 Hi friends, very good morning to all of you. So in the last video, we have covered the first exercise of ellipse, right? So today we are going to take questions from the next exercise that is exercise number two and we will solve the questions. Okay, so let's start. Okay, so here we go. The question number one of exercise two from the chapter ellipse, right? So see what the question is saying the number of values of C such that the straight line y equal to 4x plus C touches the curve. x square upon 4 plus y square equal to 1. Okay, so basically the given straight line is the tangent to the given curve. Okay, so let me write the question of the curve. So it is x square, x square upon 4 plus y square is equal to 1. Okay, and the given straight line and the given straight line is a tangent is a tangent to this curve. So what is the straight line? It is y equal to 4x plus C, right? Now for this line to act as a tangent to this ellipse, the value of this C square condition of tangency. I hope everyone is aware of this. So what is the condition of tangency? Condition of tangency is the C square, the C thing, the C square must be equal to a square m square plus a square, right? So as per given condition, what is our a square? If you see from the equation of this equation of the ellipse, the a square value is 4, right? And the value of b square is 1, right? And if you observe what is the slope of this line? The slope of this line is 4, right? So let's put the value of this a square, b square and m in this equation, okay? So what we will get? C square is equal to a square m square, means 4 into m square that is 4k square. And what is the value of b square that is 1? So basically we get C square is equal to 4 into 60, that is 64 plus 1, that is 65, 65. So the value of C is coming out to be plus minus under root of 65, right? So this is the value of C, means what? The question is saying the number of values of C. So we are having basically two values of C. Two values of C is possible, two values of C possible. For this straight line, for this straight line to behave as a tangent to the given ellipse, okay? So the important thing is this condition of tangency should be known to all of you, right? This C square is equal to a square m square 4 plus b square, right? So let's move to the next one. Question number two. It is given if any tangent to the ellipse. So ellipse is our standard ellipse, x square by a square plus y square by b square equal to 1. Cuts of intercept of length h and k on the axis. Then a square upon h square plus b square upon k square is equal to 1. Okay, so basically we have to write the equation of tangent to the given ellipse. Okay, so let us write. So the given ellipse is x square upon a square plus y square upon b square is equal to 1, right? Now I have to write the equation of tangent on this ellipse. So what will be our tangent? What will be the equation of tangent that we know we will replace this x square by xx1 and y square by yy. So this would be xx1 upon a square, right? Plus yy upon b square is equal to 1. This will be the equation of tangent, right? Now it is, this tangent is cutting an intercept of length h and k, length h and k. So for finding the value of h, right? For finding the value of h, what I have to do? I hope everyone is able to imagine that trying what the question is saying, right? This h will lie on the x-axis, right? And this k will lie on the y-axis, right? So for finding the coordinates of h, should I draw that right? Okay, let me draw it. It will not take that much time, right? So let me draw that line. So basically this is our ellipse, suppose this is our ellipse. And I am drawing the x-axis also. So now I am now drawing the tangent. So let me draw the tangent, okay? Now this thing, this is the tangent, right? This straight line in white color, white color. This is the tangent. Suppose this is the point P whose coordinates are x1 and y1. And this straight line is make, this point P is the point of contact basically of this tangent to the ellipse. This is our standard ellipse, x square upon a square, right? Plus y square upon b square is equal to 1. Now this length, if I say this point as a, right? So it is basically h comma 0. Cuts of intercept, no? So this length of intercept will be h comma 0. And if I assume this point as b, its length will be 0 comma k, right? So I hope this is clear to all. So now this equation of tangent, right? This equation of tangent, the straight line in this white color. I have written that, the equation of that straight line. Now to find the coordinates of a, for coordinates of a, for a, what I have to do for finding the x-intercept, I will put y equal to 0 in the ellipse, right? So from there, if you put y equal to 0, what you will get? The x will be a square upon x1. Is it okay? So we can say this is the value of h. This is equal to h, right? So this, the coordinates of a is like a square upon x1 comma 0. Now this, since the intercept is equal to h, so this thing will be equal to h, right? Now for b, if you see, for the point b, for finding the coordinates of p, the x-coordinate will be 0, right? So putting the x equal to 0 in the eval equation, what I will get? I will get the value of y as b square upon y1. Is it okay? Now this thing is equal to p as per given information in the question, right? Now what I have to find? I have to find the value of this thing, a square upon h square plus b square upon k, right? This is what the question is asking. So okay, a square related to b as it is, put the value of h. So our h square will be equal to e to the power 4, right? Upon x1 square, or we can say x1 square. And b square related to b as it is, what is k square? k square will be e to the power 4 upon y square, that is y1 square. Now this, if you see, it will cut, so it will become 2 and it will become 2. So finally we are having x1 square, right? Upon a square plus y1 square upon b square. Now if you observe the equation of ellipse, if you observe the equation of ellipse, this is our equation of ellipse. Now this x1 point and y1 point, this point p is lying on the ellipse also, right? This point p is lying on the ellipse also. So the value of this, so the value of this should be equal to 1. Why? Because the point p lies on ellipse also, right? So this point p should satisfy the equation of ellipse. So x1 square, therefore you can say this x1 square upon a square, right? Plus y1 square upon b square, this should be equal to 1. So this is what we need to fight. So the value of this thing will be equal to c, right? So I hope it is clear to all. What we have done in this question, nothing, we have just written the equation of tangent, right? Then we have found the coordinates of a and b. And as per the questions, the x-intercept is equal to this thing. The y-intercept is equal to k. And finally, we have put the value of this h and k in the required expression. And we got the value as, right? Now, see the next question. The equations of tangents to the ellipse this making equal intercepts on the axis. Making equal intercepts on the axis. So basically equal intercept means what? Equal intercept means what? Because this is our ellipse, okay? And the equation of tangents who are making equal intercepts on the axis. So this may be our one of the tangent, right? This may be one of our tangent. So equal intercept or this can be also one possibility, right? And if I say this line, if you see this line, so this is also one possibility because the intercepts made by this green line will be equal, okay? And this is also one possibility, right? So basically equal intercept means the slope is equal to plus minus one, right? Slope can be plus one. Like for this straight line for this line slope is equal to one, right? And for this line slope is equal to one. Now for this yellow colored line slope is equal to minus one. And here the slope is equal to minus one, right? So let me assume the equation. Let the tangent be, since I have to find the equation of tangent. So I am assuming the equation of tangent to be y equal to, right? Mx plus c, okay? And what is the ellipse given here? It is 3x square plus y square is equal to 3. Now divide by 3 on both sides. I will get x square upon 1 plus y square upon 3 is equal to 1, right? x square upon 1 plus y square upon 3 equal to 1. Now for this straight line to represent a tangent for this ellipse, this c square, c square thing must be equal to a square m square plus b square, right? And we know for having equal intercept we can have these two cases, right? Either m is equal to 1 or m is equal to minus 1, okay? So considering m is equal to 1, I will find the value of this c square. So this c square will be equal to a square plus b square. Now what is a square? If you see here, the value of a square is 1 and the value of b square is equal to 3, right? Comparing with the standard form of the ellipse. So a square equal to 1. That means this c square is coming out to be 3a1 plus 3, right? Equal to 4. So we got the value of c as plus minus 2, right? So what will be the equation in this case? Equation of tangent, it will be x plus 2, right? Or y equal to x minus 2, right? So these two equations I am getting when I am considering m is equal to 1. And when I will consider m equal to minus 1, our c square will be equal to, c square will be equal to, again it will be equal to a square plus m square 1. Why? Because this m square thing will be 1 only. So this will be a square plus b square, okay? So c square will be equal to 4 and the value of c will be equal to plus minus 2, right? And so in this case, what will be the equation? m is minus 1. So it will be minus x plus 2 or y equal to minus x minus 2, okay? So these equations, these are the possible equation of tangents which are making equal intercept on the axis, right? So seeing the option, this option a is correct, right? y equal to plus minus x plus minus 2. So I hope it is clear to all. The only point to note is any straight line which is making equal intercept can have slope either plus 1 or minus 2, okay? Now coming to the fourth question, this, if x y a plus y by b equal to root 2 touches the ellipse, x square upon a square plus y square upon b square equal to 1, then its eccentric angle theta is equal to, okay? So one equation of straight line is there, okay? And that straight line is touching the given ellipse, given standard ellipse. So we have to find the eccentric angle theta, okay? Centric angle theta. So how can we do? We can use the parametric form, right? We can use the parametric form because the question is asking about this eccentric angle theta, right? So theta, we will be able to have the theta in parametric form. So okay, let's proceed. So I'm writing the ellipse, equation of ellipse that is x square upon a square plus y square upon b square is equal to 1, right? And let me assume one point b on this ellipse, which is a cos theta, right? a cos theta comma b sin theta. So for assuming any point on the ellipse or on the given curve, we normally take the help of parametric form. So here also I'm assuming the point b in parametric form that is a cos theta comma b sin theta, okay? Now what I will do? I will write the equation of tangent at this point. I will write the equation of tangent. So let me write it. It will be basically x, x1. So what is x1 here? It is x into a cos theta, okay? Upon a square and yy1, that is y I have written in place of y1, I will write this b sin theta. So yy1 upon b square is equal to 1. Is it okay? So this is the equation of line, which is tangent to this ellipse at this point b, right? Now as per given condition, what is given in the question? It is saying this x by a, right? Plus y by p equal to or root to represent a tangent to the ellipse. So basically these two are same lines, right? Or these two represent the same tangent, right? So if that is the case, what can we confirm? Like what can we say about these two equations? We can say the coefficient of x must be equal to coefficient of y and that should must be equal to the constant. Since both these lines represent the same line, right? So I will use the same funda. If you observe here, this a thing and this a can get cancelled out, this b and this b can get cancelled, okay? So what is the coefficient of x here? It is cos theta upon a, right? Cos theta upon a. And what is the coefficient of x here? It is basically 1 upon a, right? So this must be equal to the coefficient of y. So coefficient of y here, sin theta upon b. And here it is 1 by b. And this must be equal to constant term that is 1 upon 1 here in this case and root to b, right? So this a and this 1 by a thing will get cancelled out, this b and this b will be cancelled out. So what we got? We got this thing. This sin theta or okay, let me write it as sin theta. Sin theta is equal to cos theta, right? And that is equal to 1 by root 2. So what does it mean? What will be the value of theta? So theta will be equal to Pythagoras degree or you can say what? Pi by 4. Pi by 4 radian, right? So this is what the question is asking. The question is asking to find the angle theta, right? This is centric angle theta. So this will be our value of theta. So I hope it is clear to all. We have assumed the point b in parametric form because we were able, we need to find this a centric angle. So we have no other options left than to assume the point in the parametric form. After that, we have written the equation of tangent and we have compared both the straight lines since they are representing the same line, right? So from here we got the value of theta as 49 degree, okay? So this was the question number 4. Now let's have a look at question number 5. The number of values of theta between 0 say 2 pi, for which the line 2 is cos theta plus 3y sin theta equal to 6 touches the ellipse, this 4x square plus 9y square equal to 36, okay? So the given ellipse is our 4x square plus 9y square is equal to 36, right? So divide by 36 on both sides. I will get x square upon 9 plus y square upon 4 equal to 1, right? So this is our ellipse in standard form. And the given line, if you see, given line is touching the ellipse. So a tangent you can say, the given line is the tangent. So the equation of tangent given is 2x cos theta, right? 2x cos theta plus 3y sin theta equal to 6, 3y sin theta equal to 6. Given in the question, it is given in the question, okay? That this line is touching the ellipse. So we can say that this line is the tangent to the ellipse. Is it okay? Now, if that is true, the number of values of theta, then what can be the values of theta? Okay, so let me, should I draw this sketch? Okay, let me draw quickly. So basically, this is our ellipse, okay? These are the axis of the ellipse. These are the axis of the ellipse, right? And this theta thing, so I should draw the auxiliary circle also. So draw with different color. So this is basically our auxiliary circle, right? And this equation of tangent is given. So we can draw one tangent also. Okay, so let me draw that tangent. Suppose this is the tangent, okay? Now coming to the question. So this is any point, suppose any point P, okay? P theta means the body rates of P will be A sin theta and A cos theta. Sorry, B sin theta. So and what is this theta? I have explained that in the last class also, but let me explain here. Here also, so I am extending this point P to the auxiliary circle and I will join that point to this center of the ellipse. Okay, so what I am meant to say, I have taken this point P, okay? And I have extended it, I have extended it on the auxiliary circle and then I have joined that point. Suppose I am saying it as Q, so I am joining this point Q to the center. So this angle is basically theta. Okay, so suppose this is our tangent, okay? This is, sorry, this is our ellipse and we don't know the equation of tangent. So let me write, let me write the equation of tangent at point P, like equation of tangent, okay? Equation of tangent at P theta. So how can we write this xx1, right? x1 will be A cos theta, xx1 upon 9, right? And plus yy1, yy1 will be B sin theta upon 4 equal to 1, right? Now, if you see here, A square is 9. So we can say A is equal to 3 and P square is equal to 4. So we can say B is equal to 2, okay? So putting the value of A and B here, what I will get? 3, A means what? 3, right? So 3 and 9. So basically 3 cos theta into x upon 9 plus B is 2. So 2y sin theta upon 4 equal to 1. Now, this will be 3 times, this will be 2 times. So LCM will be 6, right? So 2 cos theta x, 2 cos theta x plus 3 sin theta y. Check the calculation, 3 sin theta y equal to 6, okay? So we got this equation, 2x cos theta plus 3 sin theta y equal to, or 3y sin theta equal to 6. So this is the same. What is given in the question? Okay, so how can we define the number of values of theta? You can take any theta, no? You can take any theta on this ellipse. Okay, let me say this is our tangent, no? This is our tangent. So the theta value can be taken anywhere on this, what do you say, on this auxiliary circle. So basically theta can be, theta can be anywhere, can be anywhere on the auxiliary circle. Or in short, you can say this point P can be anywhere, no? On the ellipse. This point P, you can say point P can be anywhere, can be anywhere on the ellipse. So basically there will be infinite number of tangents on this ellipse. Okay, that means there will be infinite number of values of theta. That is not defined. You can't say that this is the fixed theta for this tangent. It's not like that. The P can move anywhere on the circle or on the ellipse. So infinite number of tangents are possible, right? So there will be no fixed values of theta for this, right? So basically no fixed values for theta. Now, coming to the next one. Okay, let me see whether I'm recording the video or not. Yeah, recording is going on. Yes. So this is our question number six. The common tangent of x square plus y square equal to four and two x square plus y square equal to two. Okay, basically we are provided with two curves. The first curve is this x square plus y square equal to four. So this is basically a circle, right? And the second curve is and the second curve is x square to x square plus y square is equal to two divided by two. So we will have x square by one plus y square upon two equal to one. So basically this is the circle, right? And this is our ellipse. This is our ellipse. Now, the question is saying the common tangent. We have to derive the or write the equation of common tangent for both these curves. Okay, so let me assume, let the tangent be as usual. Let tangent be y equal to mx plus c, okay? And for this line y equal to mx plus c to be a tangent to the circle. What is the condition of tangency? Condition of tangency for this. This c square must be equal to a square under root one plus hemisphere, right? Hope you remember all. Hope you all remember this condition of tangency, right? So c square should be equal to what is a square? A square is four here and under root of one plus m square. Is it fine? Is it fine? So what can we say? This c square is equal to four, one plus m square. Or you can write it as c is equal to plus minus two under root of, am I doing some mistake? This is the condition of tangency for circle. This c square is equal to a square under root of, what do you say? Under root of one plus m square. Or c square is equal to a square into one plus m square. I think that is the correct. Okay, anyhow we will be able to figure it out. I also, I think I forget or what? But c square is equal to the c value of c coming to be one under root of a. I think it should be a one plus m square. Like this under root sign should not be there, right? This is what? Okay. So I am removing this under root thing from one plus m square, right? So this will be c square must be equal to a square into one plus m square. I hope it is clear now, right? Yeah, because in ellipse what is the condition a square m square plus b square. Here circle is also a special case of ellipse. So this will be a square means what? Four times one plus m square. Or you can say c equals to plus minus two under root of one plus m square. Is it okay? So what will be the equation of tangent? Basically equation of tangent will become mx, mx plus c. And what is c? That is two under root of one plus m square, right? So this is the equation of tangent on the circle. Now this same straight line is tangent to two. This straight line is tangent to ellipse also, right? This is a tangent to ellipse also, right? So what is the condition of tangency for ellipse? Condition of tangency for ellipse is c square is equal to a square m square plus b square, right? Now what is the value of c square? c square is nothing but four times one plus m square. Four times one plus m square. Is it okay? And this must be equal to a square. What is a square here? a square is equal to one and b square is equal to two. So this will be basically m square. This is equal to m square plus two. So what we got here? Four plus four m square minus m square, okay? Minus of two equal to zero. So from here we get three times m square. Four minus two is two. Coming to the right hand side, it will be minus two. Or what we got? We got m square is equal to minus two by three. But wait, can a square of any slope can be equal to negative, right? Can a square of any number, it can't be equal to negative number. It must be either greater than or equal to zero, right? So I think there is no tangent possible, right? What does it mean? There is no tangent possible, common tangent instead. Rather you say common tangent is not possible for both these curves, right? Because the value of m square is coming out to be minus two by three. That is not possible. Since no such value of m is possible. If no such value of m is possible, no such tangent is possible, right? So none of these can be the answer because we are not getting any particular equation, right? We are not getting any tangent, common tangent for both these curves. So I am taking this as the correct option, none of these. Okay. Now, see the next one, question number seven. If the normal at any point, p on the ellipse, x square upon a square plus y square upon b square equal to one, meets the axis in g and g respectively, then pg dot pg. Okay. Okay. It should be basically is two because you see the answers are in that ratio. This must also represent a ratio, right? This pg is two pg. Okay. So I will draw a quick sketch for this, right? So this is our ellipse. Okay. This is the, this is the axis. And now what I will do. Okay. It is talking about normal. It is talking about normal on the ellipse. So I have to draw one normal. Okay. Okay. I will draw one normal. This is our normal suppose. And this normal is meeting the axis in g and g. Okay. Okay. Now I will sketch. I will write the names. Okay. So this yellow colored line, this yellow colored line is normal, right? Normal to what? Normal to this ellipse, normal to this ellipse whose equation is this is our standard ellipse x square upon a square plus y square upon b square equal to one. So this yellow color line is normal to these things. Now this normal is meeting x axis. This is our x axis. And this is our y axis. This is meeting x axis in g. What does it mean? This point is g basically. This is our g point. This is our g point. And this is our g point. Since the normal is cutting x axis at capital G. And it is intersecting our y axis at a small g, right? Then PG. And this is our point p, right? This is our point p. So we have to find the ratio of this PG, this ratio of distance of PG upon p into g. Okay. p into g. We have to find the ratio of these distances. Okay. So let me assume this point p in parametric form, right? That will be helpful, I think. That will be helpful. And in particular, in general, if you see, if we have to assume any point on the, on the curve, we normally prefer the parametric point, right? So this makes our life easier, a lot easier than assuming the point to be like x1 and y1, right? Because in parametric form, we generally have the only one unknown, which is theta. And in Cartesian form, we have, we make two assumptions, right? We assume the coordinates of p as x1 and y1. So there are two coordinates. So two unknowns and in parametric form, there is only one. So we generally prefer that one thing. Okay. So what I will do, I will assume the point p as what a cos theta. Okay. a cos theta comma b sin theta. And if you remember, you must remember the equation of normal, equation of normal at point p, normal at p, how can we write that? We write it as a x upon cos theta, a x upon cos theta minus b y upon sin theta. This is equal to a square minus b square, right? So this is the equation of normal at the point p. Okay. Now, this is the important thing which you must do, right? This is important. This equation is important. So this, I assume or I request all of you to remember this equation of normal, right? So now what will be the coordinates of this point g? If I say for g, like for g, the y coordinate will be 0, right? Y coordinate will be equal to 0. Okay. So put the equation, value of y 0 in this, what I will get? I will get a x upon cos theta minus 0 is equal to a square minus b square, right? So from here what we get? We get the value of x as, the value of x as a square minus b square into cos theta, right? a square minus b square into cos theta upon a. Okay. So this will be basically x coordinate of g, right? So you can write it in this way also. This will be g k x coordinate, right? And what will be the g k y coordinate? g k y coordinate will be 0, no doubt because this g point is lying on the x axis, right? So similarly, for point this g, small g, the x coordinate will be 0, right? x coordinate will be 0. So put x equal to 0 here. What I will get? I will get this minus b y upon sin theta minus b y upon sin theta is equal to a square minus b square. Is it okay? Is it okay? So we can take the value of y as, means the value of y is coming out to be this b square minus a square. I'm taking minus common so that this minus will get cancelled with this. So this will be b square minus a square into sin theta upon b. Is it okay? So this will be basically g k y coordinate. g k y coordinate, right? And what will be the g k x coordinate? g k x coordinate will be equal to 0, no doubt. So these two things we have got. So basically what we got? We got the coordinates of p, what we have is 0. We got the coordinates of g and we got the coordinates of g. So now the finding the length will be easier now, right? So what will be the length of pg? If you see, what will be the length of pg? So x1 minus x2 pole square plus y1 minus y2 pole square the under root, right? So pg, right? So x coordinate of p is a cos theta, a cos theta and minus this thing. a square minus b square cos theta upon a, right? So x1 minus x2, this in subcut pole square, right? And plus y1 minus y2, that is, what is y1 here? So the y coordinate of p is basically b sin theta and y coordinate of this g is 0. So minus 0 pole square, right? Or the length pg will be basically under root of this. So I'm squaring it. I don't want to put under root. So pg square will be this. And okay, you can simplify it basically. So after simplification, what it will come? This will be, let me take LCM as a. So a square cos theta, okay? Minus a square cos theta plus b square cos theta, b square cos theta, okay? In subcut pole square, right? Or plus b square sin square theta. So I hope I'm not making any mistake. So this a square cos square theta minus a square cos square theta gets cancelled out. And we are left with, we are left with what? B of power 4, B of power 4 cos square theta, okay? B of power 4 cos square theta upon a square and plus b square into sin square theta. Is it okay? Is it okay? Okay. So what can we do for that? This is our pg. So pg means square. This is our pg square. pg square. So similarly, we can find this p small g also, right? If you see, let me write that also. So nothing is left, I think in this question. Just calculation part is there. So this is our pg, pg chi square. What will be that? P chi x coordinate, that is a cos theta, okay? And minus 0, x coordinate of g is 0. So this is square, right? Now this b sin theta and minus y coordinate of the small g. That is this thing. B square minus a square sin theta, right? Sin theta upon b square whole square. So this will be basically our pg square. pg square, okay? So we can simplify it as a square cos square theta. And what I will do, I will take b, it will be b square sin square theta. Minus b square, not sin square theta, only sin theta. Minus b square sin theta, okay? Plus a square sin theta. Complete the whole square. So this b square sin theta minus b square sin theta will get cancelled out. We are left with a square cos square theta plus a power 4, a power 4 into sin square theta. a power 4 into sin square theta upon b square. Is it okay? Is it okay? So is anything getting common here? So if you see here, what we can do, we can do one thing. b square, no? So if you multiply it by b square or a square, right? Can do some manipulations here so that something gets cancelled out while we take ratio of these two. So basically this will be, or what can we do? b square we can take common, no? Here if you see, b square upon a square we can take common, okay? So if you take b square by a square common, what will be there? Here it will be b square cos square theta, okay? It's fine, no? So b square cos square theta, after multiplication b power 4 cos square theta upon a square, right? And here we will have b square is already there, no? b square is already in the outside. So we will have sin square theta. But what about this a square? We have to multiply by a square, right? We have to multiply by a square. Yeah. So this whole thing will be our pg square. Or you can say the pg will be equal to under root of all these things, okay? So why I have done? I am able to say like something will get cancelled out. Now what I will do here? I will take this thing, a square upon b square common. Or here I will take this common. So if I take a square upon b square common here, we will have this b square cos square theta. Is it okay? And a square sin square theta. a square into sin square theta. If you take its under root, it will be our pg, right? So see this b square cos square theta plus a square sin square theta, we are getting common. So now what I will do? The question is asking for this ratio. This pg upon pg, right? So basically it will be b square upon a square. What is there? Yeah. So this will be, okay, after coming out of the root, it will become b upon a into this b square cos square theta. Plus a square sin square theta upon, here we get outside of root, we get a upon b under root of same thing. b square cos square theta plus a square sin square theta. So this thing will get cancelled out, right? And we will have this, what? b square upon a square. So this is the ratio we are getting. Okay, b square upon a square. So this will be our answer. Hope it is there in the option itself. Let me see. Yeah, it is there. Option b, right? No, we got b square upon a square. No, so this will be option b. Yeah. So we are done with this. Coming to the next one. This is our question number eight. So number of distinct normal lines that can be drawn to the ellipse. That can be drawn to the ellipse. So normally we can draw maximum of four normals, right? In case of parabola, if you remember, in case of parabola, from any point we can draw a maximum of three tangents, sorry, three normals. But in case of ellipse, it's not like that. In case of ellipse, a maximum of four normals can be drawn from any point. But I don't know how many normals can be drawn in this question from the given point. We have, we can do, we can, we can do this question. So basically if you see, draw, let me write the equation. Let me write the equation of the ellipse first. So this is our x square upon 169. Okay. Plus y square upon 25 is equals to, right? So this is our ellipse and the question is asking, like, how many number of distinct normal lines can be drawn from the point 0,6? So this is our point, which is, those coordinates are 0,6. Okay. So how can we write the equation of normal? In the last, I think last or before that, we have seen the equation of normal at p. Okay. Equation of normal, right? Equation of normal. We can write it as ax, ax upon cos theta, right? Ax upon cos theta minus by upon sin theta is equal to a square minus b square. Now, what can we do for this ellipse? Our a square is 169 or we can say a is equal to 13 and b square is 25. So our b is 5 for this ellipse. Okay. Now put the value of a and b. So we will have this 13. a is 13. Yeah, 13 ax upon cos theta. Okay. 13 ax upon cos theta minus 5y. 5b is 5. So 5y upon sin theta and is equal to a square minus b square. a square is 169 and b square is 25. Okay. So let's solve it. So I'm taking LCM as sin theta cos theta. So cos theta is there. It will be 13, what? 13 sin theta x. 13 sin theta x. Okay. And minus of 5 cos theta y. 5 cos theta y. Okay. And that will be equal to 169 minus 25. That will be 144. Right. So what can we do for the, what can we do for the, it is coming out to be a trigonometric problem. Okay. One thing we can do now. One thing we can do. This normally is passing through this point now. Normally is passing through this point. So we could have avoid this system. Right. Okay. Anyhow. I've written it. So this point P, if you say point P should satisfy the equation. The equation of tangent. Am I right? Not tangent equation of normal. Why? Because I have a written the equation of normal and that normal should must pass from this point. So it should satisfy. So now I'm just by forgetting the code here. And I'm going here. So if you say X is zero. So basically what we get, we get this thing minus of five Y upon sine theta minus of five. I'm writing here also this point should satisfy this. Okay. Zero comma six. So minus five Y upon sine theta. This must be equal to what? 144. Am I right? 144. No. Okay. Okay. In place of why we should put six. No. So it will be minus 30. Anyhow. Let me write in another instead. Minus five into six. That will be minus 30 upon sine theta. This will be equal to 144. What you can see here. Minus five into six. That will be minus 30 upon sine theta. This will be equal to 144. Or what you can say this, the value of sine theta. We will have as the value of science. It has minus 30 upon 144. That is. Six. Minus five. Six. Two. And four. Okay. So the value of sine theta, we got the value of sine theta minus five upon 24. Is it possible? Can we have this value? Yeah. Sine theta, sine theta ranges between minus one to one. Right. So it is somewhere between zero to minus one. Yeah. It is acceptable. Means it is acceptable. The sine theta value is acceptable. Okay. Now. What is the range of this theta basically? What is the range of this is centric angle? It is between zero to two pi. Right. So let me let me. Draw one rough figure of sine theta. So this is our sine theta. Right. This is a pie. This point is two pie. And this value minus five upon 24 will be somewhere here. So if you observe between zero to two pi, we are having two values. Means two theaters. We can have two theaters here. Look at these points. So two. Thetas are possible. So two theaters are possible means two angles are possible. What does it mean? We can have two normals here. Because you place the, you put the value of theta in this equation and you will have the normal. Since we are having two theaters, we can have two normals. Okay. So no doubt there will be two normal, but there is one more catch in this question. If you observe the point, what is the point? This is zero comma six. So no doubt there two normals. Here I am accepting two normals are okay. But if you observe more closely, you will able to see there is one more normal possible. Right. So let's see. So for that, I will need the help of a sketch. Okay. I will try it quickly. So suppose this is our. This is our ellipse. Okay. This is. This is the coordinate axis. Okay. Now. Name this point. This vertex is a since a is greater than b. So this is our vertex a. This is also. Next one. Second vertex a dash. Now this is our point b. Okay. And this is our point b dash. So what will be the coordinate of this a? It will be a comma zero means what 13 comma zero. This will be 13 comma zero. Similarly, this minus a dash will be minus 13 comma zero. And this B will be what five comma zero or zero comma five, not five comma zero. Zero comma five and this will be zero comma minus five. Now, I. I am saying to observe where this point P is like. So basically this point P is lying somewhere here. Which coordinates are zero common six. Okay. So from here, from what we have done earlier. Here we are getting the two numbers. They can watch about this. This Y axis. This is our Y axis. No. Basically, this Y axis is also behaving as a normal to this ellipse from this point. Since this point is lying on the Y axis. So this Y axis is also behaving is also behaving as normal. Right. Is also behaving as normal to this ellipse. This ellipse. So basically. Basically, we are having three normals. Three normals are possible from this point on this ellipse to from here and three from one from here. So two plus one. Three normals are possible. Three normals. From point P from P zero comma six are possible. Are possible on this ellipse. Right. So, I request all of you to note this point. Okay. This point on Y axis. So basically, we can have three normals on the ellipse. Is it okay? Okay. So let me start the next question. Next one. Yeah. So if a tangent of slope to the ellipse X square upon a square plus Y square upon B square equal to one is normal to the circle. Okay. So basically. Two curves are given here. One is our ellipse in a standard form. And one is our circle in general form. And it is saying that the tangent to the ellipse know tangent to the ellipse is behaving as normal to the circle. Then the maximum value of AB. The question is asking. So, let me write it. This is our ellipse X square upon Y square. Sorry, X square upon a square plus Y square upon B square is equal to one. Okay. Let me let the tangent be let me assume the tangent to be what Y equal to MX plus C MX plus C for this line to be tangent to this ellipse. We all know the condition of tangency. What is the condition of tangency? Y square. Sorry. This C square. This C square must be equal to. A square M square plus B square. Right. What can you say? What can you say this C must be equal to plus minus under root of a square M square plus B square. Is it okay. So this is our tangent. Right. Y is equal to MX plus minus under root of a square M square plus B square. Now, this is the tangent to the ellipse. This same line behaves as same same line behaves as normal to the behaves as the same line behaves as normal to. We have says normal to the given circles. Okay. Now, what is our circle? Circle is X square. Plus Y square plus four X plus one equal to zero. Now, what is the center of this circle? The center of this circle is minus two comma zero because no Y coordinate. Sorry. No term having Y. So center will be minus two comma zero. Now this center should must satisfy this equation. This center should must satisfy. Must satisfy this equation. Right. So if it is satisfying, I'm writing in the next slide. So let me put it. Y will be equal to zero equal to what? MX. X is what? So minus two M plus minus under root of a square M square plus B square. Okay. Now we can take this two M to this side and we can square it. We will have four M square. Four M square is equal to a square M square plus B square. Sorry guys. Where are we? Yeah, we have written the equation. Okay. So what we got? Yeah, we know the value of M, right? Given that, given the slope of this is M, no? So we could have replaced it earlier only. But anyhow, this M is equal to two. So the slope is equal to two. It is given. So I'm writing now. We could have written it earlier. So instead of writing M, we could have taken two. Anyway, so put M equal to two. So this will be two times four. That is 16 is equals to 16 is equals to a square M square. That is what? Four. That is four a square plus B square. This is what we got. This is what we got as per the given condition, right? Now what we need to know after doing all this, we have to find the maximum value of AB. We have to find the maximum value of AB, maximum value of AB. So how can we do? Now this is known to us and we have to find the maximum value of AB. So one very nice concept is utilized here. If you observe, we can use the AMGM inequality here. How? Look at this. This four a square, four a square and this B square thing. These are two positive numbers, right? Two positive numbers. Why? Because it is in the square form. This A square multiplied by the positive four. This B square, these are positive numbers. So for this, since both are positive, we can say this arithmetic mean of these two numbers should be greater than or equal to geometric mean of these two numbers, right? This is the important concept which we are using here. So now what will be the arithmetic mean of this? It will be four a square plus B square upon two. This must be greater than or equal to what? Four a square into B square per hour, one by two. Is it okay? Is it okay? I hope it is clear to all. So now we can put the value of this four a square plus B square here. So it will be basically 16 upon two should be greater than or equal to four root that is two AB. Is it okay? This thing will become two AB. Now this 16 upon two into two is greater than or equal to AB. That is four, right? This will be four is greater than or equal to AB. So we can say this AB maximum, the maximum value of AB can be equal to four. Right? So a very nice question in this coordinate geometry, like in this chapter, we are using this property also. So normally J also asks question based on this AMGM inequality. So it's a very important concept. You should have practice on these type of questions, right? So what can we say the maximum value of AB is four, right? So this is our question number nine. We are done with that, right? Just give me a minute. Just give me a minute. Yes. See the next one. Question number 10, right? If the normal at the point P theta to the ellipse X square upon 14 plus Y square upon 5 equal to 1 intersect it again at point Q to theta, like this theta and 2 theta are the centric angle, right? At the point B and Q. Then cos theta is equal to. So basically our ellipse is X square upon 14, right? X square upon 14 plus Y square upon 5 equal to 1. This is our ellipse, okay? And the point B is theta. We can write it as A cos theta, okay? A cos theta comma B sin theta. Now, what is A and B for this ellipse? A for this ellipse is a root 14, right? Because A square is equal to 14. And B is equal to B is equal to 5. So B will be a root. Is it okay? So I will write the equation of normal at P, okay? I will write the equation of normal at P. So this is basically A X upon cos theta. A X upon cos theta minus B Y upon sin theta. That is equal to A square minus P square, right? Now in place of A, we can write a root 14. So this will be the root 14 X, okay? Upon cos theta. Then in place of B, we can write root 5. Then Y upon sin theta. And this is equal to A square minus B square. What is A square minus B square? This is 14 minus 5, right? 14 minus 5 is 9, okay? Now, if you observe the question, the same normal is again intersecting, again intersecting at point Q, okay? So if you see what will be the coordinates of this point Q whose eccentric angle is 2 theta. So it will be, Q will be A cos of 2 theta. Is it okay? This is of theta and writing 2 theta because its eccentric angle is 2 theta. If the eccentric angle of B is theta, it's 2 theta as per the given information. We have not done any out of the box thing. So A cos 2 theta and the Y coordinate will be B sin 2 theta. B sin 2 theta, okay? Now this must satisfy this equation. This Q must satisfy, right? This Q must satisfy the above equation. So in place of X and in place of Y, I will put this thing, the coordinates of our Q point, okay? So this will be root 14. X means what? A cos 2 theta. So A is root 14 and cos of 2 theta, okay? So root 14 X upon cos theta minus root of 5, root of 5, 1, root of 5 into Y. Y is what? B sin 2 theta. So root of 5 sin 2 theta. Is it okay? B sin 2 theta upon sin theta. Is it okay? And that is equal to, that should be equal to 9. So this will become 14 times cos 2 theta, right? Cos 2 theta upon cos theta minus 5 into sin 2 theta upon sin theta, that is equal to 9. So let's take this LCM cos theta into sin theta. Or can we further simplify it? What can we do? Cos 2 theta and sin 2 theta, no? Okay. So this will be 14 into cos 2 theta, cos 2 theta into sin theta. Is it okay? Now sin theta is, so minus 5 sin 2 theta into cos theta and that is equal to 9. Or cos 2 theta in place of cos 2 theta, we can write it as, it's trigonometry part, right? So sin 2 theta, we can replace it by 2 sin theta cos theta. And this cos 2 theta, 2 cos square theta minus 1, we can write it. So instead of writing this, instead of writing this state, what I will do, I will come directly here. This is 14 into cos square theta is 2 cos square theta minus 1, right? 2 cos square theta minus 1 upon cos theta. And this thing will be 5 into 2 sin theta cos theta upon sin theta, okay? And minus 9 equal to 0. So from here, what we can do? This will be 28 times, 28 times cos square theta, right? Cos square theta minus 14, minus 14, then what? Okay. 28 cos square theta minus 14 upon cos theta. And here we have minus 10 cos theta. This sin theta, sin theta, we can cancel it. And minus of 9 equal to 0, okay? So can we take LCM now? Okay. Let me take LCM. So this will be 28 cos square theta minus 14 minus 10 cos square theta, right? Cos square theta minus 9 times cos theta equal to 0. So basically it will become 18 times cos square theta. This and this, then minus 9 times cos theta and minus of 14 equal to 0, right? So this is the final equation what we are getting. So basically this step what I have done here, you can omit this. Okay. I'm striking it off, okay? So instead of that, we have changed this cos 2 theta into cos theta. So we got a quadratic, right? We got a quadratic in cos theta. So let's try to figure it out. What can we do here? So this will be basically let this cos theta thing is equal to, let I'm assuming this cos theta to be equal to x. This cos theta is equal to x. So we can write it as 18 x square minus 9x minus 14. Why is it hanging? I don't know. Okay. Minus of 14 is equal to 0. So 18 into 14, right? 18 into 14 will be 252. If you take prime factor function, it will be 1, 2, 6, 1, 2, 6, then 6, 3, then 3, 2, 1, 3, 7. So this way, yes, if it's minus 12, will it work? 21 minus 12. Yeah. So this will be basically 18 x square and minus 21x 12 plus 12x. Is it okay? Plus 12x minus 14 equal to 0. I don't know. The board is hanging on what I'm not able to write properly. So okay, let me take this. Sorry guys. Yeah. Let me take this 3x common from here. So what I will be left with is 6x minus 7. 6x minus 7 and plus if I take 2 common here, 6x minus 7 here. I think I have come down on this board. That's why this is happening. So we are finally what we get 3x plus 2. Okay. Into 6x minus 7 equal to 0. 6x minus 7 equal to 0. What does it imply? Either x is equal to 7 by 6 or x is equals to minus 2 by 3. Right? X is equals to minus 2 by 3. So what is x? Basically x is cos theta. So cos theta is equal to 7 by 6. Right? So it is greater than 1 and cos theta cannot go above 1. Right? So this value is rejected and this cos theta, we can accept it. Why? Because it is less than, less than minus 1. Not less than minus 1. Greater than minus 1 between minus 1 and so minus 2 by 3, we can accept it. So we got the value of cos theta as minus 2 by 3. So what was there in the question? What was asked there? Yeah. So we had to find the value of cos theta only. So how much it is coming? Cos theta is coming up to a minus 2 by 3. Right? Minus 2 by 3. So this will be our answer. This will be our answer. So question number 10, we are done. Then, okay. This is our question number 11. So hopefully we don't disturb now. Okay. So the line this 5x minus 3y is equal to 8 root 2 is a normal to the ellipse. Okay. So one ellipse is given here. One equation of line is given here and it is saying this line is normal to this ellipse. If theta with a centric angle of the foot of this normal, then theta is equal to. Okay. So let me write the equation of ellipse. This is x square upon 25 plus y square upon 9 is equal to, okay. And let me assume a point P, A cos theta comma B sin theta. Is it okay? This is how we are assuming a parametric point on the ellipse. So what will be the normal equation at P? Normal at P. We can write it as A x upon cos theta, A x upon cos theta minus B y upon sin theta is equal to A square minus B square. Now what is A here? A is 5. So 5x upon cos theta minus what is B? B is 3 minus 3y upon sin theta is equal to A square minus B square means 25 minus 9. That is 16. Now this is the equation of normal at P. And as for given condition, like as in the given question, this line, this line 5x minus 3y is equal to 8 root 2 is normal. So both these lines represent the same line. Both this equation represent same line, represents same line. So we have done questions on the similar logic in this exercise earlier also. So what I will do, I will compare that, I will make their coefficients equal. So this 5 upon cos theta, okay, whole divided upon 5. This must be equal to what? Minus 3 upon sin theta minus 3 upon sin theta whole by minus 3 and then this 16 upon 8 root 2. So what we got from here? We have 1 upon cos theta is equal to 1 upon sin theta. This 5 and this 5 will get cancelled out. This minus 3 minus 3 will get cancelled out. So this will be equal to 2 upon root 2, right? So this will be equal to root 2. So from here we got cos theta is equal to sin theta is equal to taking the reciprocal, reciprocating the terms. This will be 1 by root 2. So easily we can say the theta must be equal to pantherless technique, right? Theta must be equal to 45 degree. So 45 degree means pi by 4. So this is the answer. Now this is question number 11. I think 3, 4 questions more left in this exercise. Okay, anyhow. If the tangent from that point is on the parabola, y square is equal to 4x. Okay. So the question has introduced some concept of parabola also. So anyhow, we have to draw tangent at the point lambda is from a 2 lambda on this parabola. It's same. Okay. The tangent to the parabola is behaving as the normal to the ellipse. Okay. So we will draw, we will write the equation of tangent for the parabola and we will make that tangent as the normal to the given ellipse. And then I think we will have some clues for finding this value is lambda. Okay. So the parabola is y square is equal to 4x. Okay. Y square is equal to 4x. It is always standard parabola. And the given point is P whose coordinates are lambda is square comma 2 lambda. Now what I have to do, I have to write the equation of tangent at P. Tangent equation of tangent at P. So what we do, we do yy1 minus this 2 times x plus x1 is equal to 0. This will be the equation of tangent t equal to 0. Right. And now put the value of x1 and pi1 here. So y into y1 is 2 lambda and minus 2 and what you say x plus x1. X1 is what? Lambda is plus lambda is equal to 0. Right. And this tangent is same as the normal drawn at a point this. Okay. So further we can simplify it. What can we write it as 2 lambda y. Is it okay? Minus 2x minus 2x and minus 2 lambda is correct. The same equation. I have just opened the bracket. So 2 lambda y minus 2x minus 2 lambda equal to minus 2 lambda is equal to 0. So this is the tangent at P at point P on the parabola. Now what I will do, this equation of let me write the equation of lips. So this is our ellipse x square upon 5. Right. x square upon 5 plus y square upon 4. Is it okay? x square upon 5 plus y square upon 4 equal to 1. Right. And the point P. Okay. P I have taken it there. So I'm taking this point as Q whose coordinates are root 5 cos theta comma 2 sin theta. Is it okay? So I will write the normal equation of normal at P. This is what is given normal drawn at point. Okay. So I will write the equation of normal at Q. So what will be the equation of normal? So this will be basically a x. Right. So a. So a x upon cos theta. A x upon cos theta minus P y upon sin theta is equal to a square minus B square. So further we can simplify it as what is a is root 5 fight. So this will be a root 5 root 5 root 5 x upon cos theta. Then what is B? B is 2. B is 2. So 2 y upon sin theta and is equal to what is a square minus B square a square minus B square will be 5 minus 4. That is 1. Right. So let me call this as equation 1 and this as equation 2. Now as per given condition this equation 1 and equation 2 are same line. 1 and 2 represent same line. Right. Represent same line. This is what is given in the question. So we can say that we can say that this thing. So minus 2 coefficient of x here minus 2 upon what is coefficient of x here a root 5 root 5 into cos theta. Okay. This must be equal to coefficient of y here to lambda to lambda upon what is the coefficient of y here minus 2 minus 2 upon sin theta and this must be equal to the constant term. So 2 lambda square. So I'm writing it as 2 lambda square upon 1. Is it okay? Right. So what can we do for the what can we we have to find the value of this lambda and theta. Right. We have to find the value of lambda and theta lambda and theta. So how can we do how can we do if the tangent to the parabola is same as the normal. Okay. So I have written the equation of tangent and written the equation of normal. Okay. So let's let me proceed like I think we will get something. So this will be minus 2 cos theta. Okay. 2 cos theta upon a root of 5 is equal to this 2 lambda square upon 1. And what else we are getting? What else we are getting? Okay. This normal this if you see this equation of normal must satisfy this point. So root 5 cos theta comma 2 sin theta. So x and y we can place here. So let me see whether we are getting any foot full thing from there. So this point q this point q must satisfy it. Right. So this will be the root of 5 into x is what root 5 cos theta upon upon cos theta and minus 2 y y is what 2 sin theta 2 sin theta upon sin theta. So 5 minus 4 equal to 1. Yeah. This will be there. So no, I think we will have something from this equation. I'm just raising it. Yeah. This one and two if represent the same line, their coefficients must be equal to must be equal. So I'm equating that. So this thing coefficient of x is minus 2 upon root 5 cos theta. Okay. This one thing I'm getting here and what else I'm getting 2 alpha sorry 2 lambda sin theta upon minus 2 is equal to 2 lambda square upon 1. So this 2 lambda get cancelled out. Okay. And we get the value of lambda as what we got the value of lambda as sin theta upon p and we have something extra this lambda minus no minus sign will come. So minus sin theta upon 2. Is it okay? So minus 2 lambda equal to sin theta. So this is what we are getting. No, we have to find the value of lambda and theta. Right. So we got a relation between lambda and theta. Right. No. We got the relation between lambda and theta. No. We what can we say for this like lambda. Can we say anything? This 2 square this must satisfy this point must satisfy. So here if you say this will be 4 lambda square 4 lambda square that will be equal to 4 lambda square. So nothing can be derived from there. Okay. So these two relations we got these two relations we got the sin theta. This sin theta is equal to minus 2 lambda. Okay. Sin theta is equals to minus 2 lambda. And from here what we can say we can say cos theta is equals to this 2 and 2. We can cancel it out. So we can say from here cos theta is equals to root of 5 lambda square root 5 lambda square is it okay with negative sign with negative sign cos theta is equal to root 5 lambda square with negative sign. So I think we have to utilize these two things for you to find the value of lambda. Now this sin square theta this means what 4 lambda square this 4 lambda square and plus cos square theta that is 5 lambda power 4 this is equal to 1. Right. So later lambda square is equal to x assuming this what we can say this 5 x square 5x square plus 4x minus 1 equal to 0. Right. So 5 minus 1. Right. So 5x square plus 5x minus x minus 1 equal to 0. So taking 5x common x plus 1 minus 1 x plus 1 equal to 0. So what we can say from here either lambda square is equal to minus 1 lambda square is equal to minus 1 which is not possible or this thing 5 lambda right lambda 5 lambda 5x plus 1 equal to 1 means 5x equal to 1 or x is equal to 1 by 5 for lambda square is equal to 1 by 5. Right. So from here we get the value of lambda as plus minus 1 by root 5. Is it okay? We got the value of lambda as plus minus 1 by root 5. So I think this is clear to everyone. So once we got I'm writing it here clearly. Okay. So we got the value of lambda as plus minus 1 by root 5. So once the value of lambda is known to us, we can put here. So we will have cos theta is equals to minus root 5 minus root 5 into lambda square. What is lambda square? So lambda is where is 1 by 5 right lambda is where is 1 by 5 or we can say cos theta is equals to minus 1 by root 5. Is it okay? Or theta we can say as cos inverse minus of 1 by root 5. So this will be the value of theta and this will be the value of lambda. It is a bit lengthy. So anyhow we invested one to a minute here in this calculation part. So this is done. This is the value of our lambda and theta. Okay. I'm coming to the next one. This is our question number 13. If the normal at any point B of the ellipse meets the measure and minor in G and H. Okay. And C is the center of ellipse. So basically this type of questions we return in this exercise. I've already done but okay. We will do it once again. So this is our ellipse. Okay. This is the excess of the ellipse and a normal at the right. We are drawing a normal at P meets the major and minor axis. Okay. So let me say this is our normal. Okay. And it is meeting the major axis and G. Let me write this point at G. Okay. And it is meeting the major axis at G and minor at H. Okay. Now question where this H was given as a small g that ratio we had to find. This is our point P. So similar type of question. This is also, this is also same, same logic based question. And C is the center. Okay. So definitely this is the center of the circle will be 0, 0. So center of circle will be origin. Right. And let me take this point as P as A cos theta comma B sin theta in parametric point. So I will write the equation of normal at P. Right. Equation of normal at P. That will be A X upon cos theta. Okay. Minus B by upon sin theta is equal to A square minus. Is it okay? Is it okay? So for having the coordinates of G or G, we know for having the coordinate of G, I have to put Y equal to 0. Right. For G, I have to put Y equal to 0 in the, in equation one. So what I will get, I will get the value of X and that will be equal to X will be equal to A square minus B square. Right. Into cos theta, into cos theta upon A. Right. And similarly for having the coordinate of this H, I will put X equal to 0. And from there, from there, I will have Y coordinate of H as Y coordinate of H as X will be equal to 0. So this will be B square minus A square. Right. B square minus A square into sin theta and whole divided upon by B. Now, what is asking the question? This A square into CG square. A square into this distance CG square and plus B square into CH squared. Okay. So what is CG? This is nothing but this thing only the X coordinate of G. Right. And what is CH? CH is nothing but the Y coordinate of H. So let me write it as A square into, this is basically equal to CG. We can write here also and this will be equal to basically CH. So CG square. Right. A square minus B square whole square cos square theta upon A square. This will be our CG square. Right. And plus B square into CH square. That will be B square minus A square whole square sin square theta upon this thing B square. Is it okay? Now this A square A square will get cancelled out. This B square B square will get cancelled out and we will be having A square minus B square whole square. Okay. We can take common from both these terms and we will be having cos square theta. This minus A square, you can take minus common from here. So this will become minus of A square minus B square whole square. Now again, squaring will have no effect at all. So I'm taking this thing common and plus sin square theta. Right. So this will be equal to A square minus B square whole square. Right. So this is what we were required to prove. So you can say in stroke. Okay. In stroke. So this is our question number 20. Now question number 14. If the normal at the point P theta to the ellipse this intersecting again at point Q to theta. Oh, same question we have done like so that cost it equal to minus. Okay. Okay. This question is repeated. If you see, we have done same question. So I think I'm not wrong. Yeah. This is the question. See. Point P is theta. Then it is intersecting again. Okay. Let me see the equation of ellipse. The X square upon 14 plus Y square upon five. The square upon 14 Y square upon five. Yeah. So it's, it's the same one. I'm not going to write it again. Okay. So that was question number 14. Okay. This is the last one. This is the last question I think. Okay. Let's finish it. The tangent and normal at a point P of an ellipse this cuts it major axis in Q and are respectively. Okay. If Q R is equal to a group that is centric ankle of the point B is given. Okay. So basically we have to take a point P and we have to write the equation of normal and tangent at the same point. Okay. So just making a rough sketch for that. So suppose this is our, this is our point P. Suppose I'm taking this as point P. Okay. Major axis is given. So I think I have to draw the major axis also. And I have to draw normal at this point and I have to draw tangent at this point. Right. So I think this will be enough. So this is our point P, right? This is our point of P. And this is the normal at point P. Right. And this is the tangent at point P. Right. So the tangent and normal at point P of the ellipse this cuts its major axis in point Q. So the tangent is cutting the major axis at point Q. So basically this is our point Q. Okay. And normal is cutting it as point R. So this is our point R. Now if Q R, the length of Q R is given to be a. Okay. So this Q R thing, this length of Q R is given to be a. So we have to prove that a centric angle of point P. Okay. A centric angle means theta. A centric angle of point P is given by. So what is a centric angle? Angle is theta only. With theta, we used to call it as theta. So let me assume this point P as a cos theta. A cos theta comma B sin theta. Right. Now what I will do, I will write the equation of normal and tangent at this point. So first let me write the equation of equation of normal. Okay. Equation of normal at P. So what will be the equation? A x upon cos theta. Right. Minus B y upon sin theta is equal to A square minus B square. Is it okay? Now this normal is cutting the major axis at point R. So okay. So if you see, this is the major axis. No. So what is major axis? Y equal to zero. Y equal to zero is the major axis for this standard ellipse. Okay. So I will put Y equal to zero and I will have the point R, the coordinates of point R as what will be x? A square minus B square, A square minus B square into cos theta, right? Into cos theta upon A comma zero. This will be the coordinate of R, no doubt. And now I will write tangent, equation of tangent. Okay. At the same point B. So that we can do by x x upon A square, right? This is how we used to write notes. X x upon A square plus Y y upon B square is equal to one. Okay. So, so basically now how can we write this x one means what this point is a cos theta. And so x into a cos theta. Okay. X into a cos theta upon A square plus Y into what is why one B sin theta B sin theta upon B square is equal to one. So this A, this B will get cancelled out and we will have x cos theta upon A plus Y sin theta, Y sin theta upon B is equal to one. So this is the equation of tangent, right? Now the tangent is meeting the major axis at point Q, right? The tangent is meeting at point Q respectively and normal data. So now for Q, for Q, again we have to put Y equal to zero, right? So our coordinates of Q will be, our coordinates of Q will be what? Coordinates of Q will be X, no? So Y will be zero. So this thing will be zero. Our coordinates of Q will be A upon, A upon cos theta, is it okay? A upon cos theta and Y coordinate will be zero, right? Now this QR distance is given as A. So what will be QR basically? What will be QR? QR will be this minus this whole square, right? This A upon cos theta. Okay, A upon cos theta, X1 minus X2. What is X2? This thing, X coordinate of R, A square minus B square cos theta. Okay, A square minus B square cos theta upon A. So X1 minus X2, this whole square, okay? Under root, so I'm writing this as QR square. Instead of QR, I'm writing it as QR square. Now what is QR? QR is basically given as A square. So this is equal to A square. And if you take A cos theta, right? A cos theta is the same here. So what we will have? A square, right? A square minus cos theta thing will be there. So minus A square cos square theta. A square cos square theta. And plus B square plus B square cos square theta. Is it okay? Cos square theta. Okay, this thing will not come up. Why I'm writing this whole square? No whole square. Why whole square? This QR will be under root, no? QR will be under root. I am squaring QR. Means I'm squaring then that under root will be gone off. Why to bring this square here? Right, no? So QR is A. So QR square will be A square, right? And here we will have A cos theta taken LCM. So A square minus A square cos square theta plus B square cos square. I think it's fine only, right? So what can we do for that? This will be A cube. A cube, what? A cube cos theta will be equal to A square. Same thing, A square cos square theta. And plus B square cos square theta. Is it okay? Now if I divide whole thing by this A square, I will get, I will have this A cos theta. A cos theta is equal to 1 minus cos square theta plus B square upon A square cos square theta, right? Or what we can do? And whether it is going in the right direction or what? I have to find the eccentric angle. Okay, okay. So let me take this thing here. This cos square theta part, I'm taking here in the left-hand side and I will take cos square theta common. So I will have 1 minus B square upon A square, right? And here in the right-hand side, I will have 1 minus 1 minus A cos theta. Is it okay? A cos theta. Now this cos square theta. And what can we say this 1 minus B square upon A square is nothing but, is nothing but A square, is it okay? Is equal to 1 minus A cos theta. 1 minus cos phi A cos phi A cos theta. I'm taking as theta, the question is given in the file. So E square cos square theta minus 1 plus A cos theta equal to 0, right? So I think this one was the last question. Okay, and okay, here is our thank you slide. So yeah, thank you everyone. I hope you have listened all the questions like carefully and you have practiced it also. So this video series will be only helpful once, if you practice the four questions first. If you are not able to solve or if you are struggling in view of the questions, you can directly come to that question. You can see the solution to that. So yeah, happy New Year to all of you. And keep rocking, keep studying. Okay, so take care, take care, goodbye.