 So hello students, a very good morning to all of you. So in the last video we have discussed the exercise too of parabola chapter, but I think we were not able to cover all the questions. So we are here today with the remaining questions of that exercise that is exercise to and we will try to complete this exercise right so without wasting any time. Let's see the question. This is question number eight means we have covered up to question number seven in our last video. Okay. So, question is saying prove that any three tangents to a product to a parabola whose slopes are in harmonic progression encloses a triangle of constant area. So, no parabola is given here so we will take our standard parabola that is why square is equal to 4x. Okay, and we will solve this question. So, let me draw one rough sketch for this. Let's take this as parabola. And what we will do we will draw three tangents on this parabola right. And let's take this tangent. Okay. Let's take this one. And we will draw one tangent on this side. Okay. So, this is what we got this triangle ABC suppose I'm naming it as this is point as a right this point as a this point as B and this point as C. Okay, and this parabola what we have taken this parabola is y square is equal to 4x. Now the question is saying that we have to prove that this triangle will be of constant area. Okay, if the slopes of these tangents. Okay. The slope of these tangents this suppose I'm taking this as point P. Okay, this point P on which we are trying tangents this point as Q. Okay, and this point is on parabola these points P Q are are on the parabola. Okay. Now the slope of this tangent at even tangent, sorry, tangent at P and tangent at Q and tangent at are are in harmonic probation. Okay, so let me take this P point as T one. Okay. I'm assuming the parametric points. Okay, so the point P as T one, the point Q as T two. Okay, and this our point as T three. What does it mean, taking this T one as parametric point means the coordinates of this P will be 801 square comma 281 similarly the coordinates of Q will be 82 square comma 282 and this T three same 83 comma 83 square comma 283. So, now what we will do, I will write the tangent at P equation of tangent at P in parametric form. Okay, so let me write it. So I'm writing the tangent at P in parametric form. Okay. So what it will be basically it will be our T y right T one y you can say T one y is equal to x plus 81 square x plus 81 square right and what will be the equation of tangent at Q equation of tangent at Q will be T two y is equal to x plus 82 square. And what will be the equation of tangent at R similarly this T three y plus T three y is equal to T three y is equal to our x plus 83 square hope this is clear to all. Now, what I will do, like we have to find the area of triangle ABC right so basically we need to have the coordinates of this points. If we are, we will have the coordinates of this AB and C, then we can we can easily find that area. Okay, so for finding the coordinates of this ABC, what we have to do. Basically, this point a this point a is the intersection of tangent at T one and tangent at T three. Okay, so for solving for having the coordinate of a, we have to solve this tangent at P tangent at P and tangent at R. Okay, so let me solve these two equations, so that we can have the coordinates of it. Okay. So, let me name it as equation one, equation two, and this has equation three. So, I will do one minus three. Okay, equation one minus equation three. Then what will I have it will be basically why okay T one minus T three T one minus T three is equal to this x will be cancelled out and we will have a times T one square minus T three square. So, it will be a T one minus T three into T one plus T three I think it is okay to all of you. Okay, now this T one minus T three term will get cancelled out and we will we will have the y coordinate of a as this a into T one plus T three. Is it okay. This is the y coordinate of a. Okay, now putting this value in equation one, put in equation one, we will put the equation value of this y in equation one, we will have T one T one into y means a into T one plus T three right a into T one plus T three is equal to this x plus a T one square. Now from here if you say a T one square plus a T one T three is equal to x plus a T one square. Okay, this x a T one is where a T one is where a T one is where we will get cancelled out and our x coordinate of a will be a into T one into T three. So, this is the coordinate of our point a right this is the coordinate of our point a. So, let me divide this board. So, we got the coordinate of a as we got the coordinate of a as what this a T one T three a T one T three comma a T one plus two. So, I request all of you to remember this basically. If we can remember this it usually makes our task easy while solving the question. Okay, so whenever up to tangents at point to tangents of parabola meet at any certain point the coordinates of it. A times the parametric this T one into T three and a times T one plus T three it's x coordinate is given as this a T one T three and why coordinate is a T one plus T three. So, now what will be the coordinate of be based on this. This analysis what will be the coordinate of be it is the basically the intersection of tangent at Q and P right so T one T two are the parameters so basically the coordinates of be will be a into T one into T two. And the Y coordinate will be a into T one plus T two. Is it okay and what will be the coordinates of see it is the meeting point of tangents at R and Q means T two and T three are involved. So, it will be a into T T two into T three. Okay. And the Y coordinate will be a into T two plus T three. From here we got the coordinates of this a B and C. Okay. Now, one more thing is given in the question if you observe like the tangents. Okay, the tangents at this PQ are their slopes are in harmonic progression right their slopes are in harmonic progression. So, what is the slope here. The slope of this slope of this is one upon T one right one upon T one we have to divide by T one. So, it's a slope is one by T one it's a slope is one by T two one by T two and it's a slope is it's a slope is suppose I'm taking a slope of P tangent at P this has Q and this has R. So, the slope is one by T three. Now, as per question as per question this one by T one. Okay. And one by T two and one by T three these slopes are in HP. What does it mean we can say this T one. Okay, this T one T two and T three are in AP. Okay. Now, we will use this information further in the question. So, let's now focus to this triangle ABC right area of triangle ABC. So, if you see area of triangle ABC. How can we give this, how can we get this area. It will be x one y one solving this determinant x two y two one and x three y three where the x one y one x two y two and x three y three are the coordinates of the vertices of the trend. Okay, so now let's put the value of x one y one. So, what is our first coordinate like suppose I'm taking this a. So, this will be basically a T one T three a T one P three. It's why coordinate is a times T one plus T three. Then a times T one T two a times T one plus T two one and this will be a times T two a three a times T two plus T three. Okay, and what now we have to solve this determinant. Okay. We have to solve this determinant how to solve. We can basically what we can do, we will try to make as much zeros as possible so I will do one operation okay what I will do. I will do operation on this R one by R one minus R two and I will do one operation on this R three. And we will keep this R two unchanged. Okay, so let's see what this determinant is coming out to be. So, basically it will be R one minus R two. So, a T one a T one will be common right a T one will be common T three minus T two T three minus T two and from here if you see a T three a T one a T three so a T one will get cancelled out it will be a times T three minus T two. Okay, T three minus T two, and this position will be zero. Okay, and we are not disturbing this row to so let it be as it is a T one T two a T one plus T two. Okay, and one and this thing will be R three minus R two no so a a T two will be common right a R three minus R two. Okay, so a T two will be common what will be left out T three minus T one T three minus T one. Similarly, here if you see a T two will get cancelled out it will be a times T three minus T one and one minus one will be zero. Now we will expand this matrix by this third column, expanding by third column. So, what will happen if we expanded on third column. Okay, so this will coming out to be this what you say zero into some two by two metrics it will be it will be cancelled out since we are multiplying by zero and this minus one. What is its position I the second row and third column so minus one will be there minus one will be there and we will be left out this with these two metrics that is a T one T three minus T two okay. And this element a T three minus T two and this thing a T two T three minus T one. Okay, and this will be a into T three minus two. Okay. So, on expanding this, we will get the, get the area of the training. So, let's see how can we do it. So this will be minus sign. Okay, let it be as it is we have to multiply this no AD minus BC, this and this. So, what we can do a square it will be basically a square a square T one a square T one and T three minus T two into T three minus T one we are getting right. So, basically T three minus T two. Okay, let me first multiply a square T one then T three minus T two into T three minus T one. Okay, and minus of a square minus of a square T two a square T two into this thing T three minus T two into T three minus T one. And this whole thing will be with one negative sign is it okay. Yes. So, let's see what we are getting T three minus T two and T three minus T one. So we are getting this thing completely common right, we are getting this a square. First negative side, and this a square and T three minus T two T three minus T two T three minus T one, we are getting this common from both the terms right. So what we are left out with this T one, okay T one minus T two T one minus T two. Okay. So, basically, what we got this T three minus T two into T three minus T one. Okay, it is representing a D basically if you see here, this T one T two T three are in AP know. So, we can say the common difference of this AP will be T two minus T one. Or we can say it will be same as T three minus T two. Is it okay. So let's put this as D in terms of common difference. So it will be minus of a square into T three minus T two is D. So into D T three minus T one T three minus T one. So T three minus T one will be basically 2D, two times of common difference and T one minus T two. T one minus T two will be minus of P. It will be minus of P. Is it okay. So what we got. Okay, this whole thing will be multiplied by half also. Okay, while writing the area, I forgot it. This will be half of this determinant. Okay. This will be half of this determinant. So I'm just taking this thing. I have solved this thing. So this let me take it as D. Sorry, delta. So I'm solving delta and then I will multiply. So this area will be basically this area will be basically half of this delta, half of delta means what this will be half of this is our delta. No. So a square into a square into two times DQ. This minus and minus will get cancelled out. So it will be two times DQ into a square. So this two and two will get cancelled out. We got the area of triangle as a square into DQ. So it is basically constant. This is constant. No. A square into a DQ. So this is what we were required to prove. So hence, we got that the tangents drawn at three points on this parabola. Okay. And like area of triangle formed by the intersection of these three tangents will have a constant area. Will have a constant area. Okay. So basically this is not dependent on these parameters T1, T2 and A3. Okay. No. So that's why the area of this triangle will be constant. So it's this question is a bit lengthy, but we learned a lot like we got to know the how to find this intersection point of tangents. Then after that we applied the formula of finding the area of triangle ABC in determinant form. Okay. So I think we are done with this. Let's take the next question. Question number nine. Okay. So it is saying a part of parabola y square is equal to 4ax, subtents are right angle at the vertex. Find the locus of the point of intersection of tangents at its extremities. So parabola is given y square is equal to 4ax. Okay. So this is our parabola, subtents and one chord is there. One chord is there. Let me name it as PQ. Okay. This chord is subtending a right angle at the vertex. This part is subtending a right angle at the vertex. So let me draw the axis of parabola first. Okay. So this is our axis of parabola. And this chord is subtending a right angle on the vertex. Okay. This will be basically this will be our right angle. Okay. Now, find the locus of the point of intersection of the tangents at its extremities. What does it mean we have to draw tangent also at these points. So let me draw. Let me draw this tangent at this point. And also at this point. So hope this sketch is clear to all. Okay. Let me explain first. This is our chord PQ. Okay. We have drawn this chord PQ. This chord PQ is subtending a right angle at the vertex. This is our vertex of the parabola. Whose coordinates will be 0 comma 0 since parabola we are taking is of the form y square is equal to 4x. This is our standard parabola only. Now, we have drawn the tangents at the extremities of this part. Means we have drawn the tangents at point P and point Q. Okay. Now, we have to find the point locus of this point of intersection. So let me say these two tangents meet at point M whose coordinates are H comma K. Now we have to find the locus of this point M. Okay. So let me take this point P as T1 in parametric form and this point Q as T2. Okay. Now, in the last question we have already seen that the tangents at point P and point Q. Okay. When we draw tangents at point P and point Q, they intersect at what? A T1 T2. If you see in this question what we have done. What is the intersection point? This A T1 T3 and A T1 plus T3. So this is what I'm going to show here. So basically the coordinates of M will be. The coordinates of M will be A times T1 T2. Okay. A times T1 T2 comma A times T1 plus T2. Okay. This will be the coordinates of M. Okay. This is representing our H and this is representing our K. Okay. So basically we got the coordinates of this M. Now one information what is given here. The slope of this. Right. The slope of this PQ or we can say PV. The slope of PV into slope of VQ is equal to minus 1. Since both are perpendicular, this PV and VQ are perpendicular, the slope of their product of their slope will be minus 1. Now what will be the product of this? What will be the slope of PV? Okay. What is this point T1? This is nothing but P is A T1 square comma 2 A T1 and this Q is A T2 square comma 2 A T2. So basically slope of PV will be 2 A T1 2 A T1 upon A T1 square into this will be 2 A T2 into sorry by A T2 square. Is it okay? This will be equal to minus 1. So this A A will cancel out. This T1 T1 T2 T2 will get cancelled out. From here we got T1 T2 is equal to minus of 4. Right. From here we got T1 into T2 is equal to minus 4. Now how can we utilize this information? This is our H coordinate of M. Okay. So our H if you see H is equal to A times T1 T2. Okay. From here we got this. Okay. And this is our K coordinate. So I don't think we require this K coordinate. This Y coordinate we don't need. So this H is equals to H is equals to A into we got the value of T1 T2 is minus 4. Okay. So H is equals to minus 4A. Now for writing locus what we do? We replace this H by X. So this we will replace this H by X and we will got X plus 4A is equal to 0. This will be the required locus. Right. This will be our required locus. This is the required locus of the point of intersection of these two tangents. Okay. So this information we got from the last question but this is what I am saying. It's better to remember like when tangents at this point P and Q intersect. They the coordinates of that intersecting point will be this A T1 T2 comma A T1 plus T2. So this will make our life easier. So I think this is done. Let's see the next question. Find the equation of the normal to the parabola Y square is equal to 4X. Okay. So we have to write the equation of the normal to this parabola Y square is equal to 4X. Okay. Which is parallel to this line and which is perpendicular to this given line. So basically we are having this parabola Y square is equal to 4X. Okay. And we have to the given line I am doing this part A. The given line is Y minus 2X plus 5 equal to 0. Is it okay? Y minus 2X plus 5 equal to 0. Now we have to write the equation of normal. We have to write the equation of normal which is parallel to this. Which is parallel to this. Okay. This is our first part of the question and the second part is basically. Let me bypass this screen. And the second part is saying we have to write the equation of normal which is perpendicular to this line. So the line given line is 2X plus 6 ohm plus 5 equal to 0 and we have to write the equation. Okay. So if you see if you see equation of normal no we have to write equation of normal here which is parallel to this line. So equation of normal on this parabola on this parabola. Okay. Y square is equal to 4X. So from here we got this A is equal to 1. Okay. This A is coming out to be 1. So in slope form if you see in slope form. We write the equation of normal as Y is equal to MX. Okay. Minus 2AM minus AMQ. Okay. This is the equation of normal which we write on the parabola of the form Y square is equal to 4X. Okay. In slope form. Why I am writing in slope form. If you see we know the slope of this tangent. Sorry. We know the slope of the normal. Why? Because the slope of this line. Suppose what is the slope of this line? Y is equal to 2X minus 5. So slope of this line is 2. Right. Slope of this line is 2. So slope of normal. Slope of normal should also be 2. Slope of normal should also be 2. Okay. So we can say this value of AM will be equal to 2. Okay. So what will be our equation of normal? Our equation of normal will be replace and I put the value of 2 in place of M. So it will be 2X. Okay. Minus 2A. What is A? A is 1 here. In this case. So 2 into M. M means what? 2 and minus 1MQ. Like this will be 2. Okay. So we got the equation of normal as 2X. Y is equal to 2X minus 4 minus 8. Minus 2. This will be our equation of normal. Right. Now, this is the equation of normal which is parallel to this line. Now we have to again write the equation of normal which is perpendicular to this given line in second part of this question. Okay. So we will apply the same approach here also. Like in slope form how we write the equation of normal for this parabola Y square is equal to 4X. Same Y is equal to MX minus 2AM minus 2AM minus of AMQ. Okay. Now if you see what is slope of this line? Slope of this line. This is basically if you rewrite it as 6Y is equal to minus 2X minus 5. Okay. So our Y will be equal to minus 1 by 3X and minus 5 by 6. Okay. So the slope of this line, slope of this line is coming out to be minus of 1 by 3. Is it okay? Slope of this line 2X plus 6Y. So 6Y we can write it as minus 2X minus 5. So minus 1 by 3. So what we got? Slope of this line is minus 1 by 3. So our slope of normal, our slope of normal should be 3. Right. Because the product of these two slopes should be equal to minus 1. So we got the slope of normal to be 3. So I will put this value in this equation. So our Y will become, our equation of normal will become 3MX means what? 3X minus 2 times M means 3 and minus MQ, MQ will be what? 27. It means 3Q. So it will be Y is equal to 3X minus 33. Okay. So this will be the required equation of normal. Now some student might say like why you are writing in the equation of normal in this slope form. Anyway, if you wish to write it in parametric form, then also your result will be same also, same only. Okay. So no need to worry on that. You can write the equation in whatever format or in whatever way you wish. There will be no disturbance on the final answer. Okay. We are going to get the same answer from all the ends. So suppose if you see, suppose if you see, if we write it in parametric form. Okay. So this is our in slope form. This was in slope form. Okay. This was the equation in slope form. Let me write it in parametric form. Okay. I am just doing to show you that there will be no effect on the final answer. So how we write the equation of normal in parametric form? It's basically TX plus Y is equal to AT cube, 280 plus AT cube. Right. 280 plus AT cube. If I am not wrong, this is the equation of normal in parametric form. Okay. So from here if you see our Y will be equal to minus of 2X plus 280 plus AT cube. Okay. So what is here? Minus T means minus 3 is the slope here. Slope here and that is equal to what is the slope 2. So from here we got T is equal to minus 2. Right. So put this in this equation. So put this value in this equation. We will get this minus 2X plus Y is equals to 2T means minus 2 and T cube means plus or minus 2 cube. So from here we get what Y is equal to 2X. Okay. And minus 4 and minus 8 minus 12. So see both answers are coming out to the same. So it's your wish. Okay. It's your wish you can write it in slope form or parametric form as per your wish. Similarly you can write the equation for this also. So I think this question is done. So let's move to the next one. Question number 11. It is saying the ordinates of point P and Q on the parabola this are in the ratio 1 is to 2. Okay. The ordinates of point P and Q on the parabola Y square is equal to 12X are in the ratio 1 is to 2. So point the locus of the point of intersection of the normals to the parabola at P and Q. Okay. So basically when parabola is given here Y square is equal to 12X. Okay. Let me assume this as parabola. So and we are dropping we are having normals right normals at P and Q. So let me draw this suppose this is our point P. This is our point Q we have drawn normal at this and they're meeting at some point. Now we have to find the locus of the point of intersection of normals. Okay. Okay. So basically this is the axis of the parabola. Okay. And this is our point P. This is our point P and this is our point Q and both we have drawn normals at this point P and Q and both these normals are meeting at point M suppose. Okay. And we have to find the locus of this point in right now find the locus of the point of intersection of the normals to the parabola at P and Q. We have to find the locus of this point in and this parabola is basically Y square is equal to 12X. So comparing with the standard form we get the value of AS 3 here. Right. The value of AS 3 here. And one more information is given here. The ordinates of the point P and Q are in the ratio one is to two. What does it mean? Suppose I'm taking this point P as I'm taking point P as AT1 square. Okay. AT1 square comma 2A T1 and I'm taking this point Q as AT2 square comma 2A T2. Is it okay? We can take it in parametric form. So as per given information in the question, the ordinates means what this 2A T1, the ordinate of this point P and Q, their ratio is one is to two. So basically this 2A T1 upon 2A T2 is given out to be one is to two. What does it mean? This 2A, 2A will be cancelled out. We got the value of T2S. We got the value of T2S 2 times T1. Is it okay? We got the value of T2S 2 times T1. Right. And the parabola is given as Y square is equal to 12X. So from this information, we segregated this. Okay, that the T2 is equal to 2 times T1. We have to find the locus of the point of intersection. We have to find the locus of the point of intersection of the normals to the parabola, this. So how can we do that? How can we do that? We can write the equation of normal from point P. Okay, that one thing we can do or find the locus of the point of intersection of the normals. So this T2 is given out to be two times T1. How can we proceed? We can write the equation of this PM and QM also. And all right, we can take the intersection point that we also, we can proceed right now. So what I'm thinking, let me draw the equation of, let me write the equation of this PM. So PM, right. And we'll write the equation of PM. Since definitely we can write, it will be basically in parametric form. I'm writing the equation of normal, okay. So this will be basically Tx plus y is equal to 2A T1, right, 2A T1 plus A T1 Q. And we can write the equation of this QMS. Okay, this will be T1x. So this will be Q2 will be QM will be T2x plus y, okay, is equal to 2A T2 plus A plus A T2Q. Hope I'm writing the correct one. So this will be the equation of these two normals, okay. Now what I will do, I will solve these equations, okay, to find this intersecting point M. So I will subtract this 1 minus 2, H2 will be basically T1 minus x times T1 minus T2 will be 2A T1 minus T2. And then plus A times T1 Q, T1 Q minus T2 Q, is it okay? I think we are going in the right way. So x times T1 minus T2 will be 2A T1 minus T2. Now open this. This will be A and AQ minus BQ means what? T1 minus T2 then T1 square plus T2 square and plus T1 T2. Is it okay? Now I will, we can take the complete comment from you. This x into T1 minus T2. This T1 minus T2 is common from the both terms. So it will be 2A, okay, 2A plus A times this thing. T1 square plus T2 square plus T1 T2. So this T1 minus T2 will get cancelled out. What we get is, okay, we also know the value of A, so we can put here. So x will be equal to basically, x will be equal to 6. 2 into 3, it will be 6 plus 3 times, okay, this T2 we can write it as 2 times T1. So this thing will become, okay, let me write it once again. So 6 plus 3 into this T1 square plus T2 square can be written as, we can write it as 4 times T1 square. Is it okay? And this thing will become T1 into T2 means 2 times T1 square, 2 times T1 square. So I have written all the things in terms of T1. So this will be x minus 6 upon 3 will be equal to 15, 15 to 7 into 21. Right? Or you can write it as x minus 6 upon 3 is equal to 7 times T1 square. Is it okay? Or from here we can let it be as it is. x minus 6 upon 3 is equal to 7 T1 square. Now, this is what? This is the x coordinate, okay, means h. This is equal to h. Okay, so basically this is the x coordinate. This is nothing but our h, what we have considered. What we have considered h. So this is our x. And one more thing we can say. Now, since we got this value of x, we can have the value of y also. Right? Can have the value of y also. So let me put that in here. So let us simplify this. Okay. It's no need to write in this form. What I will do? I will write this x as, I will write this x as 6 plus, okay, 6 plus 7 T1 square. So 21 T1 square. This is what the x coordinate. And this is nothing but equal to as per our assumption. We have considered the intersection point coordinates as h format. Now I will put this value of x in this equation one. Put this value in equation one. What we got this T1 into 6 plus 21 T1 square. Okay. Plus y is equal to 2 T1 to a T1 means a is 3 now. So this will be basically 6 T1 and 2 T1 cube plus 2 T1 cube. I hope is it, it is clear to all. So this will be basically 6 T1. So this 6 T1, 6 T1 will get cancelled out and we will be left out with 21 T1 cube. 21 times T1 cube plus y is equal to 2 T1 cube. Okay. So from here we got this value of y s minus 9 times. Right. So 2 T1 cube. And this thing will be having. We put the value correctly. The x will be 6 plus 21 T1 square. So the 6 T1 will get cancelled out. And no, no, no. This will be three. No, the value of a is three. So what am I written here to it will be three basically. It will be three. So if you observe, we found the value of why is on the value of why is minus minus of 18 T1 cube minus of 18 T1 cube. Okay. And this is nothing but our K. So we have to find this value, a locus of M. No. So we have to replace this. We have to remove this parameter T1 from this equations of x and w. So how we can do. If you say from here we can say from here we can say T1 T1 is equal to x minus 6 upon 21 car under it. Right. And we will put this value here. We will put this value here. So what we will get, we will get y is equal to minus of 18 times T1 cube. No. So T1 is what or we can simply write it as we can write it is in this fashion also. Like from here we get T1 cube. T1 cube is equal to minus of y upon 18. This is what we got. Now put the value of T1 here. So it will be x minus 6 upon 21. Okay. Car power 3 by 2. This will be equal to minus y upon 18. Okay. So from here if you see, I will square it. Okay. So on squaring if you see on squaring we will get x minus 6 upon 21 whole cube. Okay. I'm squaring it. It will be equal to y square upon 18 times square. Is it okay? So on squaring this 2 in the denominator we get cancelled out. Okay. So this is what we got. So basically this will be our equation of locus. This will be locus. Right. No. This will be our locus. So if you wish you can simplify it, how can we simplify this x minus 6 the whole cube? Okay. Upon this 21 into 21 into 21. This will be equal to y square upon 18 into 18. Is it okay? So by 3 we can cancel it. This will go 6 times. This will go 7 times. Again, this will go 6 times. This will go 7 times. Okay. Then this will go 2 times. This will go 7 times. So finally this will be our answer this x minus 6 whole cube. Okay. A whole divided upon what is left out 7 into 7 into 7. That is 49 into 7. And this will be equal to y square upon. This will be equal to y square upon 6 into 2. That is 12. So hope this is clear to all. Okay. This will be our required locus. So, yeah. Now let's see the next question. It is given here. Okay. Question number 12. The normals at PQR on the parabola y square is equal to 4x meet in a point online y equal to C. So the normals at PQR. Prove that the sites of triangle PQR touches the parabola x square is equal to 2 C by means this is based on the concept of our co-normal points because we are drawing normals at three points this PQR. And these three normals are meeting at one point like they are intersecting at a common point. So let's try to make a rough sketch for this. Okay. So this is our parabola. Okay. And let me assume this to be excess and we are drawing what three normals. So this is one normal. This is second normal. And this is our term. Resume it in this way. So this is our point P. Okay. This is our point Q. And this is our point R. And let me assume they are meeting at point this in. So the normals at PQ and R. And this is our parabola y square is equal to 4x. So the normals at these three points meet in a point on the line y is equal to C. Okay. So this line basically we can draw it manually only. So basically this is our point y equal to C. This is our point. This the y coordinate of this will lie on y is equal to C. Okay. And prove that the sites of the triangle PQR PQR. Okay. Now we have to go on triangle also. So let me try it PQR. The question is asking for this triangle PQR. Okay. We have to prove the sites of this triangle PQR. What is the parabola x square is equal to C. I mean say anyhow we can if we can prove that for this parabola x square is equal to C. Why this side of this triangle is tangent. We can say yeah. The sites of triangle PQR touches the same thing we have approaching in that in that way. So what we can do here. So if you see if you see the coordinates of this M. Let me assume it as H comma K but H comma C because it's y coordinate is lying on this C y equal to C. So H comma C will be the coordinates of this M. Okay. So we can write the equation of normal in slow form. What is that y equal to Mx minus 2am minus 2am. Okay. Minus of am true. Okay. Now this will pass through this point M. Right. No. This is intersecting at point M. So the coordinates of this point M must satisfy it. So this will be equal to C is equal to M into H right image minus 2am minus am Q. Okay. So let me make it as a cubic in M. Okay. So this will be a M Q. Okay. And then this to a minus H to a minus H into M. I'm bringing all the things on the left hand side. So am Q plus 2am minus image plus C plus C is equals to zero. Is it okay. Now, actually, this is a cubic in M, which will have three rows, which will have three rows. Let me assume the roots to be this M1 M2 and M3. Okay. So what does it mean? This mean that the slope of this PM, the slope of this tangent PM is M1, the slope of this tangent, sorry, normal, normal MQ is M2, and the slope of normal MR is basically M3. Is it okay. Now, since the coefficient of M square is missing in this cubic equation, we can say the sum of the roots of this cubic, this M1 plus M2 plus M3 will be equal to zero. Right. And what will be the product of roots? The product of rules will be M1 M2 M3 and that will be equal to minus C upon it minus C upon it. Okay. One more relation we have M1 M2 plus M2 M3. So that also we can write this M1 M2 plus M2 M3 and plus M3 M1 that will be equal to 2A minus H 2A minus H upon. So these three relations we got from this cubic. Okay. Now, what I will do, we will write the equation of any of the sides of parabola, sorry, any of the sides of this triangle PQ. So let me write the equation of this PQ. Okay. Let me write the equation of this PQ. So basically, how can we write this? We know the coordinates of this point P and Q. Right now, we know the coordinates of point P and Q. What is that? The coordinates of P is, what is the parabola? A square is equal to 4AX. So it will be AM1 square. Okay. We are assuming the slope to be M1. Okay. So AM1 square minus of 2AM. This will be the coordinates of P and what will be the coordinates of Q? It will be AM2 square comma minus 2AM2. Okay. So these will be the coordinates of P and Q. So easily we can write the equation of PQ. So this will be Y minus Y1. Y minus 1 means Y minus Y1 is equal to this Y1 minus Y2 to minus 2AM1 then minus 2AM2. So this will be minus minus between plus and this will be AM1 square. AM1 square minus AM2 square and X minus X1. That is X minus AM1 square. Is it okay? So from here, we get Y plus 2 times AM1. This will be equal to, if we take this 2A thing common, what we will get? M2 minus M1. M2 minus M1, right? And from here, we get M1 minus M2. M1 minus M2 and M1 plus M2. M1 plus M2. So basically if you take minus 2A common from here, so it will become M1 minus M2. So it's better to do in that way so that this M1 and M2 will get cancelled out. M1 minus M2. And this will be whole multiplied by X minus AM1 square. So this M1 minus M2 terms gets cancelled out. And we are left with, okay, this A will also get cancelled out. And we will, we are left out with Y into M1 plus M2. Okay. And this thing multiplied by this M1 plus M2 means what? 2A M1 square. 2A M1 square plus 2A M1 M2. And that will be equal to minus of 2X minus of 2X plus 2A M1 square. 2A M1 square. Again, this 2A M1 square we term will get cancelled out. So finally, we got Y into M1 plus M2. Okay. Then this thing plus 2A M1 M2 and plus 2X is equal to 0. Okay. Now what I will do, I will do one manipulation here. Like we know the value of this M1 plus M2 plus M3, right? And we know the value of this product also M1 into M2 into M3. So we will utilize these two informations. What I will do? I will do this Y M1 plus M2. I will add one M3 and I will subtract one M3. Okay. And here I will multiply and divide by M3. Hope this is clear to all. And I will remain this as it is. So plus 2X is equal to 0. What will be benefit of this? This thing. This thing is equal to 0, right? And we know the value of this thing as minus C by A minus C by A. So we will put it. So it will basically become Y into minus M3. Okay. And this will become 2A upon M3 2A upon M3 into minus C by A. So this A term, A will cancel out and we will have plus 2X is equal to 0. Now, if you observe this. So if you observe this, okay, it will be a quadratic in M3. How? Let's take this M3 common. Sorry, M3 LCM. So this will be minus Y into M3 squared, right? Minus Y into M3 squared. And what is left out here? A got cancelled. So minus 2C. So minus 2C and plus 2X M3 plus 2X M3 is equal to 0. So we got the quadratic in M3. How? This will be Y into M3 squared and minus 2X M3 plus 2C is equal to 0. This is the quadratic in M3. Okay. Now, if anyhow, this is the equation of PQ. We are writing the equation of P3. Okay. So if anyhow we can prove that like this is the equation of PQ. So this PQ, I have to prove this PQ to be tangent to this. Given parabola, right? Okay. So for this line, for this line to be tangent on that parabola, it should have a single value of M3. Okay. Single value of M3. That means the determinant of this should be, the determinant of this quadratic should be equal to 0. So what will be the determinant of M3? The determinant of this quadratic should be equal to 0. So what will be determinant? So this will be basically B squared means this 4X squared, B squared minus 4 times A. What is A? Y into C. What is C? C is 2C. This must be equal to 0. So from here we got 4X squared is equal to 8YC. Or we can say X squared is equal to this thing, 2YC. 2YC. Okay. This was the equation of parabola. So basically we find out the equation of this PQ means one side of the parabola. Okay. Sorry, one side of this triangle PQR. And we have put D equal to 0 because we have to make this line as the tangent to that parabola. So on solving this D equal to 0, we got this equation of parabola. What does it mean? This line is a tangent to this parabola. So this is what we were required to, we were asked to prove. So we can say hence this proved. Hence it is proved. This is clear to all. So these are some lengthy questions, but we are having some good concepts in these questions, right? And this question was based on this co-normal points concept. So I think this is, this one is the last question. So let's finish it. The normals are drawn from point this to the parabola Y squared is equal to 4X. So that lambda must be greater than 1. Okay. So this is one thing which is asked in the question. And further it is saying one normal is always the X axis. And find lambda for which the other two normals are perpendicular to each other. So this question is also based on that concept of co-normal points. So let me draw one rough sketch. So this is our parabola Y square is equal to 4X. And let me assume this to be R axis. And we are drawing three normals, right? We are drawing three normal straight. We are drawing three normals. Here we should have three normals. So this should be three normals instead of normals. Three normals are drawn from this. Okay. So now what we will do, we will draw the normals. Okay. So this is our first normal. This is our second one. And this is the third one. Okay. And let me assume this point set PQR. So this is our point P. This is our point Q. And this is our point R. And let me assume this point as M. Now, its coordinates are given. What? Two lambda, zero. From this point, we have drawn three normals to the parabola. Whose equation is Y square is equal to 4X. Okay. So from here, we got A is equal to 1. So this we will utilize whenever required. So the coordinates of this M is two lambda, zero. The parabola is Y square is equal to 4X. Now what the question is saying so that the lambda must be greater than one. Okay. So for drawing means for having three normals on parabola from this three point M, the X coordinate of M, that is this two lambda, that this two lambda must be greater than A. Okay. Okay. This two lambda must be greater than A. Okay. Or what is the X coordinate of this point M? H must be for if you see for a standard parabola, for a standard parabola, this Y square is equal to 4X. Okay. And we need to draw three normals from a given point. So this H must be greater than two. This is the condition for this parabola. Okay. Means coordinate of X means X coordinate of this point M. Okay. So not this thing. Not this thing. This H must be greater than A. Okay. So basically this will be what is H in this case, H is equals to two lambda here. Okay. So H is equals to two lambda. So basically this two lambda must be greater than two A. And what is A? A is one. So this two lambda must be greater than two. A is equal to, we have shown here. So this lambda must be greater than one for having three normals. This lambda must be greater than one for three normals. So this was the first part of the question. Now what it is saying next. It is saying one normal is always the X axis. One normal is always the X axis. So first let me write the equation of normal. So we can write the equation of normal in slow format. Y is equal to MX. Okay. Minus two AM minus AMQ. This is what we have done in the question also. So now this is passing through point this two lambda comma zero. Okay. So it should satisfy. So this will be zero equal to two lambda M, two lambda M. What is A? A is one. So two lambda M minus two M minus MQ. Or we can simply write it as MQ. MQ plus two minus two lambda, right? MQ plus two minus two lambda into M is equal to zero. So from here we can take M common from both the terms and we will have M square plus two minus two lambda is equal to zero. So from here we get M is equal to zero or M square plus two minus two lambda will be equal to zero. So here you observe this is a cubic in M. So as I already told you, as I already told you the slope of this PM will be M1. The slope of QM will be M2 and the slope of RM will be M3. Right. So one of the slopes, one of the slopes is zero. One of the slopes is zero. What does it mean? One of the normal will be one normal will be always X-axis. So one normal will be always X-axis. Will be always X-axis. In drawing it is not evident but this is the rough sketch you know. So one normal will always be X-axis. Okay. So this second part is also true. Now what is a question is asking in the third part? It is saying find lambda for which the other two normals are perpendicular to each other. So like other two normals have to be perpendicular to each other. So from here if you see this is quadratic here. Okay. So this is quadratic here. M square is equal to M square plus two minus two lambda. Okay. This is equal to zero. This is what we got here. Okay. So what will be the product of roots? Suppose I'm taking this as M1. Okay. I'm taking this as M1. So our M2, the roots of this I'm assuming as M2 into M3. So the product of roots will be basically minus of C by A. That will be nothing but minus of two minus two lambda by one. Right. Now the question is asking to find lambda for which these two normals are perpendicular. Means this should be equal to minus one as per question. Right. So it means this minus one must be equal to minus of two minus two lambda. Right. So this minus and minus gate canceled out. We got two lambda is equal to three. Okay. For this value of lambda, we will be having the rest two normals to be perpendicular to each other. We have applied that condition here. This M2 into M3, the slope of both those normals will be equal to product of the slope of their normals will be minus one. From here, we got this value of lambda. So hope this is clear to all. Today, whatever the questions we have discussed that were very good, that were lengthier but we're more conceptual. So I think this is all. Okay. This. Thank you. A slide. I have incorporated newly. So this is all for today. And we are done with this exercise. We will soon come back with the next exercise of parabola. Till then take care. Okay. Take care of your health and study. So thank you. Thank you.