 So, I welcome once again you all and a very good morning to all of you. So, basically today is the 22nd, December. So, I am recording this video on the state. And, like, today is the birth anniversary of our legendary Indian mathematician. And we celebrate this day as the national mathematics day. So, I wish you all a happy national mathematics day. So, in the last video we have discussed about parabolic right exercise one. So, exercise one we have finished we have discussed all the questions there. So, today we will take the next exercise this exercise two of the same chapter this parabola. Okay. So, let's take the first question. It is saying if 2x plus y plus lambda equal to zero is a normal to the parabola. Okay. So, parabola is given as y square is equal to minus 8x. And the equation of normal is given here and we have to find the value of lambda. So, what is our parabola, it's y square is equal to minus 8x. Okay. And equation of normal is given here. If you see equation of normal is given here. In terms of and this equation contains this lambda term. This 2x plus y plus lambda equal to zero. We need to find this lambda. Okay. If you observe, this is our parabola right. So, we can write the equation of tangent at any point P for this parabola. And once we know the slope of tangent, we can receive to create with negative sign and we will get the slope of normal. Okay. So, if you see, I'm writing the equation of tangent for this parabola. So, it will be basically y y one and replacing y square by y y one and x by x plus x one by two. So, this will be minus four times this x plus. Y square is equal to minus 8x. Okay, then that's okay. Then minus four into x plus x one. Okay. I'm assuming any point on the parabola, which coordinates are x one and y one. So, from here if you see, this will be y y one. Okay. Is equal to minus of 4x and minus of 4x one. So, from here we get y is equal to minus of 4 by y one into x minus 4 times x one upon y one. Okay. So, basically this is the slope. This is the slope of the tangent. Okay. This is the slope of tangent. Slope of tangent. So, what will be our slope of normal? What will be our slope of normal? Slope of normal will be y one by four. Right. Y one by four. Basically, I'm reciprocating this slope, this slope of tangent and I'm changing this side. So, this will be y one by four. Now, one thing you observe here. Here, if you see the slope of normal is given to be slope of normal. If you observe it is given to be minus two. Right. Y is equal to minus two x minus lambda. So, from here we are getting slope of normal as minus two. What does it imply? It imply that this value must be equal to minus two. Right. This value must be equal to minus two. So, from here we get y one is equal to minus eight. Right. Y one is equal to minus eight. Now, this point x one and y one is this point x one y one is lying on this parabola. So, it must satisfy this. So, basically, if you observe this y one square must be equal to minus eight times of x one. Now, we know the value of this y one. So, we will be putting here. So, from here we get y one square is 64 is equal to minus eight times x one or from here we get x one is equals to minus eight. Okay. X one is coming out to be minus eight. What does it mean? Our point is basically minus eight comma minus eight. The point P is minus eight comma minus eight. Right. So, on this point we have to draw the equation of normal. Okay. We have to write the equation of the normal. So, anyhow we know this point. Right. We know this point through which the normal is passing. And we also know the slope of normal. So, right. Slope of normal. So, slope of normal is what? Slope of normal is nothing but minus two. So, we can write the equation of a straight line. This will be y minus y one is equal to a min two x minus x one. This is our slope. Okay. So, y minus y one minus y one will be plus eight. Then it will be equal to minus two x minus x one x minus x one will be x plus eight. Is it okay? So, basically it will become y plus eight. Okay. Is equal to minus two x and minus of 16. So, our equation of normal comes out to be y plus two x y plus two x plus 24 is equal to zero. This is our equation of normal. Right. Now, this normal and this normal are same basically. So, comparing this, comparing this, if you see our lambda is coming out to be, comparing this and this equation, our lambda is coming out to be 24. Okay. So, this will be our answer. This is what we were asked to find. So, lambda will be equal to 24. So, let's come to the next question. This question number two. It is saying that the slope of a chord of the parabola y square is equal to 48 which is normal at one end and which substrates a right angle at the origin. Okay. So, basically this standard parabola is given here, y square is equal to 48. So, let me draw it. Okay. So, this is our parabola y square is equal to 4x at sexes is real. This is at sexes and one normal we have to draw. Right. So, let me draw that normal at this point P. So, this is our point P. Okay, where we have drawn this normal. So, basically if you see at point P, the angle between this tangent and the angle between. This tangent. Let me draw it nicely. Okay, anyhow, it's okay. So, at point P, this angle between this tangent and normal is 90 degree. Right. So, this angle is 90 degree. So, at point P, we have drawn a normal which subtends a right angle at the origin. Okay. Okay. So, what is origin? Origin is basically vertex of this parabola. This parabola is nothing but y square is equal to 4x. So, origin is the vertex of the parabola. And this line PQ is our normal. Okay. The slope of God of y square is equal to 4x, which is a normal at one end, it is normal at point P and which subtends a right angle at the origin. So, this PQ is actually subtending a right angle at this origin. Right. So, let me join these two points. So, this is also 90 degree. This is also 90 degree. Okay. We have to find the slope of PQ. Our question is to find the slope of PQ. This is what asked in the question. Now, let me assume a point parametric form, a point P in parametric form. So, y square is equal to 4x. We can assume this point as this, what you say, at square. Okay. At square upon 280. Okay. We have assumed this point at square comma 280. And we have to assume this Q point also. So, better write it as 801 square and comma 2801. Okay. And I'm assuming this point QS. I'm assuming this point QS, 802. 802 square comma 2802. Is it okay? Now, we have to find the slope of this line PQ. One information what we have provided with is this angle PQ. Sorry, called this chord PQ. chord PQ is obtaining an angle of 90 degrees. So, basically, if you say the slope of, if I say OP, right, the slope of OP into slope of OQ is equal to minus 1. Right. Since these two lines are at 90 degrees. Okay. So, what will be this slope of OP? It will be 2801 minus of 0. And 801 square, right? 801 square minus of 0. This will be the slope of OP. And what will be the slope of OQ? It will be 2802 upon 802 square. 802 square. This must be equal to minus 1. So, from here if you say this A will be cancelled out. One T1 will vanish. Okay. Here also same. So, we are left with this 4 upon T1 into T2. 4 upon T1 into T2 is equal to minus 1. Right. Or we can say this T1 T2 is equal to minus 1. We can write this T1 T2 is equal to minus 1. Now, what we need to find? We need to find the slope of PQ. Okay. Slope of PQ. So, basically slope of PQ will be this 2801 minus 2802 upon 801 square minus A T2 square. Right. This will be the slope of PQ. If you take 2A common from here, it will be T1 minus T2. And I am taking A common here. It will be T1 square minus T2 square. That I am writing it as T1 minus T2 and T1 plus T2. Is it okay? After taking A common, we will be left out with this T1 square minus T2 square. That I am further factorizing it as T1 minus T2 and T1 plus T2. So, this T1 minus T2 will be cancelled out. This A will be cancelled out. Finally, we are left with 2 upon T1 plus T2. Is it okay? 2 upon T1 plus T2. So, and we know this information. This T1 T2 is equal to minus 1. So, if you see, if a normal at a point P, okay, I am talking in general and talking in general. So, if we draw a normal at P, like P is the foot of the normal. And if we extend that chord to further intersect the parabola at any point Q in this case. So, there is one relation. There is one relation between this T1 and T2. What is that relation? That relation is basically T2 is equal to minus of T1 minus 2 upon T1. This is the relation which you must know, right? Which you must know. So, what is that? This T2 is equals to minus of T1 minus 2 upon T1. Now, if you see, we know this value T1 into T2. What is that? T1 T2 is equals to, okay, this will be minus 4. This will be minus 4, not minus 1, right? T1 T2 will be equal to minus 4. So, basically what I will do, I will try to find this value. This T1, if you see, this T1 into T2, instead of T2, I will write this, okay? Minus T1 minus 2 upon T1 is equals to what? Minus of 4. So, this will be minus of T1 square, okay? Minus of T1 square plus, no, minus of T1 square minus of 2, right? So, let me erase it first. Minus of T1 square minus of 2 is equal to minus 4, okay? So, from here we get T1 square is equal to, from here we get what? T1 square is equal to 2. Is it okay? 4 minus 2 to T1 square. So, we are getting the value of T1 as plus minus root 2, okay? Plus minus root 2. Now, what will be the value of T2 if we got T1? So, it will be basically minus of considering this T1 as, suppose I am considering T1 as plus of root 2 in first case. And one more case will be there then when T1 will be equal to minus of root 2. So, when T1 is plus root 2, what will be our T2? It will be minus of root 2, right? And minus of 2 upon root 2. Is it okay? So, it will be minus root 2 and this thing will become root 2. So, minus of 2 root 2. So, when T1 is plus root 2, our T2 is minus root 2 and when T1 is minus root 2, our T2 will become 2 root 2. Our T2 will become 2 root 2. You put the value of T1, you will get this value, okay? So, this is one case and this is one case where T1 is equal to root 2, our T2, we got T2 as minus 2 root 2. So, now this slope of PQ, if you observe, this slope of PQ will be this T1 plus T2 will be root 2 and minus 2 root 2. So, from here, we get the value as 2 upon minus root 2, right? So, minus of root 2. So, this is one possible slope of PQ and there is one more possibility that is when we take T1 as minus root 2, when we take T1 as minus root 2 and T2 as plus 2 root 2, okay? So, it will be basically 2 upon 2 root 2 minus root 2 will be root 2, that is 2 root 2, okay? So, these two possibilities are there for the slope of this PQ. So, an option, this, okay, both are present. So, B and D, both will be our answer. This B and D, both will be our answer. So, hope this is clear to all of us. The only important thing which I guess you should be aware of is this thing, like if a normal drawn at P intersects further on the parabola, the coordinates of this means this parameter T2 can be expressed as minus T1 minus 2 upon T1. If this thing is known to you, if this thing you are aware of this, then you can easily solve this question. So, this is question number 2. Now, let's move through this question number 3. It is saying that the common tangent to the parabola, y square is equal to 4x, okay? So, here is our parabola, y square is equal to 4x, and x square is equal to 4a1. So, the question is saying the common tangent to both the parabola. So, let me first write the tangent for this parabola. So, let me first write the tangent for this parabola, okay? So, what will be tangent for this parabola in slope form? We will write it in slope form. So, basically, y is equal to mx plus c. Suppose I am taking this as a tangent through this. So, what is the condition of tangency? Condition of tangency. I am not going in derivation and all, since you have already covered this chapter. I hope this is known to you. So, condition of tangency is basically c is equals to a by, for this line, for this line y is equal to mx plus c. To be a tangent to this parabola, this c must be equal to a by m. So, our tangent will be y is equal to mx plus a by m. Now, the same line, if you see, the same line is tangent to this parabola also, this parabola also, right? Tangent to that parabola also. Since it is a common tangent. So, what I will do? I will solve this equation. I will solve this equation of line and this parabola, okay? So, and I will get the intersection point. So, basically, I will put the value of y in this equation of parabola. This will be mx plus a by m, okay? So, what we will get? We will basically get a quadratic in x. So, this will be x square minus 4amx, okay? And minus of 4a square minus of 4a square by m, okay? This is equal to 0. So, for tangent, for this line to be tangent, this must have the, or we can say for tangent or for tangency, okay? There must be only one value of x, right? This quadratic must have only one root and that will be equal, like common root, both roots will be equal and it will be real, right? So, for tangency, the discriminant, the discriminant of this quadratic must be equal to 0. Now, what is discriminant here? So, this is basically b square minus 4ac. So, this will be 16a square m square. This is b square minus 4 times a is 1. And what is c? c is minus 4a square upon m. This must be equal to 0 for tangency. So, from here, we get 16a square m square is equal to minus of 16a square by m. Am I doing some mistake? Please check it. 16a square m square is equal to 16a square by m. I think it's okay. So, this 16a square, 16a square will be cancelled. From here, we got m cube as minus of 1. So, we can say the value of m is minus 2. So, what does it mean? The slope of this line y is equal to mx plus c is minus 1. Hence, what will be our equation of tangent? So, equation of tangent will become y is equal to mx plus a by m was there. This was the equation of tangent. Now, put m equal to minus 1. It will be minus x and minus a. Or we can say y plus x plus a is equal to 0. So, this will be the common tangent to both parabola y square is equal to 4a x and x square is equal to 4a y. So, y plus x plus a, that is option a. Option a is correct. Now, let's see this next question. The circle x square plus y square plus 4 lambda x equal to 0. Okay. Lambda belongs to real number touches the parabola y square is equal to 8x. Okay. So, basically, one parabola is there. Okay. One parabola is there. So, x is this one. And one circle is there. One circle is there. Okay. This is our circle. Both are, this circle is touching the parabola. And we have to find the value of lambda in this case. So, basically, if you see, this is our parabola y square is equal to y square is equal to 8x. So, it is of a standard form only. What does it mean? This will be our x axis. This will be our x axis. The axis of this parabola will be x axis. And this circle is given by y is x square plus y square plus 4 lambda x is equal to 0. Okay. So, no y terms means its center will lie on the x axis itself. And what will be the center of this circle? The center of this circle will be 2 lambda with negative sign that is minus 2 lambda comma 0. And what will be radius of this? What will be the radius of this circle? It will be g square means 4 lambda square plus f square no f square no c. So, radius will be equal to 2 lambda. So, this is our center. This is our center of circle whose coordinates are minus 2 lambda comma 0. Okay. And this is our focus. So, if you see, this is our parabola. So, if you see from here we get a is equal to comparing with a standard form. 4a is equal to 8 means a is equal to 2. Right. a is equals to 2. So, this will be our focus 2 comma 0. I have just written. I don't know whether we will need this information or not. So, focus for this will be 2 comma 0. So, what we were asked. Okay. This circle touches this parabola. Okay. It is touching here at the origin we can say they are touching each other at the origin. So, I can draw this thing also. Sorry. This what you say. I can draw this y axis also. So, this will be our y axis basically. Okay. So, now what will be the value of lambda? So, basically this center of circle should lie on the negative x axis. Right. Because once it will this circle will cross this origin and it will come on the positive x axis it will cut it will intersect the parabola at two distinct points, but we don't need that. Okay. So, the center of circle must lie on the negative x direction negative x axis. Right. So, this minus 2 lambda thing this minus 2 lambda thing the center of this circle should be less than 0. Right. So, if you multiply it with negative sign what does it become? It will become lambda will be greater than 0. Right. So, the value of lambda will ranges from this 0 to infinity. Right. 0 to infinity. So, is there any option 0 to infinity? Yeah. Option A is there, but I'm thinking of one more case like one more case is possible here if you say what if what if this is our let me let me draw first. So, this is our parabola and one more case may arise where this circle will be like this the circle will be sorry. So, one more case may arise where circle will be like this. Right. Like it will touch the parabola at two points at two points. So, suppose this is our origin this is our origin and this is the center of circle. This is the center of circle. So, in this case basically what will happen? Like this suppose I'm making one tangent at this point P. Okay. And here also if you say here also we can make a one tangent to both. This point P and Q at point P and Q basically we can have a common tangent to both this circle and parabola. Okay. And the coordinates of C is given as minus two lambda comma zero this we know. So, basically what we will do we will make one we will take a parametric point at this P that is 80 square comma 280. Okay. And we will write the equation of tangent and from there that tangent that tangent will be behave as tangent to this circle also. So, we will take the distance from center to that tangent and we will equate that to the radius of the circle. Right. So, basically this Cp this Cp thing this Cp will be equal to radius of circle. Okay. And we can write this tangent at P tangent at P. Right. The equation of tangent at P in terms of parameter also we can write this ty is equal to x plus 80 square or in terms of slope also you can write it as y is equal to mx plus a by m where m is the slope. And then in that case the point of this coordinates of P will be what will be its coordinate a by m square comma 2 a by m. So, anyhow from there we can find this lambda. But as per this question what I am getting this will be your answer. Right. From here we get lambda is greater than zero. So, the answer to this question will be this option K. But this is also one case which is possible. So, let's take the next question. Question number five. So, it is saying if the normals at two points p and q of a parabola this intersect at a third point are on the curve. Then the product of the ordinates of p and q. Okay. So, our standard parabola is given here y square is equal to four ax. Okay. So, this is our parabola y square is equal to four ax. And it is saying if the normals at point p and q. So, let me take this as p. And I'm drawing one normal here. And let me take this as R. I'm drawing one more. So, as per question, this is point p. This is point q and this is point R. And R is on the parabola like the normals at p and q intersect at a third point are which is on the curve itself. So, this is our parabola y square is equal to four ax. Okay. This is our points. Now, what we need to do, we have to find the product of ordinates of p and q. So, let me take, let me take this point p as a t one is where let me change the color. So, this a t square or you can say a t one is where comma two a t one. So, this is the coordinate of p and let me take coordinate of q as a t two square a t two square comma two a t two. Okay. And we can also assume this coordinate of R also since it is also on the same curve y square is equal to four ax. So, let me take it as a t three square a t three square comma two way t three. Okay. So, question is asking to find the product of the ordinates of p and q. So, what is the ordinate of p? It is two a t one. Okay. And what is the ordinate of q? It is two a t two. This product question is demanding what will be this product. So, this is our target to find. So, let me talk about this. This this p r is a normal. Okay. At p. So, basically this t three, we can write it as we have discussed the same concept in I think question number two. So, let me repeat it. We can write this t three s we can write this t three s what t three know. Yeah. And t three we can write it as a minus of t one. Okay. And minus two upon t one. Right. And one more way, like we can write this t three in one more way out. This we are drawing this q r is also a normal at q. Okay. And this is also meeting at the same point t three. So, from here t three will be equal to minus t two minus two upon t two. Okay. Is it okay? These two things. If it's fine to you, then we are done. We can easily solve for that. Now, if you see, we can write it as let me take it as equation one. Let me take it as equation two. So, from one and two, you can write minus of t one minus two upon t one is two minus of t two minus two upon t two. Okay. So, let me take this t two to this left hand side. It will be t two minus t one t two minus t one and let me move it to right hand side. So, it will be two upon t one and a minus two upon t two. Is it okay? So, t two minus t one or or t two minus t one is equal to this t one t two is our LCM. T one means two t two two t two minus two t one. Okay. So, from here, if you see, you take two as common. So, we are left out with this thing two times t two minus t one upon t one t two. So, t two minus t one will get penciled out. And from here, we get t one t two is equals to two. Now, this is what is asked here. So, if you see this product is nothing but let me name it as let me name it as a. So, let me name it this product as a. So, our a is nothing but four a square this four a square into t one t two. They put this value of t one t two here. So, it will be two into four a square. So, this is nothing but eight a square. So, this is our product. This is our answer. Okay. So, this is what this is what question is asking, but let me let me tell you some other facts also for this type of question, like this t three will be. So, for this t three means this normal at P and Q meet at R. Okay, this relation is valid everywhere. This T one into t two will be equal to two. So, one thing which we learned here this T one T two will be equal to two. So, this is I'm writing as learnings. So, this T one into T two will be equal to two when when product there when the normal drawn at P and Q on the parabola meets the parabola again at point R. In that case this relation will be valid and one more thing this T three will be equal to this T three will be equal to minus of T one plus T two. Okay, this you can verify from these two eight patients. Okay, this you can easily verify. So, these two things I advise you to remember. Okay, but this normal at P and Q must meet at R and R must be on the parabola. Then only this thing will be valid. So, I think we are done with this. So, let's take this next question question number six. The normals at three points P Q R of parabola vice versa equal to four X meet in H comma K. Okay, the centroid of triangle P Q R. So, basically, this is our parabola. Okay. This is our parabola, this is our axis. So, question is saying the normals at P Q and R meet in H comma K. So, let me draw. This is our one normal. This is our one normal. This is our one normal. So, this point is basically P. Okay, this point is P. This point is Q and this point is R. And these three normals are meeting at a point. Any point suppose I'm taking it as M, okay. And the coordinates of M is H comma K. Right, this is what given in the question. Now the question is saying the centroid of triangle P Q R. Okay, so let me draw the triangle P Q R first. So, P Q and R. Okay, this will be P Q side. This will be that. Okay, so this triangle P Q R. What I have drawn in this green colored line. So, question is asking where does the centroid of this triangle lie. Okay, the centroid of triangle P Q R lies on. So, this and what is this parabola? This parabola is nothing but our standard parabola y square is equal to 4 ax. Okay, so from point M, right, from point M, we are drawing three normals. We can say in that way also from point M, we are drawing three normals to this paragraph. So, basically what is the equation of normal in slow form equation of normal equation of normal in slow form. For this parabola y square is equal to 4 ax. It is basically y is equal to y is equal to mx minus 2 a m minus of a m q where m is the slope of the normal. Right, m is the slope of the normal. Here, m is the slope of the normal slope of normal. Hope you are aware of this thing. So, this normal is passing through this point M, which is called h comma k, sorry. So, it will satisfy it. So, I am writing it as k is equal to k is equal to mh minus 2 a m minus a m q. Okay, now I will rearrange this equation as a m q a m q plus 2 a m. Okay, minus mh or h m minus h m plus a is equal to 0. I have just shifted this right hand things to the left hand side. So, a m q plus 2 a m minus h m plus k is equal to 0. Now, basically this is a cubic in m, right? Let me rearrange it. So, a m q plus 2 a minus h, okay, into m plus k. This is cubic in m. This is a cubic in m. So, it will have three roots. It will have three roots m1 m2 and m3 whose roots are, it will have three roots. So, I am assuming it as m1 m2 and m3. Okay. Now, what does it mean? What are this m1 m2 and m3? This m2 m, like this will be slope of these normals. Like, suppose this is our m1 slope of this pm, pq slope is m2 and this slope of mr is m3. Is it okay? Is it okay to everyone? So, now let us observe this cubic. The coefficient of m square is missing. Coefficient of m square is missing here. Right? Coefficient of m square is missing here. So, what will be the sum of roots? What will be the sum of roots? m1 plus m2 plus m3 will be equal to 0. Is it okay? This sum of roots will be equal to 0. Now, I have assumed this equation of normal to be this, right? So, what we get? Minus 2 a m minus a m2. Right? So, this m1 plus m2 plus m3 is equal to 0. And we can write other relations also, like product of roots, like m1 m2 m3 will be equal to this minus k upon it. But we do not need here. Right? We do not need here. The coordinates of p, q and c, if you observe, the y coordinates will be minus 2 a m1. Okay. I am not bothered about the x coordinate. And what will be coordinate of this q? This will be minus 2 a m2. And what will be the coordinates of r? It will be minus 2 a m3. For finding the centroid of triangle p, q are centroid of p, q are. Okay. I am representing that by g. I do not know about the x axis, like x coordinate. But this y coordinate will be basically minus 2 a m1 plus m2 plus m3 by 3. Is it okay? Now, this m1 plus m2 plus m3 is equal to 0. So this thing will be equal to 0. So the centroid of this, so the centroid of this triangle p, q are will lie on y equal to 0, y equal to 0. This is confirmed that x, this x coordinate may vary, but this y coordinate is 0. So what does it mean if some point is there? So centroid will be somewhere between somewhere on this line. This is our axis. This is our axis for this parabola y is equal to 4x. This is x axis. Its equation is nothing but y equal to 0. Okay. So this centroid of triangle p, q are centroid of triangle p, q are will definitely lie on. Lie on x axis. That is y equal to 0. Okay. So this is our, this option is correct y equal to 0. And this, this is one of the property of parabola also. We used to remember this in our school days as a properties. Okay. And these properties are equivalent to questions also. Like J also used to create questions directly based on properties of parabola. So this is just a minute. Yeah. So this is our question number six. Let's take the next one. This question number seven. What it is saying the set of points on the axis of parabola this from which all the three normals to the parabola are real. Okay. In last question also, in last question also if you see this p, q are just I'm giving note this p, q are basically called co-normal points p, q are are called co-normal points. Okay. Now let's say this next question. The set of points on the axis of parabola this from which three normals to the parabola are real. So one parabola is given here, y square minus four x minus two y plus five equal to two. So let me try to change it in some standard form. So this will be y square minus two y plus one. Okay. And minus four x plus four equal to zero. So this will be y minus one whole square is equal to four times x minus one. This will be four x minus four. Four x minus four. Yeah. So this is our parabola y square is equal to four a x. Okay. With shifted vertex. Basically this is our parabola with shifted vertex. So we will compare this with our standard parabola y square is equal to four a x. So what will be a vertex basically what will be vertex. So this y minus one will be equal to zeroes. So from here we get y equal to one and this x minus one. Am I doing correctly y minus one for x minus one. So x minus one will be equal to zero. So from here we get x equal to one. So this will be vertex of this parabola. So let me draw one graph is sketch for this. Okay. Yeah. So this will be our vertex basically which coordinates are what one comma one. Yes. One comma one. Okay. So and what will be the focus. What will be focus of this parabola if you compare it from here. So this is done with the way we are done with the vertex here. Now 48 is equal to four. Right. Well comparing with the standard form. So a is coming out to be one. So what will be our focus. Focus will be a means this is what a is the distance between this vertex and focus. So x coordinate is of vertex is one. So x coordinate of focus will be basically two comma one. And this line is y equal to this line is y equal to one. That is the axis. The axis of this parabola will be y equal to one. Now where will our origin like our origin will be somewhere here. Let me draw with the other pin that will be it's not happening later. So basically this will be our x and y axis. This is our x axis. And this will be our y axis. And this is our origin zero comma three. Okay. So we are done with the basic informations what we have given we have plotted all the things. Now what is asked the set of points on the axis of parabola on the axis of parabola. Okay. From which all the three normals to the parabola are real. Okay. So what we have done in the last question. Same concept here also like one point he has taken the question is the question has taken one point like on parabola on the axis of parabola. Okay. And from this point, we are drawing three normals to the parabola. Okay. So this is suppose this is our normal one. Okay. You can say this is normal one. This is our first normal. This is our second normal. And this is our third normal. Okay. So at this point, this point, suppose I'm taking this point as M. So what will be its coordinate basically. It's a coordinate will be some lambda and why coordinate will be fixed that is one moment since it is lying on the axis and what is axis y equal to one. So the coordinates of M will be lambda comma one. Okay. Now this lambda may vary now for drawing for three normals for three real normals for three real normals for three real normals from the point from point M. Okay. From point M, this x coordinate of M, this x coordinate of M should be greater than two way. This lambda should be greater than two way. Okay. This is the condition. This is the condition for three real moments for three real moments. You write here three real moments. Okay. So this lambda the x coordinate of this point M should be greater than two a. Okay. Hope it is clear to all. Now what will be this lambda means what lambda should be greater than it should be greater than normally we used to learn in this way for for this. Let me write here in box. For if our parabola is y square is equal to 4x. Okay. And we have to draw three real normals. So this H must be greater than two. What is H the x coordinate of point M from where we are drawing the popping from where we are going the normals to the parabola. Okay. So this lambda thing, this lambda must be greater than two way. And what is this lambda with what will be our lambda basically. What is our x coordinate this this thing. Right. x minus a. Okay. So H basically H is nothing but for for our this particular y square is equal to 4x. What is H? H is the x coordinate. Like this X capital X must be greater than two way. Okay. So from here we will say capital X must be greater than two way. Okay. And now what is capital X? Capital X is nothing but this is our capital X right. And this is our capital Y. Okay. So capital X must be greater than two way. And what is a is one is one no. So I will put the value of a and I will put the value of capital X that is X minus a. X minus one. X minus one should be greater than this two. So from here X should be greater than three. Okay. X should be greater than three. And X what what is X this lambda thing is only. Okay. So basically lambda this implies lambda should be greater than three. Okay. Lambda should be greater than three. If lambda is greater than three then only we will be able to draw three normal three normal three normal real no. If lambda is less than three we will not be able to plot the three real no. Okay. So I think this question is clear to all. So lambda should be greater than three. So this is our correct option. This option would be. So I think I have not taken further questions. Okay. We will we will shoot we will record one more video for the remaining questions. So till now let let us like keep this exercise up to this question number seven only. I will come with the next video and I will discuss the next questions there. So till then. Okay. Good bye. Take care. Yeah. I've heard of the this Corona news. I don't know whether it is what's going on. But yeah, please wear a mask if possible while going outside. So okay. We will meet you once again with the next questions of this exercise too. So bye.