 Hello all greetings from Centrum Academy So I welcome you all to the fast-track revision session on Koenig sections for JEE main exam So those who have joined in the session I would request you to type in your names in the chat box so that I know who all are attending this session so in this particular session guys we are going to Take care of three major chapters in Koenig section that is parabola ellipse and hyperbola and We'll be doing a good mix of problems from JEE main and JEE advance So I understand Raja Jee Naga people would not be able to join this because of their Graduation ceremony today, but others please type in your names. You have joined in the session right now Good afternoon everyone. Good afternoon wishes Rohan, Ramcharan, Suresh All right, so just like the previous sessions we'll start solving some problems and again I will be giving you some time to solve it and I'll only start solving it once I get This time the bash length is less so at least I need Two response for me to start solving the problems So this is the first question for you here Let A and B be two points on the parabola y square is equal to 4ax and If the circle with chord AB as diameter touches the parabola then we have to Find mod of y1-y2 So x1 y1 and x2 y2 happen to lie on the parabola and there is a circle with chord AB as the diameter We have to find the value of mod y1-y2 Hello. Good afternoon, Sundar. Yes. Anybody. All right, guys Let me just give you some hints with respect to this problem. First of all, we all know that if you're trying to write down the equation of a circle, okay Whose diametrically opposite points are x1 y1 and x2 y2, right? So we have all learned that if you are finding the equation of a circle whose diametrically opposite ends are x1 y1 and x2 y2 okay The equation for such a circle is given by x-x1 x-x2 plus y-y1 y-y2 equal to zero right Okay now this circle happens to touch this particular parabola right and x1 y1 and x2 y2 are points on this particular parabola correct Yes or no Right, so I can say y1 square is actually for a x1 and We can say y2 square is for a x2 Okay, so these two conditions will be automatically true because x1 y1 and x2 y2 happened to lie on the parabola y square Is equal to for a x? Okay Now what I'm doing now here is I'm trying to simultaneously Solve these two Right Since the parabola and the circle meet. In fact, they touch each other. I would try to simultaneously solve these two Okay, and how would I try to solve it? first of all Wherever there is an x I would write it with y square by 4a Okay, and wherever there is an x1 I would write it as y1 square by 4a x1 Why was y1 square by 4a correct? So I can write this as Y square by 4a minus y2 square by 4a Okay, and these terms I would keep it as such now the idea here is to get everything in terms of y Okay Right the idea here is to get everything in terms of y because of my required desired expression Okay, so I can write this as correct me if I'm wrong I can write this as 16 a square I'll multiply throughout with 16 a square so I can write this as 16 a square y minus y1 Okay, y minus y2 Okay Plus this will become y square minus y1 square and y square minus y2 square okay Now I can take y minus y1 and y minus y2 as common and I'll be left with 16 a square plus y plus y1 Y plus y2 Equal to zero correct now what does it suggest it suggests that The circle and the parabola they actually meet at four points because if you see essentially This is a bi quadratic in why correct this entire expression this entire formula is Actually a bi quadratic in why Correct this is a bi quadratic expression in why and it is no surprise to me that Two of the roots will be y1 and y2 because these two also lie on the parabola So y1 and y2 would be the common roots coming up So the other two roots are hidden in this right? So in this quadratic equation the other two roots are hidden the other two roots right they are hidden in this equation 16 a square and You can write this you can expand this and write as y square plus y times y1 plus y2 Plus y1 y2 equal to zero Right in fact, let me write the constant towards the end so that it actually looks like a quadratic Plus 16 a square now this quadratic Basically should give you two real and equal roots because this Circle touches the parabola so basically two points it will cut the parabola and Two coincident points it will be touching the parabola. So this must have this equation must have Real and equal roots. This must have real and equal roots. That means the discriminant of this Okay, so let me just go on to the next page We can adjust over here itself. So the discriminant of this which is actually b square minus 4 ac This should be equal to zero correct, right in other words y1 plus y2 whole square Minus 4 y1 y2 is equal to 64 a square is equal to 64 a square correct Right, which clearly implies that I'm just going on to the next page guys which clearly implies that Y1 plus y2 whole square minus 4 y1 y2, which is equal to 64 a square clearly implies that Y1 minus y2 the whole square Will be equal to 64 a square correct Now just take the square root on both the sides. That is y1 minus y2 That's going to be plus minus 8 a that means mod y1 minus y2 will definitely be 8 a and Basically, that is nothing but Your option number To which is correct Your option number to which is correct. Is that fine guys? Fine. So for the first problem, we spend a lot of time. It's around 10 minutes that we spend solving it Okay, so let us move on to the next question. So this is a second question Y is equal to under root of 3x plus lambda is Drawn through the focus of the parabola y square is equal to 8x plus 16 If two intersection points of the given line and the parabola are a and b such that the Perpendicular bisector of a b intersects the x-axis at p Then find the length of ps Let me draw the situation for you so that things are clear in your mind now this parabola is basically Y square is equal to 8x plus 2 Right y square is equal to 8x plus 2 that means it's 4a is equal to 8 and It's vertex is that minus 2 comma 0 Right, and it's opening rightwards And this parabola will be opening rightwards. Okay, so The parabola structure would look like this Right, so this point here will be minus 2 comma 0 Now what is given to us is that there is a line which passes through the focus so this line passes through the focus now Where will be the focus of this particular parabola? Can you type guys? Where will be the focus of this parabola my bad? This is 4a is equal to 8 so a is equal to 2 okay by mistake. I had written 4a is equal to 2 so a is equal to 2 Right focus will be at the origin so this will be your focus. Okay, so focus will be at origin Now there is a line this which passes through the focus Okay, we all can see that it is a line whose slope is root 3, right? And it passes through the origin means lambda has to be 0 That means lambda has to be 0 Correct, no doubt about it Because the line has to also pass through the origin so this quantity here lambda will actually become 0 Okay Right, and this is 60 degree if I'm not mistaken, right? now Let's see the next part of the question If the two intersection points of the line are a and b so let's say the point meets The line meets the parabola at a and b Such that the perpendicular bisector of ab so let's say there is a perpendicular bisector that I draw of ab like this Okay, so this is a perpendicular bisector of ab Which meets the x-axis at a point p Okay, now they're asking you the length of ps They're asking you what is the length of ps? So I think I had given I've given you enough hint for you to proceed from here So once you have solved it please type in your response in the chat box All right guys, so what do you have to do is you have to simultaneously solve Y is equal to root 3x With the equation of the parabola This correct so if you substitute y as root 3x it becomes 3x square Okay, minus 8x minus 16 equal to 0 Now we all know that the roots of this equation would be the x coordinates of ab right So let's say a is x1 comma y1 And b is x2 comma y2 Right, so we know that the roots of this equation would be x1 and x2 correct Now we know that x1 plus x2 would be Minus b by which is 8 by 3 Right, right So what will be x1 plus x2 by 2? So can I say x1 plus x2 by 2 will be nothing but 4 by 3 Right Now it is the x coordinate of this point. What are the meaning of x coordinate of this point? It means that let me drop let me call this point as e and let me call this point as m Okay, so can I say this is nothing but 4 by 3 This point is nothing but this length is nothing but 4 by 3 right Now if this is 4 by 3 and this is 60 degree Can we say that? This length over here This length over here would be 4 by 3 divided by cos of 60 degree Isn't it Based by how hypotenuse is cos 60 degree Correct, so hypotenuse is based by cos 60 degree Which is actually 8 by 3 again Correct, so this length is 8 by 3 right Correct if this length is 8 by 3 and I want to find sp or ps. So what will be ps? Can I say again that cos of 60 degree will be uh Let's say I call this point as o e o by sp In fact, I should not call it as o because I've only called it as s. So let me just not call it as o Because I've only used s for that Yeah So it will be es by ps Correct So es by ps is nothing but 8 by 3 divided by ps ps is unknown to me So again ps is going to be twice of 8 by 3 So your answer is going to be 16 by 3 units, which is actually your option number 4 which is correct Your option number 4 is correct Right guys, so never forget trigonometry and coordinate and uh simple geometry while you're solving problems on coordinate geometry So no question with respect to this particular solution if if you do have any question, please type it on the chat box If not, we'll proceed to the next one All right, so we are moving on now to the third question I think this is a simple question to solve if the points 2a and a lies inside the parabola Then a lies in which interval. I think all of you should be able to answer this This is a very simple question based on the position of a point with respect to a conic Okay, so wishes, uh says option 2 I need just one more response Okay, so purik also says the same guys. We all know that If a point Lies inside the parabola the power of the point with respect to that curve would be less than zero, right So if I put xs 2a, so that is 4a square minus 4a minus again 4a Plus 3 should be less than zero. That means 4a square minus 8a plus 3 should be less than zero correct right And we can actually factorize this Right, so 12 means we can break it as minus 2a minus 6a plus 3 less than zero So take 2a common. So it's 2a minus 1 Uh, take a minus 3 common. So 2a minus 1 is less than zero So 2a minus 3 and 2a minus 1 is less than zero Right, if you plot the wavey curve for this, you'll have a 1 by 2 and 3 by 2 with a plus minus plus So less than zero means you have to cite the interval where you have written negative sign So your a should belong to the interval half to 3 by 2 open interval, which is clearly your option number 2, which is correct Easy question. So please do not forget this Moving on to the next question That's question number 4 for you So the question says there's a line ax plus by plus c equal to zero through the point minus 2 comma zero And this intersects the curve y square is equal to 4x in p and q such that 1 by ap Such that 1 by ap plus 1 by aq is 1 by 4 So 1 by ap plus 1 by aq is equal to 1 by 4 where p and q happen to be in the first quadrant Right then find under root of a square plus b square plus c square. You all know that abc here are coefficients of x y and the cross-center Okay, so again, you try to figure out the situation through a diagram. So we have A parabola y square is equal to 4x And we have a line which passes through minus 2 comma zero Right, so minus 2 comma zero is a point on the x axis. So there's a line Uh, which passes through this Okay, and it cuts The parabola at p and q It cuts the parabola at p and q Such that 1 by ap plus 1 by aq is 1 by 4 So I'll just give you a hint over here try to think in terms of the distance form of a line Okay, so first let's say I call this line as l. Okay try to write down the distance form of this line Right, so you know a point on it correct You know a point on it minus 2 comma zero correct And in that you also know the equation for it, but you know we can always compare our final result with this result So what I'll do here is I will try to write down the equation of this line as x plus 2 by cos theta Is equal to y minus zero by sine theta equal to r correct That means I obtained the parametric form for this line, which is uh r cos theta minus 2 And y is equal to r sine theta Now as per the question This line actually meets the parabola at p and q That means it must satisfy the equation of y square is equal to 4x So put y as r sine theta. So it becomes r square sine square theta Is equal to 4x this will be 4x Correct. Okay. So here you will see that you are getting a quadratic like this in fact, I can write things in terms of A quadratic over here plus 8 equal to 0 Okay, now what is given to you is 1 by ap plus 1 by aq is 1 by 4 You know indirectly you have been given that 1 by r1 plus 1 by r2 is equal to 1 by 4 Where the r1 r2 are the roots of this equation. So its roots are r1 and r2 correct Which means r1 plus r2 By r1 r2 is given to you as 1 by 4 Right What is r1 plus r2 In a quadratic expression equation the sum of the root is minus b by a So minus b by a will be 4 cos theta by sine square theta And this will be 8 by sine square theta. So sine square theta sine square theta gets cancelled off Right So this becomes here Cos theta is equal to half. So theta will become Obviously it will become 60 degrees correct Now if theta is 60 degree we can write down the equation of the line over here as You can directly use the slope point form as well y minus 0 is tan of 60 degree times x plus 2 Isn't it which means it becomes y equal to root 3 x plus 2 root 3 Okay, now try to compare this with this line Right. So here we have uh y is equal to minus ax by b Minus c by b right Okay, or we can directly compare Uh Exact value is not possible. So the value can be it should be asking value can be Not then is it should be asking can be because if you directly compare. I'm sorry if you directly compare Let me just uh write it as uh root 3 x Minus y plus 2 root 3 equal to 0 You can directly compare this line with this line. That means a could be compared with root 3 b could be compared with minus 1 and c could be compared with 2 root 3 So in that case this expression would be under root of 3 plus 1 plus I think this will be 12 That's going to be root of 16. That is going to be 4. So it can be 4 right That actually depends upon the proportionality constant So when you're comparing two lines, you cannot directly compare the coefficient of x y and c with one another You have to take the ratio Okay, you have to take the ratio of the coefficient of x with the coefficient of y with the coefficient of the constant term Right So here the the way in which they have framed the question is slightly, you know confusing because abc could be in any form So this could be a bonus question if you attempt this question for you will be getting the marks Is that fine? All right. So moving on to the next question now Suppose is the parabola y y equal to x square minus a x minus 1 which intersects the coordinate accesses at three points a b and c respectively The circum center of triangle abc intersects the y axis again at the point 0 comma t Then the value of t is Let me again help you with the sketch of this So situation is that we have a parabola And we have a circle since circle is difficult to draw. I'll be first drawing a circle Okay, uh Let's say there's a circle like this and abc. This will be your point abc. Let's say a b And see so first I've drawn the circum center of the triangle abc. Okay, so abc are the points where the Parabola Meets the coordinate accesses. So I'll be just roughly drawing it. Okay Now this circum center intersects Let me draw a more better diagram Not happy with this Yeah, it must be somewhat like this Yeah More or less a parabola right now this circle also cuts the y axis again at a point 0 comma t Now they're asking you the value of t And this parabola over here has the equation y is equal to x square minus ax minus 1 You can assume these points to be alpha comma zero and beta comma zero Yeah, there's a slight mistake in writing away. This is actually the circum circle Not the circum center. How can a point intersect y axis circum circle Normally, there are some typo errors that happens while typing these questions So please treat this as the circum circle Now guys don't don't treat ab as the diameter, please Or will it be the diameter? It may not be the diameter By the way, can you find out the point C first of all? Who will tell you what is the coordinate of point C? Where does this parabola meet the y axis? So let's say this is your y axis. Where does the parabola meet the y axis? The parabola will meet the y axis Where x is zero, so it'll meet at zero comma minus one. Correct. Okay now What will be my alpha and beta alpha and beta would be the points where the parabola meets the x axis Correct. So x axis is y is zero. So we have to solve for this quadratic equation Correct So when you solve for this quadratic equation by using Cedar Acharya formula, I can say my alpha will be equal to a plus b square minus four ac by two a Correct And beta will be similarly a minus under root a square plus four by two a Okay, so as of now, I'll call them alpha beta because I don't want to plug in such big expressions over here Now I would like to ask you One simple question Okay, if there are two points known to us correct right And I want to write down the family of circles passing through these two points x1 y1 and x2 y2 So if I want to write the family of all circles which passes through these two points, right? So they can be infinitely many circles How would I write the family of circles passing through these two points? Right? So what do we do is we use a simple concept We first make a unique circle which passes through a and b, right So the unique circle which passes through these two points a and b let me call it for the time being or let's say call it p and q Will be a circle which is Having its diameter as p and q. Okay, and not only that we also make We also find the equation of this line So basically once, you know the l equation and s equation You can always find out the family of circles passing through the intersection of l equal to zero and s equal to zero Right in other words if you would recall Uh the concept of family of circles we had discussed that The family of circle equation in that case would be this plus lambda times The equation of this line which we normally write it in a determinant form like this Okay, so this is the equation of a family of circles Right So here if you would see I can write down the same equation The same equation can be written very easily by writing x minus alpha x minus beta Plus y minus zero y minus zero Plus lambda times this line is actually your x axis so you can write lambda y equal to zero So this would become the equation of the family of circles Okay, okay passing through point a b right Now, how do I get this lambda by using the fact that the point the circle will satisfy this the circle will satisfy this particular A point correct. So when I put the value of x as zero I get alpha beta and I put value of y as one so I get Uh plus one minus lambda right and equate it to zero Okay, what do I get the value of lambda as I get the value of lambda as one plus alpha beta Now here alpha into beta as you would see is the product of the roots So alpha beta will be equal to c by c by is minus one So lambda will become one minus one which is zero Correct So in short, basically, what did I get the the family member which I desired was x minus alpha x minus beta plus y square equal to zero correct Now I want to know where does this particular circle cut the y axis again Correct So for you to know where it cuts the y axis again, you put again x as zero and y as t square So you get again alpha beta plus t square equal to zero which means T square is going to be minus alpha beta which is actually one So t could be plus minus one But minus one is already covered over here. So this has to be plus one That means your t value is going to be your plus one which is option number two is correct Which means option number two is going to be correct Is it fine guys? So basically to summarize what this particular problem actually demanded from you Whether you are aware of the concept of family of circles Correct So this was actually testing you on the concept of family of circles and of course in Association with the concept of some and product of the roots of the quadratic equation Okay, so as you would see finding alpha beta was useless There was no use of alpha beta. We primarily use the product of alpha and beta only Is that fine guys any question any concern with respect to this problem? Okay, so let's move forward So here you have a question the line x minus b plus lambda y equal to zero cuts the parabola y square is equal to four a x However, this is a redundant information. We always know a is greater than zero. Okay at parametric points t1 and t2 and the point the Expression b over here the Parameter b over here lies between two a to four a close interval Then we have been asked the range of t1 t2 again first, please draw the graph of the situation So this is the situation where there is a line which cuts the parabola at points p and q whose Parameters are t1 and t2 They've asked the range of t1 t2 When b belongs to my two a to four a First guys try to get the equation of the line pq Connecting two parametric points t1 and t2 and then compare it with this line So i've just given you a hint try to follow that see if you are able to find something from it So when I say point t1, I actually mean a t1 square comma 2 a t1 And point t2 means a t2 square comma 2 a t2 So if you have to write down the equation of the line pq the slope first find out the slope of pq Okay, so slope of pq will be 2 a t1 minus t2 by a t1 square minus t2 square which is nothing but 2 by t1 plus t2 So you can write down the expression to be this So which is t1 plus t2 y Is equal to 2x And this will become 2 a t1 t2. Okay So this is the equation of Now we can compare this with We can compare this with Lambda y is equal to minus x plus b So let's compare these two lines Let's compare these two lines So when you compare these two I can get t1 plus t2 by lambda Right is equal to 2 by minus 1 Is equal to 2 a t1 t2 by b Correct So from here I can say directly by comparing these two equations. I can say directly That my b is equal to minus a t1 t2 Right Right Now we have been given that b lies between minus 2 a to 4 a So I can say minus a t1 t2 belongs to this interval 2 a to 4 a correct Drop the factor of a from everywhere. That means 2 Will be less than equal to minus t1 t2 less than equal to 4 That means t1 t2 should belong to the interval Minus 4 to minus 2 which is actually nothing but option number 1 which is correct. Is that fine guys? Is it clear? Purveik, Vishis, Rohan, Sundar, Ramcharan, Kondanya Okay Guys one more thing I wanted to highlight over here This equation should be known to you in order to save time. You actually wasted 30 percent of your time deriving the equation of a chord Okay, so this things You should try to You know plug in these holes in your preparation So you should always know the equation of a chord connecting two parametric points t1 t2 of a parabola y square is equal to 4 a x This is the most widely used general form So it saves a lot of your time Right, yes Good. So now we'll move on to the next problem So here we have a question tangents p q and q r are drawn to the parabola y square is equal to 20 x plus 5 And y square is equal to 60 x plus 15 respectively Such that angle r p q is 90 degree Then the locus of p Then find the locus of p So there are two different parabolas and P q and p r are tangents drawn from external point p Such that the angle r p q is always 90 degree find the locus of p First of all try to recall that if you have a parabola y square is equal to 4 x right, we know that If y is equal to mx plus c is a tangent to this parabola Right c should be equal to a by m right correct So in other words what I'm trying to say is that y equal to mx plus a by m will always be a tangent To the parabola y square is equal to 4 a x Okay Now what we'll do here is that Since it is a case of a shifted parabola. So y square is equal to 20 x plus 5 with a shifted parabola. Remember this is your 4 a Right 4 a is 20 that means a is 5 Can I say that the same equation of the tangent will now become y is equal to m x plus 5 plus a by m Correct a by m will become 5 by m Right So this will be the equation of a tangent for this. So this will be a tangent for this in a similar way y square is equal to 60 x plus 15 Right What will be the equation of a tangent? Equation of a tangent will be y equal to m x plus 15 plus Okay a by m what will be a over here A will be 15 Okay, let me write m dash because I cannot treat the same m at both the places. So it'll have let's say m dash correct now What is given to me is that let's say tangent t1 and t2. So t1 is perpendicular to t2 That means m into m dash is equal to minus of 1 Isn't it? Yes, sir Correct. So I can write down the second equation as y is equal to Minus 1 by m x plus 15 Okay, this will be minus 15 m Okay Now from these two equations first and second equation We would try to eliminate m just try to eliminate Eliminate m because m is something that you have introduced in the problem. It was not given initially in the problem So we'll try to eliminate m Correct So how do I eliminate m? If you want you can write it in a proper form You can write this as uh, let's multiply with m throughout. So you can write it as 15 m square Okay, uh plus My Plus x plus 15 This is equal to zero and here I will get similar expression I can write this as My is equal to m square x plus five plus five Okay So m has to be eliminated from these two equations Now basically If this point you consider to be h comma k This will satisfy both these equations. Isn't it? So what we'll do is We'll write the first equation as m square h plus five Correct minus m k plus five equal to zero and Here I will write 15 m square plus m k Plus h plus 15 equal to zero Okay Nothing as if you have to solve for h and k Think you have to solve for h and k and try to relay and try to get rid of m in the meanwhile Okay, so you have to simultaneously solve these two equations If you add them what happens If you add them you would see mk k will get rid of right Okay, try adding them So when you add them you'll see you'll get m square h plus 20 Okay, and again h plus 20 equal to zero Correct, which is nothing but h plus 20 times one plus m square equal to zero Right, which clearly gives me h plus 20 equal to zero Which means x plus 20 equal to zero is a possible locus of p That means option number four is correct Right because obviously one plus m square cannot be zero. So h plus 20 has to be zero and if h plus 20 is zero, which means x plus 20 is zero By generalizing h with x and hence option number four becomes correct right So guys, you can imagine how how this question was actually framed this question was actually framed to test you on your knowledge of the condition of tangency Right and whether you can actually frame a condition of tangency when it is a case of shifted opera Bola And then they mix the concept of locus with it Okay, and of course with the concept of perpendicularity of two lines. So see how questions are framed in j So if you are weak in any one of the concepts, you will definitely be not able to solve this problem Is this okay guys any questions any concerns with respect to this Can you move on to the next one? So next question is find the locus of the centroid of triangle formed by a tangent to the parabola y square is equal to 36 x With the coordinate accesses. This is pretty simple. I believe All right. So purvic has given a response option number two Anybody else I need one more response guys So that I can start solving this Again, this is the parabola y square y square is equal to four Into nine into x. So this can be considered as your a Okay So any point on this you can actually take it as 90 square comma 280 which is 18 t correct anyways Now we first have to form a tangent at any point Right, so we know the equation of tangent drawn to any point at square comma 280 At square comma 280 the equation of the tangent is given by The equation of the tangent is given by ty is equal to x plus at square right Right, that means ty Is equal to x plus nine t square Okay Now this tangent meets the x axis where y is zero Correct So it let's say it meets the x axis at point a so the coordinate will be Minus nine t square comma zero correct It meets the y axis. Let's say at A point where x is zero so that y could be 90 correct Now let the centroid be h comma k so let the centroid coordinate be h comma k So we can write from here that H will be equal to Yeah, by the way the other point is origin So the centroid will be equal to zero minus nine t square zero by three which is minus three t square And k will be three t. I'm sorry three t not three k Okay That means t is equal to k by three so I can substitute it over here. So I will get h Is equal to minus three into k square by nine Which means three h minus k square is three h is equal to minus k square. That means k square plus three h is equal to zero No Generalizing this we get y square plus three x is equal to zero which is option number two is correct Okay, that is pretty simple question Is this fine any question any concerns? Please do ask me If not, I will move on to the next question now So read the question carefully the question says PC is the normal at p to the parabola y square is equal to 4x C being on the x axis c being on the axis Cp is produced outwards to q So that pq is equal to cp Then the locus of q is the parabola whose focus is at Okay, ram chiran says option four What about others? I need one more response Before I start solving this Okay, so poor week also goes with option four. Okay, so let me just quickly sketch the graph for this So basically we have a parabola The standard one y square is equal to 4x and we are sketching a normal at a point p So this is a point p and this normal meets at The axis at c Okay, and we extend this here to q such that cp is equal to pq correct Now we all know the equation of a normal equation of the normal To the to the parabola is y plus tx is equal to 280 plus atq Correct drawn at any point p over here right So let's say p is your point at square comma 280 So the equation of a normal would be this so we all know this is the equation of the normal Right Now in order to know the coordinates of c I will put y as zero So when you put y as zero you get it as 2a plus at square comma zero Okay, let's say this point q whose locus is to be found out is h comma k correct, so can I say H plus 2a plus at square whole divided by 2 will be equal to at square correct And I can also say that k plus zero by 2 is equal to 280 Which means k is equal to 4 at correct And from this equation I can say h plus 2a is equal to at square correct Now what I will do here is that I will try to replace my t with k by 4a So when I do it over here When I make this replacement over here I get h plus 2a is equal to a t square Which means h plus 2a is equal to k square by 16a And if you generalize this you get x plus 2a is equal to this you can say y square is equal to 16 x plus 2a In fact 16a x plus 2a my bad. Yeah, let me write it properly Okay Now we know that it's a case of a parabola whose vertex is at minus 2a comma zero correct 16a is actually equal to your 4 capital a let me call the parabola as y square is equal to 4 capital a x plus 2a okay So here your capital a is going to be 4a so you shift 4a in front That means your focus will be at a distance of 4a from this will come at 2a comma zero That means option number four is correct And the first one to answer this was Ramcharan Right, so Ramcharan you're correct But actually you did not give the answer you asked me a question. Is it option four? So the first one to legally answer this was Purvik. Okay. Anyways, I give the credit to both of you for this Okay, so a proper mix and match of locus plus your concept of normal etc is tested over here Okay So fine. We'll move on to the next problem now If a tp and tq be any two tangents to a parabola and the tangent at a third point are cuts them in p dash and q dash Then find the value of tp dash by tp Plus tq dash by tq Okay, so Purvik says one Okay, so most of you in fact both of you are again Giving me the same answer now. Let's so let's check quickly over here So let's first make the diagram from the diagram many things will be clear in your in our mind So we can take a that usual parabola y square is equal to 4x so we can consider this to be our parabola Okay, so we'll consider a point p over here Okay Coordinates a t1 square comma 2 a t1 Uh, we can consider, uh Another point q over here Let's say At two square comma 2 a t2 Okay So t is the point where the tangents drawn at these two places meet Correct So this is our point t and there's also a third point. Let me call it as r let's say At 3 square comma 2 a t3 And I've drawn a third tangent At this point which meets or which cuts Them at p dash and q dash Correct So we can say here for sure that the coordinates of t will be We all know the formula go right We all know that if you're drawing a tangent At a t1 square comma 2 a t1 And at a t2 square comma 2 a t2 the tangents at these two points The tangents at these two points tangents meet at At1 t2 comma a t1 plus t2 Correct, so t will be this Similarly p dash will be what? Okay, p dash q dash also we can find out so p dash will be At1 t3 comma at1 plus t3 and q dash will be the meeting point of At2 t3 comma at2 plus t3 now What we'll do is we'll consider this ratio to be Lambda is to let's say 1 minus lambda That means I'm considering tp dash by tp to be lambda is to 1 Correct So I'm considering things to be lambda is to 1 Lambda is to 1 Okay So if I'm considering things to be lambda is to 1 so let me Find the coordinates of p dash in terms of lambda that will be lambda At1 square Okay Plus At1 t2 By 1 Correct and this to be this is to be equated to At1 t3 So p dash x coordinate is to be equated to At1 t3 Okay, so if you drop the factor of t1 from everywhere you get and a also from everywhere So at1 at1 at1 will go off from everywhere. So you get t3 is equal to Lambda t1 plus t2 Is this fine is Oh, I made a very serious mistake. This is not one guy. This is Just one second. I'll be Making a small change over here. Yeah, I'm so sorry. Yeah, so it'll be lambda times At1 square 1 minus lambda times At1 t2 By the sum of this which is 1 This should be equal to x coordinate of p dash Okay, now I can drop the factor. So it will be t3 is equal to lambda t1 1 minus lambda t2 Okay, so now you can write this as t3 minus t2 Is equal to lambda t1 minus t2 That means lambda will be equal to t3 minus t2 by t1 minus t2 Guys similarly in this case also I can consider this to be let's say alpha is to 1 minus alpha And again, I will get the value of alpha from here in terms of the parameters as 1 minus t3 By t1 minus t2 Okay, you're gonna again apply the same concept that is alpha into At2 square and 1 minus alpha alpha into at2 square 1 minus alpha into At1 t2 by 1 Is equal to the coordinate of q dash which is actually at2 t3 So drop the factor of t2 and a from everywhere And from here you can get the value of alpha as this Okay Right now this is just like saying lambda plus alpha. So add these two so add these two quantities So when you add these two quantities You will be getting t1 minus t2 in the denominator and in the numerator will get t3 minus t2 plus t1 minus t3 So t3 t3 will go off Similarly t1 minus t2 and t1 minus t2 will cancel off giving you the answer as 1 Which is option number 1 is correct So yes again the first one to answer this was Purvik I think we have done this question in our discussion of Of parabola Okay Anyways So moving on to the next question now Yes guys any idea So the distance of two points p and q On the parabola y square is equal to 4x from the focus is given to you as 3 and 12 respectively The distance of the point of intersection of the tangents at p and q from the focus s is Okay, so let me say this is your point. This is your focus s This is your point p and this is your point q This distance is given to you as 3 And this distance is given to you as 12 And when you draw tangents at these two points Let me call this as a point t So they're asking you the distance st Okay, so vicious is saying option 2 Yeah, it should not matter though. It should not matter. Okay All right guys, so let us try to understand this situation now So basically what is given to you? Let's say if I consider this point to be a t1 square comma 2 a t1 and this point to be a t2 square comma 2 a t2, correct Then this point will automatically be a t1 t2 comma a t1 plus t2 Correct. Yes or no, okay now This 3 that is given to me is actually a plus a t1 square And this 12 that is given to me is actually a plus a t2 square We all know that the distance of any point from the focus Is a plus the x coordinate of that point Isn't it? Right, so let's say the focus is a comma 0 correct Now I'll show you something very interesting If you find st square If you find st square you would see that your st square will be a t1 t2 minus a whole square plus a t1 plus t2 minus 0 whole square Correct, which is actually a square t1 minus t2 whole square is t1 square t2 square minus 2 t1 t2 plus 1 And from here I will get t1 square t2 square plus 2 t1 t2 Correct you can cancel this off which will actually give you a square 1 plus t1 square t2 square plus t1 square t2 square, correct Now coincidentally the same expression would be obtained if you multiply these two Right when you multiply these two it will become a plus a t1 square a plus a t2 square So that will give you a square times 1 plus t1 square t2 square plus t1 square t2 square correct Which clearly means So all of these implies that st square is actually sp into sq Isn't it So st square will be equal to 3 into 12 So st square is going to be 36. So st is going to be 6 So your option number 2 becomes the right answer over here. So option number 2 becomes the right answer over here And let me see who's the first one to answer this Okay, so wishes is the first one to answer this correctly. Is that fine guys? Good. So now we'll move on to the next question Again a parabola question a parabola with the directrix x plus y plus 2 equal to 0 touches a line 2x plus y minus 5 at 2 comma 1 then the semi-latus rectum of the parabola Guys, can you see the screen? Okay, so wishes says option 2. All right, so let's quickly discuss this now So guys here we have a very important property that I want to discuss Okay, so basically we have a parabola I mean, I'm not drawing I'm not drawing this as the coordinate axis is I'm just drawing it as a normal case Okay, and this is our directrix Okay now One very important thing that we should all realize over here is that when we draw a tangent At any point We know that this tangent actually bisects This angle so this angle and this angle are same, right? Okay, so basically This distance is also the same right this distance is a plus. Let's say this point is x1 y1 Then this is a plus x1 and this is also a plus x1. This is also a plus x1 that means S is basically The image of t on the tangent The image of the point t on the tangent This is something which we need to appreciate over here. Okay So now using this concept first of all, we'll try to find the foot of the perpendicular Dropped from point p on the directrix Okay, so we'll try to find out The coordinates of t Given that p is 2 comma 1 Okay, we also know the equation of the directrix. So this is the equation of the directrix correct Now guys, I want to all of you to recall this formula Which I don't know how many of you remember this So if you have a line ax plus by plus c equal to zero and From point x1 y1 Right You are drawing a foot of the perpendicular on to this. Let's say it is x2 y2 Okay Then there's a relation that we normally use x2 minus x1 by a is equal to y2 minus y1 by b Is equal to minus ax1 by b y1 plus c by a square plus b square okay Very few people know this formula and from here you can directly find your x2 and y2 If you want to find out the mirror image Let's say the same formula now If you want to find out the mirror image of x1 y1 About the line ax plus by plus c equal to zero its mirror image is given by x2 minus x1 by a y2 minus y1 by b is equal to minus two times ax1 by 1 Plus c by a square plus b square Right now why i'm discussing this all of these from you is because now we can actually find out First of all the foot of the perpendicular about foot of the perpendicular dropped from 2 comma 1 on to the directrix So let's say this is alpha comma beta So I can say alpha minus 2 by 1 Will be equal to beta minus 1 by 1 will be equal to minus ax1 which will be uh, 2 By 1 which will be 1 Plus 2 by a square plus b square which is 2 So from here we can directly write this as I think this will be minus of 5 by 2 So I can say alpha is nothing but half In fact minus half and beta is nothing but Minus of 3 by 2 Okay, so this point t is actually nothing but Minus half comma minus 3 by 2 correct Now let us find out the focus focus is the mirror image of t about this tangent Correct So let us use the formula again to find out now this formula i'll be using Okay, so let's say this point is gamma comma delta so I can say Gamma minus of minus half which is plus half By 2 now this line has this line you will be following this line now Is equal to delta plus 3 by 2 divided by 1 Will be equal to let me just erase this part minus 2 times Yeah, when you put this minus half and minus 3 by 2 into this line you get minus 1 minus 3 by 2 minus 5 Whole divided by 5 So let us solve for gamma over here I'll be erasing a few things. So let me erase this part So gamma over here will become Okay, let me solve this first So gamma plus half by 2 becomes this will be minus 2 Now minus 6 minus 3 by 2 which is actually minus 2 times minus 15 by 2 So it will become 15 Okay, which is actually minus 3 So it's minus 6 minus 1 by So plus 6 minus 1 by 2 which is 11 by 2. Okay So this point will be 11 by 2 comma In a similar way, this will be 3 minus 3 by 2 which is again 3 by 2 Okay Now once we have got the coordinates of the focus Right, we all know the distance of the focus from the directrix. That is this distance is basically your 2a, right So all you need to do is find the distance of this point from the directrix So the distance of this point from the directrix will be 11 by 2 plus 3 by 2 plus 2 by under root of 1 plus 1 which is 2 And this is going to be 9 by root 2. So 9 by root 2 is going to be your semi lattice rectum. Okay So I think most of you who answered this answered it correctly So guys, we'll take a small break over here. Okay, so we'll take a break off. Let's say 5 to 7 minutes So we'll resume back at 6 10 All right guys, welcome back from the break. So let's start with this problem So we have just started with the ellipse Again a simple concept which is based on the location of a point with respect to a conic So find the number of integral values of alpha for which the point 7 minus 5 alpha by 4 comma alpha lies inside the ellipse Okay, so poor weak says option 1 How about the others? Okay guys, it's very simple. You just have to Put the point x and y as these two Expressions and make it less than zero right because it lies within the ellipse So a 7 minus 5 by 4 alpha square by 25 Plus alpha square by 16 minus 1 should be less than zero So If you simplify this over here it becomes 28 minus 5 alpha okay, and uh 16 Into 25 will become 400 So alpha square by 16 minus 1 is less than zero Let's multiply throughout with 400 So it'll become 28 minus 5 alpha square Plus 25 alpha square Minus 400 Is less than zero Now if you open the brackets so 25 alpha square and 25 alpha square will become 50 alpha square And 140 into 2 will be minus 280 alpha And we will have the constant terms as a 28 square minus Uh 400 28 square minus 400 will be your constant of so 28 square would be 784 correct 28 square is 784 if i'm not wrong 64 4 16 16 32 4 8 784 right so it'll be minus of 3 plus 384 Less than zero hope i've done everything correctly over here Yeah, yeah Now let's divide by Let's divide by uh factor of 2 so it'll become 25 alpha square minus 140 alpha Plus 192 equal to zero Okay, plus 192 equal to zero Now is it spittable Can we factorize this? Can we try 80 60 Okay, let's try 80 60 so minus 80 minus 60 alpha plus 192 equal to zero So I can take 5 alpha common so alpha minus 16 fact 5 alpha and this is 5 alpha minus 16 And I can take uh minus 12 common so minus 1 into 5 alpha minus 16. Yes, it works minus 16 Okay, so it becomes 5 alpha minus 16 into 5 alpha minus 12. Sorry not equal to zero It should be less than zero. I should not be writing equal to zero. It's all less than zero less than zero less than zero Okay Now if it is less than zero that means if you draw the wavy curve It will be uh 12 by 5 and 16 by 5 so it will be plus minus plus So your alpha should lie between 12 by 5 to 16 by 5 So alpha should lie between 12 by 5 to 16 by 5 Now 12 by 12 by 5 is approximately 2.4 and 16 by 5 is approximately 3.2 So there is one integer actually one integral value of alpha which lies between this so option number 2 becomes correct in this case right Yes, perfect. So you identified your mistake So guys we can move on to the next concept. This was simple one Yeah, so let's look at the next question now So here is a concept where we have been uh ask the coordinates of the vertices b and c of a triangle r 2 comma 0 and 8 comma 0 respectively And there's a vertex a which is varying in such a way That 4 tan b by 2 in fact, there is a slight Error over here. This should be Only one times They have written it twice So please strike out the this one. So this is strike out Okay, so they just written it twice Okay, so 4 tan b by 2 tan c by 2 is given to you as 1 Then the locus of a is which of the following So this is b and this is c And there's a point a which is a varying point Okay But this varies in such a way that 4 tan b by 2 Tan c by 2 is always 1 So they just try to take a clue from the options options itself is saying that the locus is an ellipse right all the expression over here is ellipse Right, that means you know internally within you that your b plus c is actually your constant, right Isn't it And this a is actually your distance between the This is actually the distance between your foci It's the distance between your foci Okay, how do I note the distance between this foci because the midpoint will be 5 comma 0 And you can see in your options. They have always used The center as 5 comma 0 So take a clue that the options are telling you that b plus c has to be A constant and that constant will be actually the length of the major axis So the only thing that I need to be worried about what is my length of the major axis Correct, and of course I can find the eccentricity if I know the length of the major axis by knowing the distance between the two foci And then I can easily write down the The generalized form of the equation of the ellipse, isn't it? Right now let's use this clue over here Um Guys, I want to ask you how many of you remember the auxiliary formula, right? I'm not Sure whether you are aware of this formula tan a by 2 does everybody know that tan a by 2 would be under root of s minus b s minus c by s s minus a correct similarly tan b by 2 is under root of s minus a s minus c by s s minus b correct and tan of Tan of c by 2 should be under root of s minus a s minus b by s s minus c If you are not aware of these formulas time to revise properties of triangles, okay now as per the given question the product of The product of this and this into 4 is equal to 1 right, so let me multiply these two So the multiplication of this will be giving you s minus a s minus c by s s minus b Into s minus a s minus b s s minus c And this is equal to 1 by 4 This is equal to 1 by 4 Now which terms will get cancelled off s minus c s minus c s minus b s minus b so you get s minus a by s is equal to 1 by 4 isn't it so s is equal to 4 s minus 4 a Let us put s in terms of a b and c so I will be going to the next page guys So if you want to copy something you can copy but anyways, it is getting recorded So you can watch it as many number of times you can So I'm moving on to the next page Okay, I'll write it over here. Yeah So we just now got s is equal to 4 s minus 4 a That is a plus b Plus c by 2 is equal to twice a plus b plus c minus 4 a Okay, so a plus b plus c is equal to 2 b plus c minus 4 a Let me just rewrite this s is equal to oh say we can write 3 s is equal to 4 a Yeah, correct So we can write 3 s is equal to 4 a So s is equal to 4 a by 3 That means a plus b plus c by 2 is equal to 4 a by 3 Have been given about a some information about a Yes, a is basically the distance 6 correct So distance between 2 comma 0 and 8 comma 0 is your a actually right the distance is 8 so distance is 6 between them so I can say your s is equal to 8 correct So you got your s as 8 that means Your a plus b plus c is equal to 16 Correct that means b plus c is 16 minus 6 which is 10 Okay, so from that given figure that we had drawn 2 comma 0 8 comma 0 We know that b plus c is equal to 10 that means 2 a is equal to 10 correct in fact the 2 a a is the let me not confuse you you were here so let me call this as x for the time me or let's say p for the time me okay So 2 a is the length of the major axis and 2 a e is given to us 6 So we can easily find out e from here, which is 3 by 5. Okay, so your a is 5 e is 3 by 5 Now what is b square b square is a square 1 minus e square So b square is going to be 25 1 minus 9 by 25 that is going to be 16 correct, so in light of all this We can conclude that We can conclude that Your equation of the ellipse would be x minus 5 the whole square by a square y minus 0 the whole square by b square equal to 1 Okay, which is clearly your option number one, which is correct. Is that fine guys Shallow So with this we are good to go to the next question So read this question carefully p q and q r are two focal cords of an ellipse And the eccentric angle of p q r Is 2 alpha 2 beta 2 gamma respectively Then we have to find tan beta tan gamma is equal to which of the following options Okay, I'll draw the diagram for this so p q Okay, this is one focal cord And q This is another focal cord right so p q and q r are two focal cords of the ellipse And the eccentric angles of this point is 2 alpha that means this point is a cos 2 alpha comma b sin 2 alpha q is a cos 2 beta comma b sin 2 beta And this is going to be a cos 2 gamma b sin 2 gamma Hope you all are aware of the equation of A cord connecting two eccentric points So the equation of a cord connecting two eccentric points theta 1 and theta 2 I'm just going to write it down in case you have forgotten it Is given by x by a cos theta 1 plus theta 2 by 2 Plus y by b sin theta 1 plus theta 2 by 2 Is equal to cos of Theta 1 minus theta 2 by 2 Now here your theta 1 theta 2 etc are written in terms of 2 alpha 2 beta and 2 gamma Okay, so in this case if let's say I have to write down the equation of p q What would the equation be? It'll become x by a cos alpha plus beta y by b sin alpha plus beta Is equal to cos alpha minus beta right Now this passes through the focus a e comma 0 and this r q cord passes through minus a e comma 0 Right, so what we can do is we can make the equation satisfy a e comma 0. So this will give you e times Cos alpha plus beta is equal to cos alpha minus beta which means E can be written as cos alpha minus beta by cos alpha plus beta Okay similarly for the other one for the equation of r q we can directly comment that um e will be minus of cos Or you can say minus of e will be cos Anyway, you want to realize it. It's the same thing cos Uh gamma minus or alpha minus gamma by cos alpha plus gamma Okay Now these two are negatives of each other these two expressions are negative of each other So I can just make one the negative of another that means I can write cos alpha minus beta by cos alpha plus beta as negative of cos alpha minus gamma by cos alpha plus gamma Now apply componento and dividendo here Apply componento and dividendo Apply componento and dividendo Okay, uh, can I just clear the screen if you have understood this Guys, can I clear the screen? So let me just write it up fresh So basically what I want to say is that when you use componento dividendo on this cos alpha minus gamma by cos alpha plus gamma So using componento dividendo we can write cos alpha minus beta at alpha plus beta minus cos alpha minus beta by cos alpha plus beta plus cos alpha minus beta as So we are doing subtraction of this will become cos alpha plus gamma You are adding it up. It'll become plus cos Okay And in the denominator you are adding it up. So it'll become cos alpha plus gamma minus cos Alpha minus gamma. Okay, so let's see. What do we get? So on the numerator here we end up getting two sine alpha We end up getting this And the denominator we end up getting this Here also in the numerator we end up getting cos alpha cos gamma and denominator we get minus two sine alpha sine gamma So minus minus two two gets off right Now we can see here that This term is a cot term So it's cot alpha and this term is a tan term Similarly here also we get Sorry, uh, this is a tan term Tan term and here we have also tan term, but here we have a cot alpha And cot gamma Okay, so now let's cross multiply this So we can write this as cot square alpha That means I'm taking I'm writing this as By cot alpha, so I'm taking it on this side and I'm writing this as tan gamma So cot alpha is tan beta tan gamma, which is actually our desired expression So tan beta tan gamma will be cot square alpha That will be cot square alpha okay, so this concept was basically testing your Whether you it's a memory-based question whether you actually remember the equation of the cod of Cod connecting any two points on the ellipse. Okay, so nothing difficult about it Of course a bit of trigonometry and component div dindo was utilized But more or less this question was you know lengthy, but not difficult So let's move on to the next question now So at a point on the ellipse tangent pq is drawn If the point q be at a distance of 1 by p From the point capital p Where p is the distance of the tangent from the origin Then the locus of the point q is So let me draw the diagram in case You are lost here So the situation is like this So you have an ellipse Okay And at a point and at a point p. Let's say we are drawing a tangent Okay, so a tangent pq is drawn And there's a point q which is at a distance of 1 by p From capital p Where a small p is the distance of the tangent from the origin. So this is small p actually Okay, then we have to find the locus of q So we have to find the locus of this point q So of course as the tangent moves on your small p will change And as your small p changes 1 by p will also change And as 1 by p changes position of q will also change And if position of q changes it will trace a path and that path equation is what is desired in this particular question Okay, first thing first. So let's say at this point is a cos theta comma b sin theta, right? So what is the equation of the tangent drawn to this point? The equation of the tangent drawn is x by a cos theta plus y by b sin theta Is equal to 1 right in other words It is b x cos theta plus ay sin theta is equal to ab correct Now if I want to find out the distance from the origin distance from the origin will be mod of ab by under root of Uh, you can say b square cos square theta plus a square sin square theta correct So reciprocal of this will be 1 by p which will be under root of b square cos square theta plus a square sin square theta by Our mod ab is just ab because you know a and b are positive in case of Uh, because they represent the semi major and minor axis is respectively. So I'm just removing the mod from there Okay Now how do I reach the position of the point q q is at a distance 1 by p from this Which form of the equation of the line will help us Exactly exactly perfect the distance form is going to help us Correct So how should I write the distance form any idea? How do I write the distance form of this? So it's at a distance of 1 by p from The given point a cos theta comma b sin theta Okay, so let me just write it down for you Uh, meanwhile, I would like to clear this off and go to the next page So let me just go to the next page Because I'll get more space there to write Okay, so let the point be h comma k. So I can say h minus a cos theta Okay divided by something k minus b sin theta divided by something Will be equal to 1 by p. Now I want to know what is that divided by something? That's actually The cos of the angle of inclination of the line, isn't it? So if you see the equation of a tangent was actually this a cos theta plus y sin theta by b correct equal to 1 right so If let's say the slope of this line was tan phi Then tan phi was actually minus b cos theta by a sin theta Correct So if I need cos phi, how would I write? Right, so you'll obviously say cos phi will be You can say minus a sin theta by under root of the square of these two, right? So b square cos square theta plus a square sin square theta, correct So here instead of cos phi I can write. I'm just writing it in Some other ink so that you can distinguish between Cos phi. So this is that cos phi expression, which I just now derived Okay in a similar way I can write here as b cos phi By again the same thing b square cos square phi plus a square sin square phi And what is 1 by p? 1 by p we just now had derived was actually b square cos square theta Plus a square sin square theta by a b Now this brings us to the simplification part now Okay, so I can always Write this this term will get cancelled with this term for both the expressions. So I will say H minus a cos theta actually I can write it as minus a sin theta by a b Correct And similarly k minus b sin theta By mistake. I wrote a phi over here. I should have written theta Yeah, k minus b sin theta. I can write it as minus plus b cos theta By a b Okay, so from here I can say h is equal to a cos theta minus a sin theta by a b And k is equal to b sin theta plus b cos theta by a b Let me do a simple act. I will now divide this by a And I will now divide this by b Now I'll square these two terms And add them Okay, which will actually give you cos square theta minus 2 sin theta cos theta a b Plus sin square theta by a square b square And here I can write sin square theta plus cos square theta by a square b square Minus 2 sin theta cos theta by a b Sorry plus plus so these two can cancel off Cos square and sin square will become 1 and here also I can take 1 by a square b square common, right? And when we finally generalize this Guys hope till this page you have understood everything Any questions till this stage? So if you generalize this I'm again going to write the expression back over here So we got h square by a square plus k square by b square is equal to 1 plus 1 by a square b square So on generalizing this You get x square by a square Plus y square by b square is equal to 1 plus a square b square, which is actually your Option number one, which is correct. So guys, what we learned in this particular Question was the use of distance formula and The use of parametric form of the equation of a tangent and of course how to work with locus So you would realize that many of the problems that we did today were actually based on locus So any question with respect to this guys, please let me know Okay So guys we'll be stopping over here because uh the time is up now almost And what I will do is I'll be sending across The remaining questions to you so that you can complete it at home And of course, I'll be also sending you the Assignment for this particular exercise Okay What are the next chapter that we are doing? Anybody any idea? I think we have been already you have been already been shared with the Fast track revision program So on saturday We are going to have progression and series Okay, unfortunately, we could not solve any question on hyperbola But there are many questions coming after this which is in on hyperbola So please do solve them and send the answer to me as soon as possible So guys, thank you over and out from centrum academy Okay, have a good night. Bye. Bye. Take care