis only. Now because there are three unknowns involved here A, B, C, there are three points provided to us. So we need to get these three points. Sorry, these three unknowns. So let us use the first point 4 comma 0. So put 4 as A and sorry 4 as y. Okay. And this will be obtained. C is obtained. Put y as 9. Okay. x as 1. So that gives you A plus B as a 5. Then put 4 comma 5. So put y as 5 and x as a 4. So this gives you I think 1 equal to 16 A plus 4 B. Now using these two, we will have to solve for A and B because C is already known. Okay. So let us solve A and B. So let us multiply the first equation with a 4 and let us subtract it. So 12 A is minus 19. So A value will be minus 19 by 12. Put it over here. So B value will be 5 minus A. 5 minus A will be this which is nothing but 79 by 12. Okay. So let us put this back in our equation y equal to A x square which is sorry y equal to not y square. Sorry, my mistake. I wrote y. Very good. Prasimha actually got it. Okay. Then B is 79 by 12 x and C is a 4. Okay. Is this fine? Any questions? Any concerns? Is it fine? All right. So I think we have covered up enough content on parabola to suffice your school needs. But as I told you, this chapter is not yet over. We have at least four times more content left to be covered. We will talk about it in near future. Okay. So thank you. Bye-bye. Good night. Stay safe. Shabba khair. Shukranthi. Bye-bye.