 Again, let us not waste time in any kind of derivation. We will directly go to questions based on this. Question is, prove that the locus of the midpoints of normal chords, locus of the midpoints of normal chords of the ellipse, x square by a square plus y square by b square equal to 1 is the curve, is the curve. So it's a huge equation. Yeah. So please solve this now. Again, a locus question. So take it very, very seriously. It's all about testing your locus finding skills. It's not about the question. It's about testing how you deal with locus. How do you tackle locus questions? No, no, no, normal chord is basically just a normal. That's it. Normal chord is just a normal to the ellipse. That's it. Normally, normal is also sometimes called as normal chord, because anyways, it's going to be a chord. Okay, guys, soon after your second term gets over, please start taking up the online tests. I hope you have all got the logins. Start attempting level two of all the chapters, whichever chapter you have not attempted so far. Start attempting level two of the online test, which was login was given to you. So complete it. Try to write one test. If this message delivered, please acknowledge, write ack. If you have heard this message that you have to keep writing online tests from the login gives is given to you. Always remember, Jee main is a very easy exam because they have to select almost 2.25 lakh people. So getting 200 plus is not difficult in Jee main. Almost 20, 30,000 people get that. So please make sure every test that you are taking, you are crossing this 200, 210 mark invariably. Don't have very, very difficult picture of this exam in your mind. It's a very easy exam. Jee advance is okay. I can stand it's a slightly level above. But Jee main is an exam for the masses, just like neat is the exam for the medical people. So don't get hyped because of this exam. It's easy exam. Just know your basic stuffs properly. Things will be very, very fine. Don't do any silly mistakes. Manage your time well. Alright. So guys have to find the locus of the midpoints of normal chords. So basically this, let's forget this word normal as of now. Let's say if I say the midpoint is h, k, then what would be the equation of the chord be? The equation of the chord will be t equal to s1. So let this be the midpoint of the chord. So the equation of the chord would be t is equal to s1. So this is your t equal to s1. Now this minus one, there is no point writing it because they will cancel from both the sides. That's why we prefer writing it like this only. Now I can clearly say that let's say this chord, which is actually a normal also, let it be a normal at a point a seek phi, sorry a cos phi, comma b sin phi. So let this be normal to this. So I can also write down this equation of the same normal chord as, as a x seek phi minus b y cos seek phi is equal to a square minus b square. Right? Now in principle they represent the same equations because they represent the same lines. So what, what perspective I have taken in this problem is I have taken the perspective of a chord whose midpoint is known, whose midpoint is known and here I have taken the perspective of a normal at a parametric point phi. Correct? And since it is the normal chord, basically both the equations are same. Can you make progress from here? Now try to eliminate phi, you will automatically end up getting your locus. So eliminate your parameter phi. You will automatically end up getting your locus. If done, please type done. Sure, sure. Take your time. Alright, so let's, let's compare the coefficients. So comparing the coefficient of x, I will get h by a square divided by a seek phi will be k by b square divided by minus b cos seek phi, b cos seek phi will be equal to h square by a square plus k square by b square divided by a square minus b square. Okay? So guys, from here I can get my cos phi, cos phi, correct me if I'm wrong, a cube by h times h square by a square plus k square by b square divided by a square minus b square. So by using these two, I can get this. Now by using these two, I can get sin phi as minus b cube by k h square by a square plus k square by b square by a square minus b square, correct? Now the moment I have got sin phi and cos phi, it becomes super simple to eliminate phi. We can just apply your Pythagorean identity. We know cos square phi plus sin square phi is going to be one. So which means, let me for the time being call this term as lambda because I don't want to keep writing it again and again. Okay? So let me call this term as lambda for the time being because they are the same expression. So cos alpha whole square is going to be a to the power 6 h square lambda square. And here we are going to get b to the power 6 by k square lambda square is going to be one. So a to the power 6 h square b to the power 6 k square lambda square lambda square is going to be h square by a square plus k square by b square whole square divided by a square minus b square whole square equal to one, right? Which means a to the power 6 by h square b to the power 6 by k square times h square by a square k square by b square whole square is equal to a square minus b square whole square. Now you can always generalize this and when you generalize this you end up getting x square by a square y square by b square whole square a to the power 6 x square b to the power 6 y square is equal to a square minus b square whole square. Is that clear guys? Clear or not? Please type CLR if it is clear. God level clear. Okay. This should be done guys. This is a very easy question. Last question of the day. Then you can go and study your CBC Maths. Show that the locus of the midpoints, midpoints of quads of quads of an ellipse of an ellipse or quads of the ellipse probably x square by a square plus y square by b square equal to 1 which pass through the quads which pass through a fixed point, pass through a fixed point is another ellipse is another ellipse. So if you draw all possible quads through a fixed point then the locus of the midpoint of all those quads will also be an ellipse and if possible tell me the center of that ellipse. Let's say the fixed point is h comma k. Let's say then find the center of the locus of the midpoint of the quads. That means center of the new ellipse that you get. First of all, we'll write the equation of the chord of contact. Sorry equation of the chord whose midpoint is known to us. So let's say the midpoint is x1, y1. So the equation of the chord whose midpoint is x1, y1 is this. Now this chord is passing through a fixed point. It passes through passing through fixed point. Let's say h comma k. So let me replace your x with h and y with k. Now guys, what else do we require? We require to find the locus of x1, y1. We need the locus of x1, y1. So we read the locus of x1, y1. So generalize it, replace it with x comma y. So when you replace it with x comma y, you get hx by a square, ky by b square is x square by a square, y square by b square. This is the desired equation. Now this equation is actually the equation of an ellipse. How will you show that? How will we show that? Simple. Bring this term to this side. Okay, bring this term to this side. So it becomes 1 by a square x square minus hx, 1 by b square y square minus ky equal to 0, equal to 0. So let's do one thing. I can write it as x minus h by 2 the whole square and subtract h square by 4. Here also y minus k by 2 whole square and subtract k square by 4 equal to 0. That means you get x minus h by 2 whole square by a square, y minus k by 2 whole square by b square is equal to 1 fourth h square by a square plus k square by b square. So which clearly represents the equation of an ellipse. By the way, I asked you a follow-up question. What is going to be the center, center for such an ellipse? What is going to be your answer? Please type it in the chat box. So what is the center for this? Exactly, h by 2 comma k by 2. Okay. So great guys. We still haven't done the chapter because we have concepts like pole and polar left. We have concepts like equation of diameter and conjugate diameters left which we'll be covering in the next class. Meanwhile, you can focus on your semester exams. All the best. Do well and please practice the two papers which have sent on the group. Thank you all for coming online over and out from outside. And yes, best of luck for Friday's paper.