 Next question, find the equation of the hyperbola whose 4 psi are at minus 4,4 and 6,4 and eccentricity is 2. Nidish, what's your response on this? Teyugia? Teyugia? Snigda? So first draw the diagram for this, everything will be very evident from there itself. So your 4 psi is at 6,4 and minus 4,4. Right? So let's say one is 6,4 and the other is at minus 4,4. Okay, so yes, this is the nature of your hyperbola. Okay, remember the center of this hyperbola will always be at the midpoints of, always be on the midpoint of the line connecting the 2,4 side. So this will be your center and center will be the midpoint which is going to be 1,4. Center is going to be 1,4. Okay, so all of a sudden I have seen the attendance the number of people watching has grown up. So it's important to call people's name few times. Right, now what, how will I get my A and B? Okay, very simple, the distance between the 2,4 psi is 2Ae. So it is basically 10 in this case, right? So the distance between let's say S1 and S2 is 2Ae. Okay? So e is given to us as 2, so A becomes 5 upon 2. Right? Now it's a case where the eccentricity is 1 plus b square by a square under root. So e square is 4, b square is unknown and a square is 4 by 25. So this implies b square is going to be 75 upon 4. Now we can always write down the equation of the hyperbola once we know a square, b square and the center of the hyperbola. So your final equation will be x minus 1 the whole square by 25 upon 4 minus y minus 4 whole square upon 75 by 4 equal to 1. Okay, now I am leaving it at this stage because I don't want to waste time simplifying it. My idea was to convey the concept. So how to deal with the equations when some critical information about the equation is provided to you. Next concept that we are going to talk about is the concept of focal distances, focal distance of a point from the 2, 4 side. So let's say I take a standard case of a hyperbola. So let me pick up a point, let me pick up a point over here. So what are the distance of this point from this focus and what are the distance of this point from this focus is what I am interested in. So let's say this is p and let's say this point is S1, this point is S2. Okay, so assuming this point to be x1, y1, let us find out S1p and S2p that's called the focal distances. These are called the focal distances. Now S1p can be found out very easily because let's say I drop a perpendicular over here and I call this as m1 and I call this as m2. S1p is e times pm1 which is going to e times x minus x1 minus a by e. That is going to be ex1 minus a. This will be e times pm2 which is going to be e times x plus a by e which is ex1 plus a. Okay, and when you take the difference of these two mod of S1p minus S2p you are going to get 2a. This is a very important piece of information for us which says that you can also define hyperbola as the locus of a point, locus of a point which moves in such a way that the difference of its distances from two fixed point, two fixed points is a constant and this constant is lesser than the distance between the two fixed points, between the two fixed points and not zero. Is that fine? So please remember this, this definition is very important in solving questions which come related to complex numbers also sometimes. Next is the parametric equation of a hyperbola, parametric equation of a hyperbola, x square by a square minus y square by b square equal to 1. Okay, now normally we represent the parametric equation of a hyperbola to be x equal to ac theta and y is equal to b tan theta, right? Where theta is a parameter which is called the eccentric angle also, theta is called the eccentric angle, it actually behaves as a parameter. Okay, now let me just show you what is this eccentric angle actually. So let me hide this. Okay, so as you can see on your screen over here, I have drawn for you a circle whose diameter is exactly the distance between the vertices. Okay, by the way, this circle is actually called the auxiliary circle. This circle is actually called the auxiliary circle, right? All of you please pay attention over here in this auxiliary circle what I'll do, I'll make an angle, I'll make an angle theta from the origin like this. So let's say this is angle theta. Okay, and from this point I'm going to sketch a tangent. Okay, so what I did, I took angle theta and I sketch a tangent at this point and this tangent I'm going to take the point of intersection over here and scale it up till it hits the hyperbola. So what I did first, I made a 90 degree tangent over here. This tangent meets the transverse axis at this point. Let me name this point. Let's say this point is the point P. From this point I took a perpendicular up to cut the hyperbola at, let's say, point Q. Then we say that the coordinates of Q will be AC theta comma B tan theta. So this is how this angle theta is related to this point Q. Many people wrongly believe that this angle is basically the angle made from here to this point, right? And the thing that this angle is theta. Please know that angle is not theta. By mistake also, please do not take that. Is that fine, guys? Now, if I take the parametric equation of, if I ask you the parametric equation of this x minus alpha square by a square minus y square, sorry, y minus beta whole square by b square equal to 1, then what would be the parametric form? The parametric form would be x equal to alpha plus AC theta and y will be beta plus beta tan theta. So this will become the parametric form. And if you have the conjugate x square by a square minus y square by b square equal to minus 1, then this parametric form would be x equal to a tan theta and y equal to bc theta. Okay? Now, guys, various type of parametric forms are also possible. These are the most commonly used ones. So I'm not saying this is the only way to represent, you know, parametric equation for a hyperbola. Okay? So let us take some questions to, you know, appreciate other forms that can also be possible. For example, let's take this as a question. Prove that a by 2, t plus 1 by t comma b by 2, t minus 1 by t, this point lies on the hyperbola and get the equation of the hyperbola as well. Get the equation of the hyperbola as well. Yeah, so here t is a parameter and you need to actually eliminate t in order to get the equation of the desired hyperbola. Correct? So first, let us look into this aspect. So A x is equal to a by 2, t plus 1 by t and y is equal to b by 2, t minus 1 by t. Okay? So I can say 2x by a is t plus 1 by t and 2y by b is t minus 1 by t. Now when you square both the sides, so you get t square plus 1 by t square plus 2. Here you get 2y by b whole square as t, sorry, this t square will not come. This becomes t square plus 1 by t square minus 2. Now let me call this as 1. Let me call this as 2. Let's subtract 1 minus 2. That gives you 4x square by a square minus 4y square by b square is equal to 4. Drop the factor of 4. So you get the equation to be exactly the standard case of a hyperbola. So even this could be used as a parametric form. So my point is don't be like under the impression that it is always ac theta comma beta and theta. Okay? Next question. If e and e dash be the eccentricities be the eccentricities of a hyperbola and its conjugate and its conjugate, then prove that 1 by e square plus 1 by e dash square is equal to 1. So please type done if you're done with this. It's an easy problem. Probably you would have done this in school as well. Alright, so only Lalitha could do it. Simple. So for the normal ellipse, we have e square as 1 plus b square by a square which is actually a square plus b square by a square. And for the conjugate, this will be 1 plus a square by b square which is actually 1 plus a square by b square which is actually again a square plus b square by b square. So when you reciprocate these two and add it, you get a square by a square plus b square plus b square by a square plus b square. That is clearly your a square plus b square by a square plus b square. That's going to be 1. So next concept that we're going to talk about is the concept of position of a point with respect to a hyperbola. Position of a point with respect to a hyperbola. Can you see the screen? Alright, so let's again make a hyperbola. Now a point can be at three locations with respect to a hyperbola. It can be at a position a, it can be at a position b or it could be at a position c. Okay, so let's say the point a here is, the point here is x1, y1. Okay, let's say x1, y1 is the point and I want to find out whether it's located somewhere within the arms of the hyperbola, that is at position number a or whether it's located on the hyperbola or whether it's located outside the arms of the hyperbola. Now guys, in case of a hyperbola, the situation is slightly different. If the point is at a, then if you substitute a in place of x and y for this hyperbola, you realize that x1 square by a square minus y1 square by b square is actually greater than 1. That is x square by a square minus y square by b square minus 1 will be positive. If it is a situation number b, then you would realize x1 square by a square minus y1 square by b square minus 1 will be equal to 0. And if it is at position number c, x square by a square minus y square by b square minus 1 would be less than 0. Again, this expression is called s. x square by a square minus y square by b square minus 1, this is called s. And when you put the point in place of x and y, this is called s1. So this is like saying s1 is greater than 0. This is like saying s1 is equal to 0 and this is like saying s1 is less than 0. So it is inside if it is greater than 0, on if it is equal to 0 and outside if it is less than 0.