 Now, based on this, we can take up some problems on the equation of a norm. Let's try with this question. Prove that the line Lx plus My minus n equal to 0 will be normal to, will be normal to the standard form of the hyperbola which is x square by a square minus y square by b square equal to 1 if, if a square by L square minus a square by L square minus b square by m square is equal to a square plus b square whole square by n square. So, I would request those of you who are sitting with a pen and paper, please start working this out along with me. So, again, I will first start with the equation of the normal in the parametric form. So, we know that if I am drawing the equation of a normal at any general point, a seek phi comma b tan phi that would look like ax cos phi plus by cot phi is equal to a square plus b square. Which means I can write this equation to be this. Now, again, it's a, it's a matter of comparison. So, what I'm going to do, I'm going to compare this with Lx plus My minus n equal to 0. So, it's all about comparing these two equations because they both are representing the same equation. So, let us compare. So, when we compare, we get a cos phi by L that is the coefficient of x from in both the equations should be equal to b cot phi by m should be equal to a square plus b square by n, a square plus b square by m. Now, from this equation, I can always get my seek phi and tan phi. So, seek phi from here would be nothing but n a by l a square plus b square. And similarly, so I compare these two and I got this equation. Now, I'm comparing these two and I would get this equation that is tan of phi is b n by m a square plus b square. Now, we can use the famous trigonometric identity seek in n square phi minus tan square phi equal to 1 which will give you n square by a square by l square a square plus b square whole square minus b square by n square m square a square plus b square whole square is equal to 1 is equal to 1. So, I can always divide by n square by a square plus b square whole square. So, it will give you a square by l square minus b square by m square is equal to a square plus b square by n square. So, this is the desired condition and hence proved. Now, moving on to the next concept which is the concept of pair of tangents pair of tangents of a hyperbola. Now, guys, these concepts are all repetitive. We have already done similar concepts in parabola ellipse circle as well. So, I'm not going to repeat this same concept over and over again. So, just to quickly recall the equation of a pair of tangents to a hyperbola. Let's say I write down the equation of a hyperbola to be our general case of a hyperbola like this. Then we know that the equation of the pair of tangents would be t square is equal to s s 1 t square is equal to s s 1. So, just replacing our t with what we have learned this is the expression for t square is equal to s this is your s and this is your s 1 and this is your s 1. So, t square is equal to s s 1 becomes the equation of the pair of tangents. Similarly, chord of contact chord of contact. So, chord of contact is a chord connecting the points of contact of the tangents drawn from an external point on to the hyperbola. So, just quickly sketching the chord of contact over here. So, let's say these are your arms of the hyperbola and you have drawn two tangents. You have drawn two tangents from a point x 1 y 1. Then this line connecting let's say I call this as Q and R. Q and R will be called your chord of contact. It would be called your chord of contact and chord of contact equation is known to us as t equal to 0. A similar concept linked to this is the equation of the chord bisected at a given point equation of a chord of a chord bisected at a given point. So, if you know the midpoint of any chord let's say x 1 y 1 then the equation of the chord is given by t equal to s 1, t is equal to s 1. Where the general meaning of t s s 1 remains the same as what we had discussed earlier. So, we will quickly take a question on the concepts which we have discussed in this particular slide. So, let's begin with a question. Prove that the locus of the midpoints, the locus of the midpoints of the chords of the hyperbola x square by a square minus y square by b square equal to 1 with subtend 90 degrees at the origin with subtend 90 degree at the origin is x square by a square minus y square by b square whole square times 1 minus a square minus 1 minus 1 by b square is equal to x square a to the power 4 plus y square b to the power 4. Now the moment it comes to a line a pair of lines subtending a certain angle at the origin we actually think of the concept of homogenization. So, the hint for this problem is we can solve this problem simply by using the concept of homogenization. I have already discussed this with you in the past but while solving this problem I will discuss with you again. So, what do you have to do is in this case we have to homogenize the equation of the line which is the chord whose midpoint is let's say h comma k. So, let's say the midpoint is h comma k. So, we know t equal to s 1, t is equal to s 1. So, this is your t equal to s 1. This is the equation of a chord whose midpoint is h comma k. So, what do you have to do is we have to homogenize these two equations. Homogenize means we have to convert this together as a second degree equation which actually is representative of a pair of straight lines which passes through origin. So, how do we do that? How do we homogenize these two equations? It's very simple process. So, what do we do is we don't disturb the second degree term because they are already into second degree. Instead of this one we actually will write one square. Instead of this one we will actually write a one square and I will show you from where we will replace this one square. So, from here we can see that we can write 1 to be hx by a square minus ky by b square divided by this. So, what I will do is let me call this as 1. Let me call this as 2. We will use the second equation in the first equation. We will use the second equation in the first equation and when we do that we end up getting x square by a square minus y square by b square is equal to hx by a square minus ky by b square whole divided by h square by a square minus k square by b square the square of this entire thing. The square of this entire thing. Now, just simplifying this guys we can write it as h square by a square minus k square by b square whole square times x square by a square minus y square by b square is equal to hx by a square minus ky by b square whole square. Right? Now, what have we learned? We have learned that in any second degree equation we know that in any second degree equation which actually represents a pair of straight lines. Okay? If the lines are perpendicular to each other then a plus b is equal to 0. So, this implies the lines are perpendicular to each other. Okay? So, basically it's passing through the origin also and perpendicular to each other which is precisely what I want to address in this problem as well. Okay? So, a plus b equal to 0 means coefficients of this is coefficient of x square and this is coefficient of y square. So, coefficient of x square and coefficient of y square should add up to give you 0. So, what is the coefficient of x square? Coefficient of x square, first of all I can get x square from this expression which is actually this term and from this expression will get minus h square by a to the power 4 plus coefficient of y square again I can get it from here. Okay? And from here which is minus of k square by b to the power 4 is equal to 0. Now, I can just simply take h square by a square minus k square by b square whole square common and in the brackets I will get 1 by a square minus 1 by b square and I will take h square by a to the power 4 plus k square by b to the power 4 on the other side. So, we can now generalize this. We can now generalize this x square by a square minus y square by b square whole square 1 by a square minus 1 by b square equal to x square by a to the power 4 plus y square by b to the power 4 which actually gives you the equation of the desired locus. Okay? So, guys again I am repeating this concept. The concept of locus is of lot of importance in J. So, please practice lot of questions on locus because all the concepts which we are discussing in coordinate geometry are somehow linked to the concept of locus. Okay? So, moving on to the concept of pole and polar. Pole and polar. So, what is the concept of pole and polar? Again, very similar to what we discussed in case of circles ellipse and parabola. So, when a point is external or internal or on the hyperbola and you draw all sort of lines passing through that point. Okay? So, let us say I take one of the branches of a hyperbola and there is a point like this and let us say I draw a line through it. So, the ends of these lines let me draw tangents. Let me draw tangents. Okay? Let us say I draw another line like this. So, let me draw tangents from this point as well. Now, all these tangents, the meeting point of these tangents will lie on a straight line and that straight line is called the polar for this pole. The straight line is called the polar for this point which is called the pole. Okay? So, not a new concept. I have already discussed with you and I have also discussed with you that when the point is on the hyperbola, the polar is basically nothing but the tangent drawn at that point. Right? And when it is outside the hyperbola, it is going to be the chord of contact. So, in all the cases, the equation of the, the equation of the polar is going to be t equal to 0. The equation of the polar is going to be t equal to 0. Now, based on this, I would like to just take up a simple problem. Again, a locus-based problem. Find the locus of, find the locus of the poles of normal chords of the hyperbola, x square by a square minus y square by b square equal to 1. So, find the locus of the poles of normal chords. So, what is the meaning of a normal chord? So, chord which is also normal to the hyperbola. So, we have to find what is the locus of the poles of all normal chords to this hyperbola. Again, it is a simple concept which is based on the understanding of locus. So, let us start with the equation. So, let us say the pole b h comma k. The pole b h comma k. So, we know that the equation of the polar, equation of the polar would be t equal to 0. That is h x by a square minus y k by b square equal to 1. And let us say this is also behaving as a normal. Let us say this polar is also a normal at the point, at the point a seek phi comma b tan phi. So, the equation of the normal would be nothing but a x cos phi plus b y cot phi is equal to a square plus b square. Now, these two equations 1 and 2 they are same. So, comparing equation 1 and 2 since they represent the same thing, I can write h by a square divided by a cos phi is equal to minus y by b sorry minus k by b square divided by b cot phi is equal to 1 by a square plus b square. So, from here we can get seek phi as a cube by h times 1 by a square plus b square. So, by comparing these two I can always get my seek phi like this and a similar way by comparing these two I can get tan phi as minus b cube by k 1 by a square plus b square. Now, the next step is very simple. We have to eliminate our phi. That means we have to make use of trigonometric identity which is nothing but seek phi minus tan phi seek square phi minus tan square phi is equal to 1 which ends up giving us a to the past 6 by h square times 1 by a square plus b square whole square minus b to the power 6 by k square a square plus b square whole square equal to 1. Now, we can further simplify this as a to the power 6 by h square minus b to the power 6 by k square is equal to a square plus b square whole square. And now we can generalize this as a to the power 6 by x square minus b to the power 6 by y square equal to a square plus b square whole square. So, this is the required locus for the poles of the normal quads of the hyperbola. Now, moving on to the next concept which is the concept of diameter, which is the concept of diameter of a hyperbola. So, first of all, let us define the diameter. So, the diameter is nothing but the locus of the middle points or midpoints of a system of parallel quads of a system of parallel quads, okay. This is called the diameter of the any conic section. In fact, in circle also the same definition will be true, in parabola also the same definition will be true, in ellipse also the same definition would be true. So, now let us move to the equation of the diameter. We can prove the equation of a diameter as y is equal to b square x by a square m, okay, where m is the slope of the system of quads, slope of the parallel quads, which is bisected by the diameter, slope of the parallel quads, okay, of the parabola, sorry of the hyperbola x square by a square minus y square by b square equal to 1, okay. So, how do we prove this? Again, the proof is very simple, it comes from the concept of the equation of a chord whose midpoint is known. So, what we will do is, since the diameter is the locus of the midpoint of the parallel quads, I will say let the midpoint of the parallel chord be let the midpoint of the parallel quads be h comma k. So, h comma k is a variable point, okay. So, t equal to s 1 would be the equation of the chord whose midpoint is h comma k. So, t equal to h 1 will be t equal to s 1 will be this. Now, I am only claiming that this equation has a slope of m because all these quads are having a slope of m, right. So, what I am going to do is, I am going to write down the slope which is in this case will be h b square by a square k to be equal to m and that is it. We have to just generalize our k with y and h with x. So, it becomes b square x by a square y equal to m which means y is equal to b square x by a square m is the desired equation of the diameter having a slope diameter bisecting all the quads having a slope of m.