 So, hello, students, a very good morning to all of you. I welcome you all once again to this platform, right? Where we are discussing this, where we are having question answer sessions from our events publication book. So, the coming two, three months are very important for us, right? Like the board exams are approaching our exam is also coming. So, I will advise you all to please take care of your health, right? Don't be involved in some unkind like unwanted activities and refrain yourself from all those activities and. Please be focused on your studies. Like give your 100% in studies, you keep on practicing the problems and give your 100%. Okay. Don't think much of the exams and all just give your best. That's all I would like to advise you all. Coming to our topic, we were discussing this parabolic chapter, right? And we have completed two exercises of that. So, today we will have discussions from exercise three of this parabolic chapter. So, this is our first question, right? So please read it. First question from exercise three. So, if M1, M2 are the slopes of two tangents that are drawn from 2,3 to the parabola y square is equal to 4x, then the value of this 1 by M1 plus 1 by M2. Okay, so one point is given and we are trying two tangents to the parabola. So, our parabola is y square is equal to 4x. So, this is our parabola y square is equal to 4x and so let me draw one rough sketch. So, this is our parabola and we are drawing two tangents, right? From this point. So, suppose this is our first tangent. This is our second tangent and this is our point B. Let's assume this point as B because coordinates are 2,3. Okay. Now, let's suppose this tangent is having slope of M1 and this tangent is having slope of M2. So, we need to find the value of 1 by M1 plus 1 by M2. Now, what I will do, I will assume, I will assume the equation of tangent to B, like, let the tangent be y is equal to Mx plus C. Okay. Or we can say, okay, we are assuming it here for tangency, we know the conditions, right, that C must be equal to A by M. So, I'm writing that y is equal to Mx plus A by M, like, for this line to be a tangent to the parabola y square is equal to 4x, our C must be equal to A by M. Now, what I will do, I will solve this tangent with the equation of parabola y square is equal to 4x. Okay. So, y square is equal to 4x. Or we can do one more thing, we don't have to solve this, we can put this point in this equation, no? Like, if this is the tangent, this must satisfy this point P. Okay. So, let me put the coordinates here. So, like, our point P is 2 comma 3, it must satisfy this equation. So, from here, if you see, we will get 3 is equal to, in place of x, we will write 2. So, this will be 2M. And what is the value of A? From here, if you see, A will be coming out to be 1, because 4A is equal to 4, that means A is equal to 1. So, this will be 1 upon M, right? Now, from here, we get 3M is equal to this 2M is square plus 1. So, basically, we are having this quadratic in M, right? 2M is square minus 3M plus 1 equal to 0. Okay. We are having this quadratic in M. So, obviously, it will have two roots M1 and M2. Okay. And those two slopes, which we are getting from this quadratic will be slopes of these two tangents. Okay. So, from here, if you see, on solving on further factorizing it, we will have this 2M is square, okay? So, minus 2M minus M plus 1 equal to 0. So, take 2M common from here. It will be M minus 1 minus 1 M minus 1 equal to 0. So, basically, after factorization, we got M minus 1 into 2M minus 1 equal to 0. So, from here, we will have two values of M. So, M1 will be equal to, let me write it as M1 and M2. So, M1 will be equal to 1 and our M2 will be equal to 1 by 2. Okay. So, these two values of M, like these two values of slopes, denote the slopes of these two tangents, right? This M1 and M2. So, now, what question is asking? It is asking the value of 1 by M1 plus 1 by M2. Okay. So, 1 by M1 will be basically 1 and 1 by M2 will be 2. So, this will be equal to 3. Okay. So, this will be our answer to this question. So, option B here I'm able to see, please give it. So, this will be our answer. Okay. Now, let's take the next question. This question number 2. We are facing the angle between the tangents drawn from the horizon to the parabola. This y square is equal to 4A x minus A. Okay. So, parabola here is given as y square is equal to 4A into x minus A. Right? So, we have to find the angle between the tangents drawn from the horizon. Okay. We are drawing tangents from the horizon to this parabola. So, like on comparison, on comparing this equation with our standard equation of parabola, we are having like the vertex of this parabola will be this x minus A must be equal to 0. That means x must be equal to A and what will be the y coordinate? It will be 0. So, the vertex of this parabola will be basically 8 comma 0. Okay. So, if we draw the parabola. Okay. So, this will be a parabola and this is the axis. Okay. And okay, we need to find the angle between the tangents. So, x axis, I'm sorry, yeah, x axis will be not x axis. Where will this center means origin line that we have to figure it out. So, this vertex is 8 comma 0 basically. Right. So, this vertex is 8 comma 0. So, our origin will be somewhere here. Okay. In this point. And we are drawing tangents from origin, right. So, this vertex is 8 comma 0. I'm assuming the origin to be here. I'm drawing tangents from here. So, let me draw tangents from here on this parabola. So, this is one tangent and this one is another. Okay. And our focus will lie somewhere here. Our focus will lie somewhere here. Okay. Let me change the color of the pen. So, our focus will lie somewhere here. Okay. I'm writing it as this. This is our vertex. This is our vertex B. Okay. And this is our origin. So, this is our y axis. This is our x axis. Since the axis of this parabola is x axis only, right. So, our vertex is the coordinates of vertex is 8 comma 0. Okay. 8 comma 0. Now, what will be the coordinates of this? What will be the coordinates of our focus? So, the distance between vertex and focus is A. Okay. On comparing this equation of the given parabola with the standard equation parabola, we get the value of A as A1A because 4A is mentioned. So, the coordinates of focus will be basically this 2A comma 0. 2A comma 0. Okay. Now, I hope everyone is understanding this. And now, what will be the equation of our directrix? So, basically this distance, the distance between vertex and directrix should be A, right. So, this origin, this y axis is behaving as basically directrix because the coordinates of this B is A comma 0 and this is our origin. So, this line will be x equal to, this x equal to 0 will behave as our directrix of the parabola, right. This x equal to 0 will behave like directrix of the parabola and its coordinate is 0 comma 0. Obviously, the origin is having coordinates 0 comma 0. So, from this point, this point O, we are drawing two tangents, this T1 and T2. Okay. And what is further asked? Okay, this angle is being asked. What will be this angle theta? This angle is what we have to calculate, right. So, how can we approach? Basically, if you see this, let me assume this line as y is equal to mx plus c. Okay, like, let the tangent be, okay, or we can do one thing. We can, we should go in this way. That will be more helpful, I think. Let me join this point A and B, okay. Let me join this point A and B, like this point of context. So, basically, what will be equation of AB? Here is this thing. Suddenly, one other method strike in my mind. So, this equation of AB, let me write the equation of AB. So, basically, it is the part of contact, right. So, it will be T equal to 0. Now, what is our S like? This is our equation of parabola y square is equal to 4Ax, okay, minus 4A square. Okay, now I will write the equation of AB as T comma 0. That is nothing but yy1 is equal to 2A into x plus x1 minus of 4A square. Okay, now what I will do, this origin will satisfy, not origin. I am writing the equation of this AB, no, T equal to 0. So, this 0 comma 0 should satisfy this equation, right, 0 comma 0 should satisfy this equation. So, what it will become, basically, 0 is equal to this 2Ax. So, 2Ax will be 0, then plus 2Ax1, 2Ax1 minus 4A square, right. So, our x1 will be equal to this 4A square, 4A square upon 2A, okay. So, this is nothing but 2A, right. This is nothing but 2A. So, basically, what we got, the value of this x1. So, what is this x1? x1 is the x coordinate of this A, okay, this x coordinate of A. And now I will put the value of x1 in the equation of parabola, okay. So, it will be basically y square is equal to 4A into x, I will put as 2A minus 4A square. That will be nothing but y square will be equal to 8A square minus 4A square, that will be equal to 4A square, okay. From here, we got the value of ys plus minus 2A, okay. We got the value of ys plus minus 2. So, basically, the coordinates of this A and B, the coordinates of this A and B are this 2A comma 2A and this is our 2A comma minus 2A, okay. So, if you observe, it is the focal chord. Why? Because it is passing through the focus, because the coordinates of focus is 2A, right. And here also the coordinates of this A and B are coming out to be 2A. So, basically, it is the focal chord, okay. And this y axis is directrix. And this y axis is directrix. So, actually, we have one property for parabola, right. If any focal chord, okay, the angle between the tangents drawn at the extremity of the focal chord, and if that tangents meet at directrix, the angle between them should be 90 degrees. So, this is one of the property, we will verify it here also, okay. So, we got the coordinates of this A and B. So, basically, what we need, we need the angle between OA and OB, right. So, what will be the slope of OA? What will be the slope of OA? What will be the slope of OA? It will be 2A minus 0, right, 2A minus 0 upon 2A minus G. So, this is nothing but 1. And what will be the slope of our OB? What will be the slope of OB? It will be minus 2A minus 0 upon 2A minus 0. So, it is coming out to be minus 1, right. And the product of this OA and OB, this product of OA, slope of OA and OB is coming out to be minus 1. What does it mean? It means both these lines are perpendicular, right. It means both these tangents are perpendicular. So, I am writing perpendicular tangents. So, angle between the tangents should be basically 90 degrees, right. Angle will be equal to 90 degrees between both tangents. And what I have informed you earlier, this is one of the property of parabola. So, you can remember it. So, any focal chord, if we are drawing the tangents at the extremities of focal chord, and then those tangents meeting at direct tracks must make an angle of 90 degrees between them, okay. So, this option A is correct. This will be our answer. Okay. Now, let's move to the next question. This question number three. If E, B is the midpoint of curve passing through the vertex of parabola, y square is equal to 4x, then some relations is given here, between A and B, we have to identify which one is correct. Okay. So, the parabola is y square is equal to 4x, y square is equal to 4x. So, first of all, let's find the value of A. So, 4a is if 4 means A is equal to 1, okay. We get the value of A beforehand. If applicable, we will use this. Anyhow, so what if A, B is the midpoint of chord passing through the vertex. Okay, so let me draw the parabola first. So, this is our parabola. And this is our vertex. Sorry, this is our axis. And one, I have to draw one chord, which is passing through the vertex. Okay. So, let me draw this chord A, B. Okay. It is passing through vertex. So, this is our V means vertex. And let me name it as U. Okay. So, this UV is the chord, which is passing through the vertex. And it's midpoint. Okay. So, its midpoint is basically, its coordinates are, its coordinate is A, B. Okay. So, we have to find some relation between this A and B. So, what will be the coordinates of the vertex here, vertex of this parabola? It will be 0, 0, right. And what will be the coordinates of focus? It will be 1, 0. Right. This is our focus is. So, since value of A is 1, it is 1, 0. Now, let me assume, let me assume the coordinates of U as AT square, AT square, 2, AT. Okay. So, coordinates of U is basically, okay, A is 1. So, coordinates of U will become T square comma 2T, right. T square comma 2T. Now, M is the midpoint of this UV. So, basically, if you see the, what will be the coordinates of M? It will be this T square, T square plus 0 upon 2, right. This will be X coordinate and our Y coordinate will be 2T plus 0 upon 2. So, our coordinates of M comes out to be this T square upon 2 and T. Now, as per question, it is given that T square upon 2 is equal to A. Okay. And this T is equal to B, right. This A comma B is given in the question itself. So, from here, we easily got this T is equal to B means what? This B square is equal to 2A. This will be the relation between A and B. Okay. Substitute the value of B here. So, T square will become B square. B square is equal to 2A. So, this will be our answer. So, I think this option D is matching here. B square is equal to 2A. So, this option D will be correct. Now, heading towards the, to the next question. The diameter of the parabola, Y square is equal to 6X corresponding to the system of parallel chords, 3X minus Y plus C equal to 0 is. So, one concept is here regarding this diameter of parabola. Okay. So, let me assume this parabola. Okay. This Y square is equal to 6X. Y square is equal to 6X. This is our parabola. This is the excess of the parabola. Okay. Now, one chord is given here. This 3X minus Y plus C some chord. So, the equation of some chords is given here. Okay. So, basically, if you see, actually we have to understand what is the diameter of the parabola, right. So, let me assume this line to be like Y is equal to MX plus C. Okay. Y is equal to MX plus C and let me assume this parabola as Y square is equal to 4X. So, basically the diameter of parabola will be the locus of the midpoint of these family of lines. Y is equal to MX plus C. Like M will be constant. All these chords are parallel. Okay. These chords which I have drawn here in white color. All these chords are parallel. And the locus of the midpoint of these chords. Okay. Will give me the diameter of the parabola. Will give me the diameter of the parabola. This Y square is equal to 4X. So, what is this? Y is equal to MX plus C if you consider this thing. M is constant for all the chords. Right. M is constant. Now C is getting varied. Like C is variable. Okay. So, suppose if I join, suppose the midpoint of this is our M. Okay. This, suppose I'm writing this as AB. So, chord AB, its midpoint is M and it's, let me take it as H comma K. So, the locus of the midpoint of all the chords. Okay. That will give me the equation of diameter of the parabola. So, I think it is clear to all. Okay. Now, coming to the question, what is given here? Our parabola is given as Y square is equal to 6X. Okay. And this AB, this AB in the question as per question, it is given as 3X minus Y plus C equal to 0. Okay. Or we can, you can write it as in this way also. Y is equal to 3X plus C. Y is equal to 3X plus C in this form. Y is equal to MX plus C. Means slope intercept form. So, now what I will do? I will solve it. Okay. I will solve the equation of this parabola and this equation of form. Okay. So, let's see what we are getting. So, we will give, we will get this 3X plus C. Okay. 3X plus C square is equal to 6X. Is it okay? Or better we make a quadratic in Y. That will be more helpful. So, if you see what we can do, this 3X minus Y plus C equal to 0, or we can solve in this way also. Like what we will get? Let's try to figure it out. Okay. So, but finally we need to have the locus of this M. No. So, let's try to make a quadratic in Y. Okay. So, from here if you see, from here, from this equation of AB, we can have the value of X. So, this will be 3X. 3X is equal to Y minus C. Or we can say X is equal to Y minus C upon 3. Okay. I will tell you the reason why I am trying to make a quadratic in Y. Okay. So, substitute this value in equation 1. I am raising this portion because it is getting a quadratic in X. Okay. Substitute this value here. So, we will get Y square is equal to 6 into X. X is nothing but this Y minus C upon 3. So, we will get 3Y square. Okay. 3Y square minus of 6Y. Okay. Minus of 6Y and plus of 6A. Plus 6A is equal to 0. So, finally we are getting this quadratic in Y. Now, what will be the sum of this, sum of the roots of this quadratic? Y1 plus Y2 will be equal to 6 upon 3. That is nothing but 2. Okay. Now, let me ask you. Let me take this coordinates of A as X1, Y1. Now, I am coming to the theory part. So, let me take the coordinates of A as X1, Y1 and this coordinates of B as X2, Y2. Now, what I told you, what will be the diameter of this parabola? It will be the locus of the midpoints of the given quotes. Given the family of this parallel quotes. So, this K, okay. This K will be equal to Y1 plus Y2 upon 2. Okay. And what will be the value of H? H will be X1 plus X2 upon 2. Right. Now, I will use this thing. I will use this, because we know the value of this Y1 plus Y2. Now, this Y, divide it by 2. So, Y1 plus Y2 by 2 will be equal to 1. Right. From here we get this. Now, what is the value of this Y1 plus Y2 upon 2? This is nothing but K. This is nothing but K. And K is coming out to be 1. The value of K is coming out to be 1. Now, in while writing locus, what we do, we normally replace this H and K by X and Y. So, this I will replace this Y by 1. I will replace this K by Y. So, finally, we get this Y equal to 1. So, basically this will be the locus. This will be the locus of the midpoint of this chord EP. Or you can consider other parallel chords also. So, this Y minus 1 equal to 0 will be the diameter. This will be the diameter of the given parable. Okay. So, hope this concept is clear to all. Like, what is the diameter? What is this word mean? And how to find the locus of that point M? Okay. For the family of chords which are parallel to each other. Okay. One more thing you should know. So, this diameter will always be, this diameter will diameter of parabola will always be parallel to X's. Will always be parallel to parallel to the axis of parabola, axis of parabola. So, if you see, what is the axis of parabola here? Y square is equal to 4X. It's in the standard form only. So, this X axis is the, what do you say? X axis is the axis of parabola. Okay. Now, this Y equal to 1. What does it mean? Something like this. This Y equal to 1 will be this. Let me try it. So, our midpoint will be something here. Somewhere here. Okay. So, this will be Y equal to 1. This is our Y equal to 1. Is it okay? So, this will be our diameter of the parable. So, I assume like everyone is clear on this. So, let's move ahead. We will take the next question. This was question number 4. So, now we will take this question number 5. From the point minus 1 comma 2, tangent lines are drawn to the parable of this. The area of triangle formed by the chord of contact and the tangents is given by, okay. Like one point is given. From there, we are drawing two tangents to the parabola. Y square is equal to 4X. Now, the area of triangle formed by the tangents, like pair of tangents and the chord of contact that we have to find the value of that triangle area of triangle. So, Y square is equal to 4X means our standard parabola only. Okay. So, this is our parabola. This is the axis of the parabola. Now, point is given as minus 1 comma 2. Okay. So, let me assume that point will be here. Minus 1 comma 2. This is one tangent I am drawing and this is one tangent. Okay. And now I have to join the, I have to draw the chord of contact also. Right. So, this will be our chord of contact. So, suppose this is our point P. Okay. From where we are drawing this pair of tangents. This is minus 1 comma 2. Let me assume this point as A and this point as, okay. So, AB is our chord of contact. AB is our chord of contact. COC. Okay. And we have to find the area of triangle. This area of triangle PAB, PAB that we have to find as per the question. So, can we write the equation of AB, this chord of contact? Yeah, easily we can write it. So, basically equation of AB if you say, this equation of AB is given by T equal to 0. Okay. This equation of chord of contact we know its equation is given by T equal to 0. So, and what is this parabola? Okay. This parabola is y square is equal to 4x. So, this will be yy1 minus 4x plus x1 by 2 is equal to 0. Now put the value of this x1 and y1. So, it will be 2y, 2y and minus 2x minus 2x minus 2 times x1. Minus 2 times x1 will be what? Plus 2, right? Plus 2 equal to 0. So, minus 2 times x1, x1 is minus 1. So, it will be plus 2. Okay. So, divide it by 2. I will get y minus x plus 1 equal to 0. Okay. So, this is our equation of AB. This is the equation of AB. Right? Now, can we find this? If we can find the coordinates of A and B, right? So, our task will be easy. So, what is the process like? How to find this value? This coordinates of A and B. I want to have the coordinates of A and B. So, that the area of triangle can be easily find out. We can easily find out the area of the triangle. So, we got the equation of AB. Okay. Should we solve with this? The equation of parabola. The equation of parabola is this y square is equal to 4x. Okay. So, this will become y square is equal to 4 times x. From here, if you see what will be our x, x will be equal to y plus 1. Right? So, put here. This will be y plus 1. So, y square minus 4 y minus 1 equal to 0. Okay. So, from here what we get y is equal to minus B means 4 plus minus under root of B square. B square is nothing but 16 minus 4 A and minus 1 upon C. C is what? Sorry. A minus B plus minus B square minus 4 AC upon 2A. So, this will be 2. So, from here we get 4 plus minus 16 plus 4 that is 20. Okay. 20 we can write it as 2 root 5. Right? 20 means 4 into 4 into 5 that is 2 root 5. Okay. So, this will be 2 root 5 upon 2. Or we can say this thing to be 2 plus minus 2 plus minus root 5. Is it okay? This is the value of y. Okay. So, this will be basically coordinates of this y coordinates of this A and B. Is it okay? So, this is what we got the value of y. Now, we have to find the x coordinate of A and B. Right? So, you once shake the calculation. Hope it is not wrong. So, once I will take the value of y1 as this 2 plus root 5. Okay. So, what will be our value of x1? So, x1 will be equal to now I am solving this. I am trying to find the value of x. So, this will be 2 plus root 5 and plus 1. Right? So, this will be 3 root 5. Is it okay? And if we take the value of y2 as 2 minus of root 5 our x1 will become sorry our x2 will become y means 2 minus root 5 and plus 1. So, this will be 3 minus root 5. 2 plus 1, 3 minus root 5. And this is our x2. And what is our x1? x1 is 3 plus root 5. Okay. There it will be plus 3 plus root 5. So, this is our x. This is the x1. This is our x2. So, basically we got the coordinates of A, B, S. We got the coordinates of A, B, A. Suppose I am taking it as x1, y1. So, this will be 3 plus root 5. Okay. And its y coordinate is 2 plus root 5. And the coordinates of B will be 3 minus root 5. And y coordinate will be 2 minus root 5. Okay. Now, what will be this area of triangle? Area of triangle P, A, B will be half base. Base is A, B and height that is suppose I am taking this height as let me draw it also. Okay. So, let me name it as M. So, half A, B into P, M. Okay. Now, what will be A, B? What will be this A, B distance? So, A, B will be equal to x1 minus x2 whole square. So, this will be root 5. Right. 3 plus root 5 minus 3 means 2 root 5. 2 root 5 is squared. And plus 2 plus root 5 minus 2 means again 2 root 5 is squared. Whole under root 5. So, it will become 4 into 5. That is 20. Right. 20 plus 20 under root 40. And what will be our P M basically? What will be our P M? Okay. So, P M will be the perpendicular distance on A, B. Right. P M will be perpendicular distance on this line. So, this is basically minus of 1 means 1 plus 2 plus 2 plus 1 upon under root of A square plus B square that is root 2. So, this will become 2, 3, 4 upon 4 upon 1. Okay. So, our area will be, area will be equal to half times A, B. A, B is under root 40. What can we write this under root 40 as 8 into 5 or 4 into 10. So, 2 root 10. Okay. Let me write it as 2 root 10. Okay. 2 root 10 and this P M will be 4 by root 2. 4 by root 2. So, what we are getting this 2, 2 will be cancelled out. Okay. So, 4 upon root 2. No. So, this we can write this root 10 as root 5 into root 2. root 5 into root 2 into 4 upon root 2. So, this will get cancelled out. 4 root 5. But we are not having any option of 4 root 5. Have we done mistakes somewhere? So, what we have done, we have written the equation of A, B, this yy1 minus 4 x plus x1 upon 2 equal to 0. Now, we have put this value 2, 2y minus 2x and this will be minus, brother. This minus 2x, this will be minus 2, no. This will be, okay, okay. This x1 I am putting minus 1, no. So, it's fine only. So, this x1 will be minus 1. So, plus 2. We got the equation of A, B, S, y minus x plus 1 equal to 0. From here, we got x is equal to y plus 1. Now, what we have done, we have solved with the, solved this straight line with the equation of parabola y square is equal to 4x. We have put the value here. y square minus 4y. This should be minus 4. Here it will be minus 4. We have to calculate once again, right. So, this will be 4 plus, okay. I am trying to do it here only. This will be 4 plus minus B square minus 4A minus 4. So, this will be basically under root of 30. Let me erase it and do it firstly. Due to this error, we have to do once again whole thing. Anyhow, what can we do? Okay. So, let me write it fatafat. This will be y equal to, y equal to 4 plus minus 16 plus 16, right, 32. Means 4 root 2. No. 4 into 8. So, 4 root 2 root 8. This will be 4 plus minus 16 plus 16 32. So, 16 plus 16 32 means 16 into 2 we can write. 4 root 2. Is it okay? So, 4 root 2 upon 2. So, this will become our 2 plus minus 2 plus minus 2 root 2. Okay. Hope this is correct now. So, now I will take the value of y1 as this 2 plus 2 root 2. Okay. And one more value y2 as 2 minus of 2 root 2. So, on taking this value of y1, we got x1 as 2 plus 2 root 2 plus 1. That is 3 plus 2 root 2. And we got x2 as 2 minus 2 root 2 plus 1. That will be 3 minus 2 root 2. Okay. So, basically we have to change this coordinates also. This coordinates of this a and b. Let me write this also. So, basically the coordinates of a will be 3 plus 2 root 2. Okay. 3 plus 2 root 2 and 2 plus 2 plus 2 root 2. And coordinates of b will be 3 minus 2 root 2. 3 minus 2 root 2 and 2 minus 2 root 2. Now, this will be the triangle of area of triangle half into AB into this PM. Okay. So, what will be AB? It will be x1 minus x2. That is 2 root 2 plus 2 root 2. That will be 4 root 2 whole square and plus again 4 root 2 whole square. Okay. That will be whole thing under root. So, this will be 16 into 2, 32 plus 32 under root. That is under root of 64. So, AB is coming out to be 8. Okay. Now, we think, I think we are going in the right direction. So, this AB came out to be 8. Now, what will be this PM length? PM length will be a drop of a perpendicular from this point on this line. So, it will be 1 plus 1 plus 2 plus 1 mod whole divided by under root of 1 square plus 1 square. That will be root 2. So, this will be 4 upon root 2. Now, area will become half into AB into PM. That is 4 upon root 2. So, 2, 4, half base into height. So, this will be your 16 upon root 2. Okay. Or you can write it as 8 into 2, right. 8 into 2 upon root 2. That is nothing but 8 into 2. Okay. So, this will be our area of line bound by the pair of tangents and the quarter form. So, basically it took a longer time due to this error, but can't tell. Okay. So, this option C is correct. This option C is coming out to be correct. Okay. Now, let's take the next one. This question number 6. Okay. This parabola. Okay. Now, this parabola is given as x square plus y square plus 2xy minus 6x minus 2y plus 3 equal to 0. And we have to find the focus of this parabola. So, this is basically a nice question based on the basic funda. So, like, till now, we normally see the equation of parabola in the standard form, right. Either in the standard form or with the shifted vertex, with the shifted vertex, we normally see. So, here we are having this type of performance, general second degree equation of a pony. So, from here, if you see, for a parabola, this h square must be equal to every right. So, if I write this equation of parabola here. So, x square plus y square plus 2xy minus 6x and minus 2y plus 3 equal to 0. So, what is the h here, basically? This 2h is equals to 2. It means h is equals to 1. Okay. So, h square is equal to AB. Now, what is AB? A is 1 and B is also 1. So, this is 1 into 1. That is, sorry, that is 1 into 1 is equal to 1 and h is also 1. So, definitely it is a parallel. It is given also. Just give me one second. Okay. So, we have verified through our earlier knowledge, like for a second degree pony equation, we must have h square is equal to AB for a parallel. So, it is following that condition, not an issue. Now, how to find the vertex of, sorry, how to find the focus of this parabola? That's an important question. How should we find the focus of this parabola? So, let me go with the basic thing. Let me go with the basic definition of parabola. Okay. So, I am considering this parabola to be this. Okay. And what I will do now, I will draw this axis. Okay. I will draw one directorates. Okay. I will take any point on this parabola and I will join this focus with that point and I will have this distance from the directorates. So, let me take this point as p. Okay. Let me take this point as p. This is our focus and this is our directorates. This is our directorates of the parabola. Okay. Let me assume this focus to be our h comma k. Okay. Let me assume this point as x1 comma y1. And let me assume this directorates to be y is equal to mx plus c. So, see, I have done nothing, I have just considered all these conditions, means all these points. So, this is a point on parabola and this is our directorates. So far, let me name it as pm. So, for this conic to represent a parabola, what is the condition this ps must be equal to pm for parabola. Okay. I think no one is having any objection on this. This ps must be equal to pm. Okay. Now, find this ps and pm. So, how to find this ps will be nothing but x1 minus h whole square. Okay. And this y1 minus k whole square, this whole thing under root. Okay. Okay. And what will be pm? What will be pm this perpendicular distance perpendicular distance from this point on this line. Now, what is this line? This is mx. You can write it in this way also mx minus y plus c equal to 0. Now, calculate the perpendicular distance from this point p on this line. So, it will be basically this mx1 minus of y1 plus c is the mod and under root of this m square plus 1 square. Is it okay? I think up to this there will be no problem for anyone. Now, I will square it. Okay. I will square it. So, let's see what we get. Let's see what we get here. So, this will be basically 1 plus m square. Right. 1 plus m square. This will be x1 square plus h square minus 2hx1. Okay. And then plus this y1 square plus k square and minus 2k y1. Is it okay? And this thing will become our mx1 minus y1 plus c whole square. Is it okay? Is it okay? Now, one thing I must do here, I must replace this x1 and y1 by x and y because this point is lying on the parabola itself. This point is lying on the parabola itself. So, I am replacing this x1 and y1 basically by x and y. So, for better understanding, better calculation purpose. Okay. I will not be putting this x1 and y1. So, I will replace it. So, is it okay? This p is equal to pm that we have equated and we have right the equation. Okay. So, now it will be 1 plus m square. Or should I open it? Should I open it or what? Okay. Let me keep it as it is. So, 1 plus m square. This will be x square plus h square minus 2hx plus y square plus a square minus 2k y. This will be multiplied. And I will, this a plus b plus c whole square formula I will use. Okay. So, this will be m square x square plus y square plus c square. Then minus of plus of 2ab means this 2mxy then minus of 2yc and plus of 2mcx. Is it okay? Plus 2mcx and yc. Okay. Now, I will put this thing. I will put all the things to the one left hand side. Okay. So, this will be basically our 1 plus m square x square. Or you can write it as the same square x square term is getting density. I am able to see that. Okay. So, let me, let me expand this. So, this will be x square or 1 plus m square. Okay. 1 plus m square. Then from here, I will get minus of m square. Okay. What I will get here? h square 1 plus m square. 1 plus m square into h square. 1 plus m square into h square. Is something getting cancelled out or what? This m square m square is getting cancelled out. m1 plus m square h square. Here, there is no h square term. Okay. This minus of 1 plus m square into 2hx into 2hx. Then 1 plus m square into y square. Right. So, this will be 1 plus m square. Here what we have? Minus of 1. Right. Minus of 1 y square. And 1 plus m square into k square. Then minus of 1 plus m square into 2k y into 2k y. And this will be equal to, this we have considered there. y square is also considered. This xy term is not considered. So, this if I write here itself. So, this will be 2mxy. Okay. And this plus 2yc plus 2yc. Okay. This plus 2yc minus 2mcx and minus of c square is equal to 0. Basically, what I will do? I will compare this equation of parabola with the given parabola. Okay. Hope this is clear to all. So, basically, if you see this considering this equation xy. Now, xy term is here. And here if you see where is the xy term here. So, basically this 2m is equal to this 2m is equal to comparing xy term. This 2m is coming out to be 2. So, from here we got m as 1. Okay. And what further we can do? If you see these things is getting cancelled out. Okay. And this equation of x square here coefficient is 1. Here also 1. What is the coefficient here? 1 minus 1 is getting cancelled out. Okay. So, from here also we are getting m as 1. And this constant term we have to take care of. Okay. So, let me put here. So, this will basically become x square. And plus 2h square minus 4hx. Okay. Minus 4hx plus y square plus y square plus 2k square and minus 4ky plus 2y. See, right. Plus 2yc minus 2mcx. Okay. Minus 2cx you can see. Minus 2cx and minus c square is equal to 0. Okay. So, this calculation is getting longer. But anyhow, what can we do? We can further compare the coefficients of x also. x and y also. So, if you see this coefficient of x, what is the coefficient of x here? So, here x term is there and here. So, basically this minus of minus of 4h plus 2c minus of 4h plus 2c is equal to into x is equal to what is here? Minus of 6. Right. So, from here this 4h plus 2c is coming out to be 6. Okay. Or you can say this 2h plus c is equal to 3. Similarly, on comparing the equation with this coefficient of y, what I will get? This minus 4k and minus 4k minus of 2y. Minus of 2c. Okay. Into y, this must be equal to this minus 2y. This must be equal to this minus 2y. So, from here if you see this thing is getting cancelled out, this 4k minus of 2c is coming out to be equal to 2. Or you can say this 2k minus of c is equals to 1. Okay. Let me just go over some. Let me have some space here. So, this is what we got. Okay. So, this thing is there. This thing, apart from that any other relation can we have? So, yeah, this thing, this constant thing, this 2h square. Okay. So, this 2h square. 2h square here. Then what we have? This 2k square plus 2k square. And this minus c square. This must be equal to the constant term here in the equation of parable. Okay. It is coming to be 3. Now, what I will do? I will replace this where, substitute the value of h and k in terms of c. Okay. So, what will be h from here? It will be 3 minus c upon 2. 3 minus c upon 2 whole square. Then plus 2 times. So, plus 2 times what? k square. What is k? k is a c plus 1 upon 2. c plus 1 upon 2 whole square. And minus of c square minus of 3 equal to 0. Okay. So, basically it will become 1 by 2. 3 minus c whole square will be 9 plus c square minus of 6c. Okay. Then here also it will be 1 by 2. c square plus 1 plus 2c plus 2c. Then minus c square minus 3 equal to 0. So, multiply it throughout by 2. We will have this c square minus 6c plus 9 plus c square plus 2c plus 1 minus c square minus 3 equal to 0. Right. So, this plus c square minus c square will get cancelled out. This will be plus c square. Okay. And minus c square will get cancelled out. What we are left with? Minus of c square. No. So, is it minus of 2c square or what? The c square should completely vanished off. Right. So, if you see here minus c square is there. Okay. So, c square plus minus 6c plus 9 minus 4c. From here what we get? Minus of 4c and plus 9, 10. 9 and 10 minus 7 plus 1. Okay. So, again I am doing somewhere mistake this. Here it must have twice of c square. Let me check once. This will be c square. 2c square it's coming out to be. Right. It's coming out to be 2c square but 9 plus 1, 10 minus 3 plus 7 plus 7. Okay. And what this c square plus c square means? 1c square we are left out with. But something is wrong. Something is wrong I think because this c square term should be vanished. So, this 1 plus x, 1 plus a b square. Okay. And this minus c square thing will be here. This 2cx 2yc minus of c square. Where I am doing mistake I am not able to identify it. This 4k minus 2c is equal to 2. So, 2k minus c is equals to 1. This is okay only. This 2c square term should come here if I am not wrong. So, this 2h square. Okay. Plus 2k square minus c square must be equal to the constant term. What is constant term here? 3. So, minus c square. I am multiplying whole thing by 2. No. My God. So, this will be 2. No. This 2c square and this minus 6. I am multiplying whole thing by 2. Okay. So, this will be minus c square c square and this will be minus 3 into 2. So, 9 plus 1. That is 10 minus 6 will be equal to 4. I am doing I don't know this minus 2c square. That will be minus 6c plus 2c. That will be minus of 4c. 9 plus 10 minus 4 means plus 4. That will be equal to minus 4. So, this the value of c is coming out to be 1. The value of c is coming out to be 1. Here also we have wasted this 5 to 4 to 5 minutes unnecessarily. So, this c is coming out to be 1. Once we got the value of c, once we got the value of c, we can easily find the value of h and k. So, this is coming out to be 1. This thing 2h is equal to 3 minus c. That is 3 minus 1. Pain is also not working. This 3 minus 1. That is 2. So, h is coming out to be 1. Okay. And again this how to find k? This 2k is equal to 2k is equals to 1 plus c. That is 1 plus 1. That is 2. So, from here also we got the value of k as 1. So, hope this is clear to all. This focus, our focus was h comma k. Okay. And in question it is asked that what is the focus only? Right. We have to identify the focus. What will be the coordinates of focus? So, this h comma k is coming out to be 1. Okay. So, it is basically 1 comma 1. But nowhere in option it is given as 1 comma 1. Okay. So, our focus is coming out to be 1 comma 1. This will be the answer. But in options it is not available here. So, I think none of this we can take none of this. So, basically focus will be 1 comma 1. I am writing here. Okay. So, again this question also took line due to my mistake only. I have taken much time in the calculation part. Any hum? Let's see this one. Why this is hanging? I don't know. Pain is also not working. So, the locus of the midpoint of that part of parabola which subtends right angle on the vertex. Okay. So, this is our parabola. Okay. And we have to find the locus of the midpoint of that part of parabola which subtends the right angle. Okay. On the vertex. So, suppose this is our axis of parabola. Okay. And this is our chord which is subtending the right angle at the vertex. So, let me assume this parabola to be y square is equal to 4x. Since nothing is given, I am going with the standard parabola only. And this is chord AB. Okay. Which is subtending the right angle at the vertex. This is our vertex whose coordinates are 0 comma 0. Okay. Now, we have to find the locus of midpoint of this part. So, let me take the midpoint of AB as A. Sorry, M and its coordinate as H comma. Is it okay? Is it okay? So, this AB chord is making an angle of 90 degree at the vertex. This is what is given in the cushion. Now, how to approach this? I think we can write the equation of AB with the given midpoint. So, if you say we can write the equation of AB as how to write the equation of chord with the given midpoint. It is basically T is equal to S1. Right. So, it will be y, y, y minus 2 times 2A x plus x1. Okay. Y, y1 minus 2A x plus x1 is equal to this S1 will be what put the value of this H and k here. So, this will be H square minus 4A sorry, k is square. k is square minus 4A H. Is it okay? Now, put x1 equal to H and y1 is equal to k here also. So, this will be y into or we can write this as k into y minus 2A x. Okay. And minus 2A x1 means what? H is equal to k is square minus 4H. Is it okay? So, finally it will become if you say k is square minus 4H plus 2H that will be minus 2A H then plus 2A x. Okay. And minus of ky is equal to 0. This will be the equation of AB. Okay. This is our equation of AB. Okay. Now, replace this H and k by y and x. So, what we get? This y is where we can, k is where can be replaced by y square. This minus 2A x. Okay. Plus 2A x minus of ky, no more. This is the equation of AB what we got, right? Now, it is obtaining a right angle on the vertex. So, this slope must pay. But how do we utilize this? This slope must pay. This slope of AB and slope of DB must be equal, product of those slopes must be equal to minus 1. So, if you see, if you see, we can do one thing. This is passing through 0 comma 0. It is passing through 0 comma, no, not passing through 0 comma 0. Okay. Let me do one more thing. Let me assume these points. Let me assume this point A as AT1 square comma 280. Okay. And let me assume this point as AT2 square comma 2AT2. Is it okay? Is it okay? I will come to this equation of AB later on. Let me try to build one relation between this T1 and T2. So, this slope of AB into slope of BV must be, product of these two slopes must be equal to minus 1. Now, slope of AB will be 2AT1, okay, upon AT1 square into 2AT2 upon AT2 square. This must be equal to minus 1. So, this AA cancelled out. This T1 will be cancelled out. This T2 will be cancelled out. So, basically we got this T1, T2 is equal to minus of 4. Okay. This T1, T2 is coming out to be minus 4. Now, what can we do for the, this AB, no, this relation between T1 and T2 we got. And equation of AB we got. Okay. So, this, we can write the equation of AB as in slope intercept form also, sorry, two point form also. So, let me write the equation of AB as this H comma K at the midpoint, no. So, this H will be equal to basically this AT1 square, AT1 square plus T2 square. Okay. AT1 square plus T2 square upon 2 and this K will be equal to 2AT1 plus T2, 2AT1 plus T2 upon 2. Okay. So, this K will be equal to A into T1 plus T2. And this H is coming out to be AT1 square plus T2 square. Okay. So, basically what we can do, we can write this H as, we can write this H as A by 2. And what is this T1 square plus T2 square, we can write it as T1 plus T2 square. Okay. Minus of 2 times T1, T2. Is it okay? Okay. Now, what I will get this A by 2 and this T1 plus T2 we know. What the T1 value of T1 plus T2 is K by A. K by A. Right. So, I am putting here it will be K square upon A square and this value of T1 into T2 also we know that is minus 4. So, minus 4 into minus 2 that will be plus of A. Is it okay? So, it will become our A K square. Okay. A K square upon 2A square and plus 8A, plus 8A upon 2 this H. Okay. Now we got a relation between this H and A. So, finally what we will have, this A and A will be cancelled out. This taking LCM as 2A, we will have this K square plus 8A square is equal to H. So, this 2AH is equals to K square plus 8A square. Okay. Now replace this H and K by X and Y. So, Y square minus 2AX plus 8A square is equal to 0. So, this will be our answer. So, basically we do not need this equation of AB. This approach was drawn. Okay. So, this Y square minus 2AX plus 8A square is equal to 0. This will be our answer. Okay. We have assumed the points this A and B to be AT square comma 2AT1 and this in parametric form and the middle point of this card is taken to be H comma K. And now as per the given condition, these slopes should be, these line AB and AB are at 90 degree. So, their slopes must be equal to, product of their slopes must be equal to minus 1. So, from there we got one relation between this H and A. So, I hope it is clear to everyone. Now, let's take this next one. This question number 8. It is seeing a ray of light moving parallel to the X axis gets reflected from a parabolic mirror whose equation is Y minus 2 whole square is equal to 4 times X plus X1. After reflection the ray must pass through the point. Okay. So, this is, this question is mainly based on the property of parabola. Our parabola is given to be Y minus 2 whole square is equal to 4 times X plus X1. So, it is compared it with our standard parabola, Y square is equal to 4AX. Okay. So, basically what will be the vertex? What will be the vertex of this? This X plus 1 must be equal to 0. That means this will be a minus 1 and Y plus 2 must be equal to 0. So, that will be minus 1 comma 2. So, this will be the vertex and what will be our parabola will look like. So, our vertex, its vertex is what? Its vertex is minus 1 comma 2. And what is A here? A is coming out to be 1. 4A is equal to 4 means A is equal to 1. So, basically its focus will be at minus 1 plus 1 that is 0 comma 2. 0 comma 2. Now, this is the axis of parabola. Okay. This is the axis of parabola. And what is this axis? Y is equal to 2. Y is equal to 2. So, as per the property of parabola, any line which is parallel to axis, any line like any ray which is parallel to axis, okay, any ray which is parallel to the axis of parabola. So, this line, this ray is parallel to the axis of parabola. After reflection through the parabolic mirror, okay, it will pass through the focus. It will pass through the focus of the parabola. This is the property of the parabolic mirror. So, this ray is parallel to the axis of the parabola. So, after reflection, it must pass through the focus. So, after reflection, the ray must pass through the point. Which point? It will must pass through the focus and our focus is nothing but 0 comma 2. Okay. So, this will be our, this will be our answer, 0 comma 2. So, this question is done. Now, what is next? Prove that the locus of intersection of tangents to the parabola, this y square is equal to 4x, which meet at an angle alpha is x plus a whole square, tan square alpha is equal to y square minus 4x. So, standard parabola is given here, y square is equal to 4x. We draw one graph, sketch. Okay. So, this is our parabola and it is asking that the point of intersection of tangents which meet at an angle alpha. Okay. So, let me assume one point here. Let me draw two tangents from that point. So, this angle is basically alpha, as per the question. This angle is alpha. And suppose this point is p. Okay. Suppose this point is p. And we have to find the locus of this point p. So, let the coordinates of this point be h comma k. And this parabola is basically y square is equal to 4x. Okay. So, let me assume, let me assume the equation of tangent. Let me assume the equation of tangent to be y is equal to mx plus a by m, right. For tangency, this c must be equal to a by m. So, if you see, since this tangent is passing through point p, so it must satisfy this, it must satisfy this k comma h comma k point. So, it will be, what you say, k must be equal to, or you can write it as m, k must be equal to m square h, m square h plus a. Okay. Now, it will be a quadratic, this m square h minus of mk plus a is equal to 0. Okay. So, this will be a quadratic in m. So, our, if you see the, what will be the sum of roots? The sum of roots will be m1 plus m2 and that will be equal to k upon, k upon h. Okay. And what will be m1, m2, the product of roots? It will be basically a upon, that is c upon a, that means a upon h. Okay. Now, it is given the angle between these two tangencies alpha. So, basically tan alpha must be equal to m1 minus m2 upon 1 plus m1, m2, 1 plus m1, m2. So, now this, what is m1 minus m2? Okay. We have to further calculate this m1 minus m2. So, okay, let me do it. So, m1 minus m2 whole square is equal to this m1 plus m2 whole square and minus of whole m1. Is it okay? So, from here, what we get? k square upon h square, k square upon h square minus of 4 a upon h. Okay. So, let me wipe out it and do it here. So, we will use this formula later on. First, let me find out this value of m1 minus m2. So, m1 plus m2 whole square k square upon h square minus 4 a upon h, that is nothing but m1 minus m2 whole square is equal to this h square means k square minus of 4 a h. Or we can say m1 minus m2 is equal to plus minus under root of k square minus 4 a h upon h, right? So, put this value here. We will get tan alpha is equal to or square it first, square it. So, it will be tan square alpha is equal to m1 minus m2 whole square, that is what we can say. This k square, okay, minus 4 a h upon h square and this will be 1 plus m1 m2, 1 plus m1 m2 will be a upon h, okay? So, this will further one simplification becomes h into what it will become h plus a, right? h plus a. So, now for finding locus replace this h and k by x and y, okay? So, it will become h into means x into x plus a, x into x plus a or you can say x plus a, x plus a, okay? So, we are squaring now. So, this will also get a square. This will also get a square. Again, I am doing mistake. So, this will also get a square and this h thing will get cancelled. This h will not appear here. This h will get cancelled out from this h square thing in the denominator. So, basically this will be x plus a whole square, x plus a whole square into tan square alpha is equal to your y square minus 4 a x. This will be the locus of parabola. That is what we have to prove. This x plus 1 whole square, tan square alpha is equal to y square n minus 4 x. Hence, it's proof, okay? Now, coming to the next question, this is question number 10. Find the locus of the middle point of the chords of parabola which passes through the focus, which passes through the focus. So, this is our parabola and this is our chord, suppose this is our chord a b and it is passing through focus, this s. And so, what will be the coordinates of s? It will be a comma 0 since the parabola is our standard parabola y square is equal to 4 x. Now, let me assume the middle point of this chord to be h comma k. Let me assume the middle point of this to be h comma k. Now, we can write the equation of a b. We can write the equation of a b. Since we are having the middle point of this chord, it will be equal to t equal to s 1. Is it okay? So, y y 1 minus 2 a x plus x 1 is equal to s 1. s 1 will be nothing but our this k square minus 4 a h. Is it okay? Now, the same chord. Let me first put it the value here. So, this will be basically k y minus 2 a x minus 2 a h is equal to k square minus 4 h. Now, same chord is passing through this focus also. So, this focus is, those coordinates are a comma 0. Much satisfied, right? So, put here. So, it will become 0. And this will become minus of 2 a square and minus of 2 a h minus of 2 a h is equal to 0 minus minus of 4 h. This k is where minus of 4 a h. So, finally, what we got? This k square minus of 2 a h minus 4 h plus 2 h minus of 2 h plus of 2 a square is equal to c. Okay. Now, replace this h and k by x and y. So, we will get y square minus 2 a x plus 2 a square is equal to 0. This will be the locus of the chords with having a middle points. This will be the locus of middle point of that part of parabola which will pass through the focus. Okay. So, this is our question number 10. I think, okay, this is the last question from the point this tangents are drawn to parabola y square is equal to 4 x. Find the equation of chord of contact and also find the area of triangle formed by the chord of contacts and the tangents. So, basically, we have done similar type of question here. I think this question number 7 or h, I don't know, but same question we have done somewhere where we have, where we found the area of triangle. Let me see where we got some calculation error known that question and saying this was there this was there. Yeah, this one. Same question because the points are also same and in the last question is it is asking about the chord of context so we have derived here we have found the chord of contact equation this equation of a b. So, this is there and the question is asking to find the area of triangle. So, 8 root root. Okay. So, all things are same. So, I'm not going to do that question. So, this is 11. Now, I think this much for this exercise. So, we got stuck in the questions and it delayed that we are 10, 5, 9 to 10 minutes behind. So, anyhow, that was the part and parcel of the question answer session. We can't help. But yeah, we can do it in a great way in the first attempt itself. So, thank you all. This parabola chapter is completed now. So we will meet you once again with the next chapter, most probably with the next conic section that is ellipse. So, till then, you take care and as I informed you earlier, please take care of your health. Don't be stressed. Okay, do practice a lot of questions. But don't, but don't stress yourself. Okay. Keep taking care of your health. So, thank you. We will meet you soon with the next chapter. Till then, good bye.