 All right, so today we'll be talking about Parabula session 3, which will be the last session for Parabula. So the concepts that we are going to cover up in this session would be the first one being the diameter of a Parabula. So we'll be talking about the first concept which is diameter of a Parabula. So how do you define a diameter in case of a Parabula? We would have heard about diameter in case of a circle. So in case of a Parabula, the diameter is defined as the locus of the midpoints of system of parallel chords. That is called the diameter in case of a Parabula. So now let us take this as a question. So let's say we have a Parabula, y square is equal to 4ax. Again, let me just load the diagram for it because I don't want to draw it again and again. So let's say we have the standard Parabula, y square is equal to 4ax. And I draw a system of parallel chords to this Parabula. Now all these chords that you see, they are parallel to each other and they have a slope of m. So let's say these chords are having a slope of m. Now when you connect the midpoints of these Parabula and join them, basically you end up getting something called the diameter. So this will be called as the diameter of this Parabula. Now guys, can you derive for me the equation of this diameter given that it connects the midpoint of all system of parallel chords having a slope of m. Is the question understood by all of you? Yes, sir. Is Kondanya there in the call? Sorry, in the session? Yes, sir. And those who are attending it through YouTube Live, I will request you to type in your names in the chat box so that I know who all are attending the session. Alright, so in order to make the task a little faster, let's say the system of parallel chords over here is having a generic, let's say the family of these chords that we have over here. Let the equation for that family of parallel chords be y is equal to mx plus c. Now remember here, m is given to you, so m is not a parameter. Whereas c is the parameter or you can say an arbitrary constant. Okay? And let us say I want to solve this equation simultaneously with the equation of the Parabula y square is equal to 4ax. Okay? Let's say I want to find out the point of intersection of these lines or these chords with the Parabula. Okay? So what I'm going to do next is I'm going to replace my x with, so let me call it as 1, let me call this as 2. So from 1 you can write x as y minus c by m. Okay? And replace this in 2. Replace this in your equation number 2 in place of x. So you end up getting y square is equal to 4a y minus c by m. Okay? Which actually reduces itself to a simple quadratic equation in y, that is, m y square minus 4a y plus 4ac is equal to 0. Correct? Now this quadratic equation basically signifies the two roots that you will obtain with the intersection of this chord with the Parabula. Let's say y1 and y2 are the two roots. Okay? So can I do this? Y1 and y2 are the two roots. Okay? So can I say that the sum of the roots, let's say the roots are y1 and y2. Can I say the sum of the two roots? Okay? Will be minus b by a minus b by a. Yes or no? And you'll be surprised to know that this doesn't contain c anywhere. That means it is independent of the c part, that is the parameter. It just depends upon the slope of these chords. That means everywhere, no matter whichever chord you are taking, the sum of the roots will be a fixed quantity because a is fixed, m is fixed. Correct? So let's say I assume h comma k to be the midpoint. Okay? So can I say k will be y1 plus y2 by 2? Isn't it? Yeah. Which means y1 plus y2 is 2k. Right? So replacing it over here, replacing it over here, I can say 2k is equal to 4a by m. Right? Which implies k is equal to 2a by m. And if we generalize this, if we generalize this, we can say y is equal to 2a by m becomes the equation of the diameter of a parabola. Equation of a diameter of a parabola which passes through the midpoint of all family of parallel chords having a slope of m. Guys, let me tell you one very important thing that what is required in JE is to understand the concept of locus via this problem rather than the concept of diameter itself. Okay? You will not find the concept of diameter being tested but you will definitely find the same concept coming in the form of a locus question. Is that fine? Yes. Okay? And you would realize that this diameter will always be parallel to the axis of the parabola. It will always be parallel to the axis of the parabola. So this line will be parallel to this line. As you can see, it's y is equal to something. Right? So it's a line parallel to the x axis in case of y square is equal to 4x. But if you take any other generic case, it will always be parallel to the axis of the parabola. Is that fine? Now, certain properties that we will take up from here. Hope there is no question. Aditya, any question with respect to it? No sir, no questions. So let's take some properties which are related to the diameter. Properties related to the diameter. The first property is the tangent drawn at the extremity of a diameter. What is the extremity of a diameter? Where the diameter meets the parabola again is called the extremity of the diameter. The tangent drawn to the extremity of the diameter of a parabola is parallel to the system of quads it bisects. Is parallel to the system of quads it bisects. Can you prove this guys? It's very, very simple. Let me just copy the figure again. So let's say these are our parallel quads. And this is our diameter which connects the midpoint of these parallel quads. So now what is this saying is that if you draw a tangent over here, not a very good one. This tangent, the red one, would be itself parallel to the system of parallel quads. That means the slope of all of them would be m. The slope of all these things would be m. It should not take you more than 20 seconds to prove it. So what is the y coordinate of the point where it touches the parabola? Is it 2a by m? Because the equation of this diameter itself is y is equal to 2a by m. So what would be the x coordinate of this point? Let's say I call this point as p. Then what is the x coordinate of this point? Put in the equation y square is equal to 4ax. So y square is going to be 4a square by m square. That's equal to 4ax. So x is going to be a by m square. So your p point will be a by m square. Isn't it? And we all know that the tangent at any point drawn to this parabola, y square is equal to 4ax is given by 2a by y. 2a by y. So slope of the tangent at p will be 2a by y. y will be 2a by m. So as to say I did not require this point but just for your information. So which will simplify to give you m. That means even this line will have a slope of m. Simple. No brainer in this. Any questions? Moving forward. Second property is tangents at the end of any chord meet on the diameter. Which bisects the chord. Okay? So again let me support this with a diagram. So let's say this is a chord. Okay? And this is a diameter which bisects this chord. Okay? And if you draw tangents at these two points they will meet on this diameter. So let's say a bisects this chord. So let's say a, b and this is b. So do they have to be concurrent? Yeah, that's what. The tangents will meet on the diameter. Sigh, is it clear? Is the question clear to you Sigh? We all know the meeting points of two tangents, right? We have discussed this in the last class. If you have two points a t1 square comma 2 a t1 and a t2 square comma 2 a t2 where do they meet? A t1 t2 comma a t1 plus t2. Excellent. A t1 t2 comma a t1 plus t2. Correct? Correct? Now you have to somehow prove that this point will also lie on on the diameter which is bisecting the chord a b at least with respect to the y coordinate. Right? If any coordinate matches it's fine. Tell me the equation of this equation of this diameter is y is equal to 2 a by m, right? So y is always 2 a by m, correct? Yes. Where m is the slope of the chord. m is the slope of the chord. Yeah. Okay. Now according to this given assumption of points, what is the slope of the chord a b? You will say simple, y2 minus x1 by x2 minus x1, correct? Which will clearly result into 2 t1 plus t2. Correct? Yes or no? Right? Which means t1 plus t2 is going to be 2 by m. Correct? And if you replace it over here do you see that this fellow becomes a into 2 by m which is actually 2 a by m which clearly indicates that it lies on which clearly indicates that it lies on the diameter which is bisecting that chord. Because the diameter will always have points where the y coordinate will be 2 a by m. The x coordinate can vary. So clearly it lies on the diameter bisecting the chord. Is that fine? Yes. Any question with respect to this? Okay. So let's take a problem based on this. If the diameter if the diameter through any point p, through any point p of a parabola of a parabola meets any chord in a, if the diameter through any point p of a parabola meets any chord in a and the tangent at the end of the chord and the tangent at the end of the chord meets the diameter in B and C. Then prove that p a square is p b into p c. Is the question clear? Can you sketch a diagram for this? Alright. So let's say y square is equal to 4 a x is your parabola. Okay. Now, this is a diameter whose end point extremity is p. Okay. And any chord, let's say chord a b it bisects. Okay. It doesn't bisect, it just cuts. Right. That's what the question says. It meets any chord in a. Okay. And if you draw tangent at a I cannot take a and b because they already used it. So let me use some other name. I'll just erase this a and b. Let's say m and n. So the tangent at these two points let's say they meet it at a and b. Sorry, meet in b and c. So this is b, this is c and this is a. So diameter through point p of a parabola meets any chord in a. So this is the point a for you and tangents drawn at the extremities of this meet the diameter in b and c. So we have to prove that p a into p a square is p b into p c. Remember, this is not a focal chord. This is any chord. Okay. All right. So let's assume that the equation of this the equation of this diameter is y is equal to 2 a by m assuming that it by six all set of chords which are having a slope of m. Okay. So what is point p just figured it out a by m square comma 2 a by m. Correct. Yes sir. What is the equation of the chord m in what is the equation of m in chord m in. Yeah, tell me. So you can assume these points to be a T1 square comma 2 a T1 and a T2 square comma 2 a T2. Yeah. So slope would be what slope would be slope of the chord m in would be 2 a T1 minus T2 by a T1 square minus T2 square which is same thing 2 by T1 plus T2 and we can easily get the equation as y minus 2 a T1 is equal to 2 by T1 plus T2 x minus a T1 square. Okay. So if you simplify this further you will get y T1 plus T2 minus 2 a T1 square minus 2 a T1 T2 is equal to 2 x minus 2 a T1 square. So this term gets cancelled off. So you end up getting the equation as y T1 plus T2 minus 2 x minus 2 a T1 T2 equal to 0. Is that fine? Now where does this particular thing meet y is equal to 2 a by m. Substitute y as 2 a by m. Correct. Yes or no? So 2 a by m T1 plus T2 minus 2 a T1 T2 is equal to 2 x. So drop the factor of 2 from everywhere. So x is going to be a by m T1 plus T2 minus 8 T1 T2. Correct. Now next a coordinate is known. So a coordinate is going to be a by m T1 plus T2 minus a T1 T2 comma 2 a by m. What about b and c? Can you similarly find b and c point? I think this is going to be a simple straightforward problem where you have to find b and c point by seeing the intersection of the tangent drawn at m with the diameter and the tangent drawn at n with the diameter. Yes or no? So can you quickly find that out and do it on your own because this is going to be a simple problem because you just have to find bc coordinates. PA is basically the difference of the x coordinates of P and A PB again is going to be the difference of x coordinates of P and B and then you can simply simplify it. There is nothing very special about it. So you can try it out similarly find b and c right? That is the meeting point of tangents m and n with the diameter verify PA square is PB into PC Is that fine? So please do that and share the information on the group Next we will talk about the concept of pole and polar So what is the concept of pole and polar in case of a parabola So it is true for all the conic sections whatever I am going to discuss now So let P be any point any point inside or outside or outside a parabola Now if we draw any line let me take this as a case here if we draw let's say P is this if I draw any chord through this parabola okay Now this parabola is cut by this chord drawn through this point P at let's say point Q and R okay If you draw tangents at Q and R tangents at Q and R okay they let's say meet at a point T right? and if we keep on doing it let's say I draw another chord passing through P let's say like this okay and again I draw tangent at these two points again I draw tangent at these two points okay and let's say it meets at T and if you keep on doing for N number of chords that you draw through P and keep on drawing tangents at the extremity realize that all these all these points that you are getting they will lie on a straight line like this they will lie on a straight line like this this line is called the polar and this point that you see over here that is called a pole again what is important for J is not whether you know what is called a polar and what is called a pole J will simply ask you a locus question that if let's say this is X1 Y1 point and any chord drawn through X1 Y1 let's say meet the parabola at Q R and if you draw tangents to Q and R to meet at T then find the locus of T that is what is going to be the polar of this pole so can you figure, can you find out the equation of the polar for this pole please type in the answer in the chat box yes sure see if you ask the polar of any point P then draw all possible chords that you can make through P okay so take for example you draw Q R chord now where the Q R meets the parabola draw tangents at the ends of these chords so they will meet at T similarly I took another chord let's say Q R and I drew tangents at the extremities of this chord to meet again at T so if you keep on doing for every possible chord that you can pass through P you would realize that the meeting point of the tangents at the extremities of all these chords would lie on a line and that line is known as the polar is the question, is the concept of polar clear to you Sai? okay it's pretty simple assuming that this point is H comma K if I ask you what is the chord of contact for two tangents drawn from H comma K you will say T equal to correct? so assuming that this is our standard case of a parabola y square is equal to 4ax you will say yk is equal to 2ax plus H right no? now this chord of contact will contain x1 y1 right so can I replace my y with y1 and x with x1 and that's it story is over and now we have got a relationship connecting H and K now generalize by replacing H with X and K with Y so you end up getting same thing y1 is equal to 2ax plus x1 right so it again becomes T equal to 0 but this time point x1 y1 is not on the parabola it is either within or outside the parabola so the same equation now which was earlier representing chord of contact as well as tangent is now used for the third time and it is now representing the equation of the polar is that clear yes or no? yes sir so what does it mean what does it mean it means that if the point P were outside the parabola if the point P were outside the parabola the chord of contact itself would have acted like a polar for it isn't it are you getting this point so what I am trying to say is what I am trying to say here is listen to this carefully if there was a point x1 y1 over here ok so x1 y1 is basically your point outside the parabola then if you start drawing if you start drawing all possible chords passing through this parabola start drawing all possible chords passing through this parabola and you start drawing tangents at these points if you start constructing tangents at these points you would realize that they start meeting on they start meeting on a line which is going to behave as so this is going to be your polar and that will also be your chord of contact are you getting this point you understand the meaning of chord of contact that means if you draw tangent from here and tangent from here this will be your chord of contact so the inside part will be your chord of contact and the entire line will be called the polar so you can say chord of contact will be a part of the polar is the idea clear yes sir now we will talk about some properties properties of properties related to pole and polar first property is the polar of the focus is the directrix the polar of the focus is the directrix so guys it is just like saying that all the focal chords which you can draw the tangents at the extremities of the focal chord will meet on the directrix didn't we know this property so what does it become the same as saying the polar of the focus of the parabola is the directrix second property is any tangent is the polar of its point of contact see because t equal to 0 is also the equation of a parabola yes or no right so basically if through the point x1 y1 if the point x1 y1 happens to lie on the parabola so if x1 y1 shifts and lies on the parabola the same equation that is t equal to 0 will actually be the equation of the tangent equation of the tangent drawn at x1 y1 so you can say the polar is behaving as the tangent is behaving as the polar of its own point of contact is that clear next property is if the polar of if the polar of x1 y1 passes through let's say I call this point as p passes through q which is x2 y2 okay then the polar of which is x2 y2 will pass through p will pass through will pass through p and such points p and q are called conjugate points now let me explain this through the help of a diagram see what it's trying to say is if you have a point p okay let's say x1 y1 the polar of it is this line let's say okay so if the polar of p that is x1 y1 passes through a point q passes through a point q then what it says is that the polar of q will pass through p are you getting this point can you prove it that means let's say you draw a chord of contact so chord of contact may be somewhat like this I am drawing it in yellow that means the polar of q will pass through p guys it is very obvious right you don't have to prove it see what is the concept of polar itself the concept of polar itself is it is going to basically be tangent at the extremities of the chord drawn through that point isn't it so all the points all the chords which are passing through p ultimately the chord of it will be a chord of contact for any point lying on this line isn't it yes or no right isn't it so whatever chords you draw through p those chords will be like chord of contacts for any point lying on the polar are you getting it so it has all these polar has to pass through p point right so there is no I mean doubt about this property by the definition it is true any question with respect to this the follow up of this property is if the pole of any line ax plus by plus c equal to 0 lies on another line lies on another line a1x plus b1y plus c1 equal to 0 then the pole of the second line in the pole of the second line which is this a1x plus b1y plus c1 equal to 0 will lie on will lie on ax plus by plus c equal to 0 ok and let me call this line as l1 and l2 and l1 and l2 would be known as conjugate lines so basically what are we calling as conjugate lines this chord of contact and this pole are I am calling as the conjugate lines just like any point p and q will become conjugate points if they lie on their respective polar of each other any two lines will be called the conjugate lines if their poles lie respectively on each other are you getting this point so this line this line pole is lying on this line right so let's take this as my ax plus by plus c equal to 0 ok and I take any one of these lines as let's say a1x plus b1y plus c1 equal to 0 right so you see you can see that the pole of this line the first line l1 lies on lies on l2 correct and the pole of l2 also lies on l1 so any two any two lines any one of the chord and this polar will become conjugate lines getting the point and any two points not any two points p and any point on this polar will become conjugate points are you getting it so p and let's say q2 will become conjugate points p and let's say q3 will become conjugate points p and let's say q4 will become conjugate points correct similarly l1 and let's say this chord becomes conjugate lines L1 and let's say this code becomes conjugate lines L1 and let's say this code becomes conjugate lines Is the idea clear to you? So let's take few questions on this before we close this topic is it clear aditya aditya mishra Yes, sir. It's clear, sir. Oh, you are there on Skype. Okay Okay Show that the locus of the poles of normal codes of y square is equal to 4 a x Is x plus 2 a times y square Plus 4 a cube equal to zero So basically the normal codes are behaving as polar So you have to say you have to find the locus of the poles of those normal codes Now this this question is actually a locus question and Jay may mention may or may not mention the word pole So don't expect Jay to mention the word pole But they will frame the question in such a way that it will make the same sense It's just like, you know the use of the word pole is Every beating the question it is making the question short Anybody has any idea are you trying guys? This is a thing Against the idea is how do you deal with locus problems? All right. So what I do is I assume that This is my pole Okay H comma k correct and From this h comma k i'm drawing a polar. Okay Now polar is going to be basically Y y1 is equal to 2a x plus x1 correct. This is the equation of the polar Now I'll assume that this is a normal to this parabola Okay, so we all know normal equation is y is equal to mx minus 2am minus am cube right Let's say m is the slope So for me here m is a parameter Because I'm assuming the slope to be m so m becomes a parameter So while solving it I have to first of all strike a relationship between h and k Okay, so from these two equations I have to get a relationship between h and k There by getting rid of m There by getting rid of m Is the Standard operating procedure clear. How do we deal with locus problems? See this idea has to set in because millions and millions of questions can be framed on the concept of locus You need to how you need to know what is the roadmap. Okay, so now you're claiming that These two equations that is equation number one and two are the same equations Okay So what we'll do is we start comparing the coefficients. So let's write one as 2a x by k Okay, 2a x by k Plus 2a h by k. So let us start comparing the coefficients if you compare the coefficients Clearly m becomes 2a by k right Minus 2am minus am cube will become 2a h by k correct Yes What else do we need we need to just eliminate m isn't it? Yes, hello. So first of all, can I just cancel off a a a from everywhere? So minus uh 2 m You may want to take a m common also. Okay, you can do that minus m square is 2h by k So now what we'll do is we'll replace We'll replace That m cube right? Oh, I'm sorry. Yeah, so we'll replace m with 2a by k. So it becomes two times 2a by k minus m cube is equal to 2h by k Okay So minus 4a by k is minus 8a cube by k cube Is equal to 2h by k You can multiply with k by 2 everywhere So when you do that you get minus 2a And you get minus 4 a cube by k square Is equal to 2h So that's clear now It's a h right Because we just h 2 by v Just h just h i'm sorry Just h okay Now take this on the other side So h plus 2a Multiply with k square and add 4a cube that should be zero Now you generalize this Now you generalize this it becomes x plus 2a times y square Plus 4a cube equal to zero. Is that clear? Again, again, I'll give you another question We have to be very good in finding locus guys because that's what is important for us these concepts of pole polar diameter They are all waste if you're not able to visualize it in terms of locus next question is the polar of a point with respect to The parabola y square is equal to 4bx Touches x square is equal to 4a y Okay, so the polar of a point with respect to y square is equal to 4ax touches x square is equal to 4a y then show that the locus Let's say I call p then show that locus of p Is a rectangular hyperbola locus of p is a rectangular hyperbola Hope you remember the condition of tangency for any line y equal to mx plus c on to the Parabola x square is equal to 4a y if not, please try to retrieve that Or while solving I'll remind you of it and if possible, uh, give me the locus equation as well In fact, while finding out you'll automatically come to know Okay, let's let's solve this anybody close to getting the answer any progress aditya naman Ram Kondinia Amog Okay, so let's say the point p is h comma k So what are the equations of the polar? yk is equal to 2ax plus h Okay, now in this case, it'll be 2b. So this will be 2b Is that fine So if you treat this as y equal to mx plus c y equal to mx plus c treat this like this Now what is the condition for condition for y equal to mx plus c To be tangent To x square is equal to 4a y The condition is c is equal to minus am square Please remember this it really saves a lot of time Okay Yes or no So c will be 2b h by k is equal to minus a which is Which is Which is a actually Why did I write b? minus a m square This will be m right? Okay So it's very clear that 2b will cancel with one of the 2b's So you'll have h by k is equal to minus a by k square into 2b One k will get cancelled. So it'll become hk is equal to minus 2ab Okay, so now when you generalize it You get x y is equal to minus 2ab That's actually a rectangular hyperbola. That's a rectangular hyperbola. Are you aware rectangular hyperbola equations? Yes sir x y equal to 0 Yeah, it's of the form x y equal to some constant Okay, is that fine Let's try one more now So we are pretty much done with the concept. So we are just trying to solve as many number of problems as possible. So Probably on a sunday. I'll I'll start with 9 30 to 11 30. I'll start with ellipse next question is in the parabola In the parabola y square is equal to 4ax prove that the locus of the poles of all chords of length to c is y square minus 4ax times y square plus 4a square Is equal to 4a square c square. This should be easy question for you. Please try this out Question is clear So locus of the poles of all chords of length to c So basically all the polar's are containing those chords Which are of length to c and you have to find the locus of all such poles form for those polar's Yes So now let us look at this problem in this way I will take the point p to be a t1 square comma 2 a t1 And q to be a t2 square comma 2 a t2 Okay And of course the pole will be at the point where the tangents at p and q meet right Because pq is now going to act as a chord of contact and we know that The pq line extended will be acting as your pole As your polar and this will be your pole Correct Now from our previous experience. We have seen that the points of intersection of the tangents The point of intersection of the tangents drawn at a t1 square comma 2 a t1 and a t2 square comma 2 a t2 is what All right, so this point has to be a t1 t2 comma a t1 plus t2 Which clearly implies h is A t1 t2 And k is a t1 plus t2 right Now, how do I eliminate this t1 and t2 together? Now we have still not used the fact that this length This length is going to be 2c right So what is the length pq or you can say pq square? pq square will be a t1 square minus a t2 square whole square and 2 a t1 minus 2 a t2 whole square Correct Yes, now in other words, I'm trying to say is that 4c square is going to be uh I can easily take a factor of a square Common from here and this I can write it as t1 plus t2 whole square t1 minus t2 whole square correct Similarly 4 a square common from here t1 minus t2 whole square. Yes or no? Yes, and I can also take a square a square t1 minus t2 square common Leaving me with t1 plus t2 whole square plus 4 correct. Yes or no? That's 4c square Now I can know I can write t1 minus t2 square as t1 plus t2 square minus 4 t1 t2 And already here we have t1 plus t2 square Plus 4 now wherever there is t1 plus t2 will replace it with k by a And wherever there is t1 t2. I'll replace it with h by a so I can say a square Even plus t2 will be k square by a square Minus 4 t1 t2 will be h by a Again here I can have k square by a square plus 4. Is that fine? Yes Right. Yes. So when you simplify it further you get 4c square as a square k square minus 4 a H by a square Into k square plus 4 a square by a square So one of the a squares will get cancelled one of the a squares will go on the left hand side It will become 4c square a square Now I'm directly generalizing over here. So it becomes y square minus 4ax times y square plus 4a square So this is our required locus Is that what we wanted to prove? Yes, so this is hence proof Is the idea setting in how to deal with locus? Okay, so I'm not going to uh, you know Start with any new topic. I'm just going to give you more and more problems On this combined topic now Now that you have studied everything So let's take combined concepts on this Next we'll do some General problem practice Especially with related to locus a variable chord a variable chord pq of the parabola y is equal to 4x square subtends 90 degrees at the vertex Subtracts the right angle at the vertex find the locus of find the locus of points of intersection points of intersection Of the normals drawn at p and q It should add normals Should I give you options? Options is 2x is equal to 8 y square minus 1 option is 2 y is equal to 8x square minus 1 2 y is equal to 8x square plus 1 And 2x is equal to 8 y square plus 1 Are you guys working on it any progress? Yes, sir. Yes, sir almost Okay, all right, so let's take a case Now can you suggest me a parametric point on this particular parabola? Can I take it as a t comma 4t square okay So, uh, let's say p and q are two such points Okay, says that this chord subtends a right angle at the At the vertex of the parabola, so this is going to be a right angle So let's say this point is t1 comma 4t1 square This is t2 comma 4t2 square Okay, so the very fact that it is subtending right angle that means The slope of OP and slope of oq should be minus 1 The product of slope of OP in the slope of oq is minus 1 slope of OP is 4t1 square by t1 4t2 square by t2 equal to minus 1 which implies t1 t2 is minus 1 by 16 Let's keep this as an information or you will need it later on Okay, now what about the equation of a normal at p and q? Okay, so let's find out the equation of a normal so Let's first find it at a generic point t comma 4t2 square So it's y minus y1 Is equal to slope What will be the slope? Slope will be minus 1 by 8 t x minus x1 isn't it? Yeah, am I correct is that fine? What is that? slight So y y square So it's 8x minus 8 by x correct So now continuing from here it will become 8ty minus 32t cube Is equal to t minus x Is that fine? Yes So you can get x plus 8ty is equal to t plus 32t cube So I can clearly say that for the normal drawn at t1 it will be x plus 8t1 y is equal to t1 plus 32t1 cube Let's call it as first equation and x plus 8t2 y Is equal to t2 plus 32t2 cube. Let's call it as equation Okay, now let's say the point of intersection is h and k So you have to find the point of intersection of 1 and 2 right? So, uh, can you find x and y and compare it with h and k? So let me just replace it with h y is with k So if you subtract it it becomes very easy to find k so 8 t1 minus t2 k will be t1 minus t2 plus 32t1 cube minus t2 cube Okay And we clearly know that t1 is not equal to t2 so I can cancel t1 t2 from everywhere So I can say 8k is 1 plus 32 t1 square t2 square plus t1 t2 right? correct Yes So k is 1 plus 32 times t1 square t2 square plus t1 t2 by 8 Yes or no? Yes, yes sir All right, I can use another information that t1 t2 is actually 1 by 16 isn't it? we had we had used this information right minus 1 by 16 So I can replace this with minus 1 by 16 over here minus 1 by 16 Now what is it? h What is going to be h? So we have to substitute it back over here. Yeah, yeah So that's cool now. Yeah, that's becoming too long It was actually it won't be because there's 8 and 8 will get cancelled Sorry? So don't be that long because that 8 and 8 will get cancelled A lot of other things can also happen. For example, if I multiply this with t1 and this with t2 So this with t2 and this with t1 and subtract Let's see what happens So when you do that you get h t2 minus t1 Will be t1 t2 Sorry, it'll automatically get cancelled over here, right? So here t1 t2 will get cancelled. So we'll be left with 32 t1 t2 t1 square minus t2 square correct That's a good news because now this will become or 32 t1 t2 will be like minus 2 right Yes This will become minus 2 And this will be t1 plus t2 Into t1 minus t2 correct So ultimately you can lose out this term. So you can say h is twice of t1 plus t2 Ah now that's something to look upon. Yeah. Hey guys, I have an idea Uh, let's not do this step. Yeah I have an idea Because I I'm in a favor of completing a perfect square over here. So can I do one thing? Just correct me if I'm wrong I can write this as 1 plus 32 t1 plus t2 square minus t1 t2 right So that you know t1 t2 so All right, so that will make it 31 1 plus 32 t1 plus t2 whole square minus 32 t1 t2 Which is actually going to be plus 2 right? Yeah Yes, it's done. Yeah, that's a great news. So What I can do next is I can write this as 8 y is equal to 1 plus 32 times Can I say Even plus e2 will be x by 2 plus 2 right square Yes, sir Yes, sir So 8 y is going to be 1 plus 8 x square plus 2 That means 8 y is going to be 3 plus 8 x square. Is that fine? Yes. Yes, sir Okay But somehow the none of the options are matching Anyways, we'll stick to this result. There must be some problem with the options Because I think we have done everything correctly, right? Yeah, so one second could you scroll up? Scroll up or scroll down or you want to go down and see all the options I think we haven't done anything wrong What happened after is that fine Yes, sir I'm sure something is so forget about these options The big ears. Yeah Moving on to the next problem Tangent is drawn at any point x1 y1 Tangent is drawn at any point x1 y1 on the parabola Y square is equal to 4x Okay Now tangents are drawn Now tangents are drawn from any point On this tangent to the circle to the circle x square plus y square is equal to a square such that all the chords of contact all the chords of contact pass through x2 y2 Okay prove that four times x1 by x2 plus y1 by y2 square is going to be zero Read the question carefully. It's a tricky question. It's easy question, but it's tricky Any idea? Yes, sir. I think I got it. It's 2 y2 is like a pole, right? Right, right. Correct. If you look at the question in an opposite way What is happening? The tangent is going to be a polar for this pole Isn't it? Yes or no? Correct So this is your tangent again and The polar will be xx2 yy2 equal to a square Okay They both represent the same line Isn't it? Yeah, so now you just equate coefficients Yeah, so equate the coefficients So let's try to equate the coefficients now So this equation I can write it as Uh 2ax minus yy1 is equal to minus 2ax1 So these two represent the same thing now So comparing the coefficients comparing the coefficients so I can say x2 by 2a is equal to y2 by minus y1 is equal to a square by minus 2ax1 So I somehow got the uh expression for y1 by y2 Now, how do I get to this expression? x1 by 1 lies on the parabola also You can use that Sorry? x1 by 1 lies on the parabola also Okay So we can have one more equation and x2 by 2 lies on the side Oh no no no no no And x1 by 1 lies on the parabola Yeah, so what does it give me? How does it help me? That provides number equal But I need x1 and x2 and y1 and y2 both over here Isn't it? Yeah, right So I can do one thing Uh From this expression, I get x2 by 2a is minus y2 by y1 is equal to minus a by 2x1 Yeah Great So I can say uh y1 By y2 is 2x1 by a correct Right Yes, sir Here what I can do is I can express everything in terms of x1 and y1 So let's say x1 by x2 x2 really what? x2 will be Minus a square by x1 Correct Yes, sir So minus a square by x1 will be this correct And y1 by y2 square will be 2x1 by a square Right So it automatically gives you the result as zero because it becomes correct 4x1 square by Minus a square and this gives you 4 x1 square by a square cancelling each other out giving you zero Okay Yes, so hence proved All right, so guys, uh, we'll call this the uh end of the session over here But I would like to emphasize on the following points You know whichever clinic you are learning At the end of the day you should be Good in manipulating you should be good in playing with the terms And you should avoid taking longer routes because it's very easy to get Just like trigonometry. It's very easy to get Lost, you know for example use parametric points wherever you can Uh, whatever concepts you are trying to apply try to think it in various directions Right, for example, the concept of polar for any point outside is basically nothing but caught of contact So that's what we use in this case as well So all these learnings are very important while you are solving question in any conic section and try to solve As many problems as you can on the concept of locus Especially where multiple, uh, conic sections are involved. For example, in this case circle was involved parabula was involved so try to deal with Multiple phonics at the same time that will be very helpful for you to understand even if you see the last j Uh question it had two conics involved in the coordinate geometry problem Okay, and rest other formulas that we have learned is just going to be acting like a tool So they just help you to understand Just help you to implement your roadmap. It is not like helping you to solve any problem So if you remember condition of tangency, it just makes your life easy Right, it doesn't help you to solve the problem Okay, so that's why many a times students who do not know the formula also are able to you know Solve the question by just using their basic understanding of locus and all those concepts So thank you very much. So we'll be ending the session sunday 9 30 to 11 30 Uh ellipse Thank you, sir over and out. Thank you. Bye. Bye. Thank you, sir. Thank you, sir