 Hello friends, greetings from Center for Academy. So we are going to resume back with the Koenig section. This is a session on Hyperbola. And if you have not watched the previous sessions, that is from one to seven, I would request you definitely please watch them. Okay, so that we make more sense on this. So I'm going to resume with the concept of equation of tangents in various forms. So equation of tangents to a hyperbola in various forms. So I had already discussed with you that if you have the standard equation of a hyperbola like this, okay, and I want to draw a tangent to this hyperbola with a slope of m. So let's say the slope is given to me, okay, which is called the slope form. Okay, so let's say I want to write down the equation of a tangent to this hyperbola having a slope of m. Then we have already discussed the condition of tangency, which is c square is equal to a square m square minus v square. Okay. And from this we get the equation of the tangent as y is equal to mx plus minus under root of a square m square minus v square. So basically what I'm doing is I'm just replacing my c from here as plus minus under root of a square m square minus v square. So please be informed that this form of the equation is called the slope form because you are given the equation of a tangent, you have to find out the equation of a tangent when you know the slope of the tangent. And I'd also discussed with you that you can only find out for those slopes where a square m square minus b square is positive, that's greater than equal to 0, which means m square minus b square by a square is greater than equal to 0, which means m could be greater than b by a or m could be less than minus b by a. That means for any value of m which lies in the interval minus b by a to b by a, you cannot find a real tangent to the hyperbola. Okay. Now we are moving on to the next form which I call as the point form, which I call as the point form. Okay. What is this point form? This point form is basically a form where you have been given a point on the hyperbola. So let's say this is your hyperbola and you know a point on the hyperbola, let's say x1, y1 and we're trying to sketch a tangent at this point. Okay. So we have already seen that in such cases, what do we do in the general equation or in any equation of the hyperbola given to us? We replace x square with xx1, y square with yy1 and our constant remains constant. Correct? So my equation would now become xx1 by a square minus yy1 by b square equal to 1. And we also know that we call this expression, we call this expression as expression T. Okay. So in future, if I'll be referring to this expression as T, please be informed that I am referring to this form. Okay. In fact, we can generalize this also, we can generalize it by saying that if you have any equation of a hyperbola of this nature as well, then we can get the equation of the tangent, then the equation of a tangent at a point x1, y1 would be nothing but a xx1 plus h xy1, x1, y plus byy1 plus gx plus x1 plus fy plus y1 plus c equal to 0. Okay. And again, in this case, your T expression would be this. So this would become your expression for T. This would be called your expression of T. This is called the point form of the equation of the tangent. Now moving on to the one of the most important forms which we call as the parametric form, which we call as the parametric form of the equation of a tangent. So parametric form is where you have been given the point in a form of a parameter. So let's say there is a point on this hyperbola which is x equal to ac phi, y is equal to b tan phi. In other words, you have a point like this and you want to sketch a tangent here. So it will be known different concept because you can just apply the same point form and use your x1 as this point and use your y1 as this point. That means you can get the equation of a tangent to be this, which means x eek phi by a minus y tan phi by b equal to 1 would be the equation of a tangent to the hyperbola in the parametric form. Now guys, I want to show you something very important over here. How does this equation become a special case of the equation of a chord connecting any two points theta and phi? Okay. So I would just like to bring your attention to the equation of a chord, right? Connecting to eccentric points, connecting to eccentric points theta and phi, theta and phi. Okay. So we can recall that the equation of the chord connecting theta and phi, which is very easy to prove. It's basically nothing but finding the equation of a line connecting two points to be this. I would request you to prove this as the equation of a chord connecting theta and phi. It's very simple to prove. Now, what I want to show you over here is that when theta tends to phi or phi tends to theta, okay? What happens to the equation of a chord? If you see that when theta tends to phi, it becomes cos 0 degree minus y by b sin of theta or phi, you can call it. Okay. And this also becomes cos of phi plus phi by 2, which is nothing but x by a minus y by b sin phi is equal to cos of phi. Right? Let us divide throughout with cos phi. When you divide throughout with cos phi, it becomes x by a seek phi minus y by b tan phi is equal to 1. Right? Now, you can compare that these two equations actually become the same. So the tangent is nothing but a limiting case of the equation of a chord where the two points which the chord is joining are basically made coincident to each other. Okay? Now, before we move on, let us take few problems based on this concept. So let's start with this simple question. Find the equation of the tangent to the hyperbola. Find the equation of the tangent to the hyperbola. x square minus 4 y square is equal to 36, which is perpendicular to the line x minus y plus 4 equal to 0, which is perpendicular to the line x minus y plus 4 equal to 0. Okay? So indirectly, I have been asked to find out the slope form of the equation of a tangent. So we know that the equation of the hyperbola written in its standard form would actually appear to be this. Okay? So clearly a square is 36, b square is 9 and since you are asked to find out a equation of a tangent perpendicular to this line, let me tell you this line has a slope of 1. Okay? So the required slope, which is m, let's say, that will be negative reciprocal of 1, which is minus of 1. Okay? So your c square is equal to a square m square minus b square. Please note that this has to be only applied to the standard form. Must be only applied to this form of a hyperbola. Please do not apply this condition of tangency if your equation of the hyperbola is not of this form. Please note this very, very carefully. So c square is equal to a square, which is 36, m square, which is minus 1 square. Okay? Minus b square, which is minus 9, which clearly becomes c square is 36 minus 9, which is 27. So c could be written as plus minus 3 root 3. That means the desired equation of the tangent can be written as y equal to minus x plus minus 3 root 3. So this is your desired equation of the tangents. So there are two tangents possible with the same slope. Okay? So I can just write it as tangents over here. So this is your desired result. Okay? So moving on to the next question. If the line y is equal to mx plus under root a square, m square minus b square touches this hyperbola at ac theta comma b tan theta. Okay? Then show that theta is sine inverse b by a m. Then show that theta is sine inverse b by a m. Okay? So this problem becomes very simple. So basically the slope of the tangent drawn at such a point should be equal to m. Okay? So dy by dx should be m drawn at ac theta comma b tan theta. Okay? That's the first and the foremost condition that needs to be satisfied. So dy by dx will be dy by d theta which is b secant square theta by dx by d theta which is ac theta tan theta. And this should be equal to m, which clearly implies that b by a sine theta is equal to m. That means sine of theta is equal to b by a m. Okay? And hence it becomes super simple problem where we can easily say theta is sine inverse b by a m and hence shown. And hence shown. Now moving on to the next concept which is the equation of normals in different forms. So we have already seen equation of tangents. Now we will be talking about equation of normals in different forms. So we will first begin with the point form of the equation of a normal. The point form is a form where you know a point on the hyperbola. And we have to find the equation of a normal at that point. So let's say we have been given a hyperbola like this. Okay? And we have to find the normal at x1 y1 point. Okay? So again this concept is very simple because we have already done the equation of a tangent and normal in application of derivatives. So slope of the normal, we all know it's minus dx by dy calculated at x1 y1. Okay? So let us differentiate both sides with respect to y. So I will get 2x by a square dx by dy minus 2y by b square is equal to 0. So 2 and 2 gets cancelled. So dx by dy becomes a square y by b square x. And since you are finding it at x1 y1, you just need to replace your xn y with x1 and y1. So this will become the slope of the normal which is negative a square y1 by b square x1. Now writing the equation of any line whose slope is known and a point is known to be y minus y1 is equal to slope times x minus x1. You realize that the equation looks like this. Okay? Which on further simplification, which on further simplification, you can convert it to this form a square x by x1 plus b square y by y1 is equal to a square plus b square. Okay? Very, very important result. Please remember this because you would be needing the equation of a normal at so many places. Okay? So you cannot afford to sit and derive this over and over again. Now using this result into the second type of the equation which we call as the parametric form. Okay? So in parametric form instead of x1 and y1, I would be asked to find the normal at a point like ac5 comma btan5. So all you need to do is in this equation replace your x1. So this will become your x1. This will become your y1. So replace your x1 with ac5 and replace your y1 with btan5. Okay? And when you simplify this, it becomes ax cos phi plus by cot phi is equal to a square plus b square. So again, this equation is very important. This is the equation of the normal in the parametric form, equation of the normal in the parametric form. So we can also find out the equation of a normal in slope form. The equation of a normal in slope form that means when you know a slope and we have to form the equation of a normal to the hyperbola with that given slope. Okay? We can always write down the equation to be this y is equal to mx plus minus m times a square plus b square by under root of a square minus m square b square. Okay? Now how does this equation come? This equation comes from the basic equation of condition of normalcy. Okay? So let us try to prove this. How does this equation come? By first proving that the condition of normalcy. So when does a line y equal to mx plus c is a normal to the standard form of the hyperbola. So let us try to first find this condition. Okay? Which obviously is the condition which I am going to state over here. The condition is actually c square is equal to m square a square plus b square whole square by a square minus m square minus b square. So how does this condition actually comes? Now the condition comes from a very, very simple comparison of equations. So let us first write down the equation of that normal in a parametric form. So we have just now seen the equation of a normal in the parametric form to be ax equation of a normal in parametric form to be ax cos phi plus by cot phi is equal to a square plus b square. Right? Now what I am going to do is I am going to first bring it to y equal to mx plus c form. Okay? So I can write this as y into b cot phi is equal to minus ax cos phi plus a square plus b square. So y is equal to minus a cos phi by b cot phi times x plus a square plus b square by b cot phi. Now what I am going to do is I am going to compare the equation of the line y equal to mx plus c with this form. So I am going to compare these two. I am going to compare these two forms. So when I am going to compare these two forms, I would realize that my m becomes minus a by b okay cos phi by cot phi cos phi by cot phi is as good as a by b sin phi. Okay? So this will be your m and c will be a square plus b square by b cot phi b cot phi. Okay? So which clearly means sin of phi could be written as minus m b by a. Right? So this is sin of phi. So what will be cot of phi? So treat it as if you are you have written over here perpendicular by hypotenuse and this has to be base by perpendicular. Okay? So perpendicular by hypotenuse so base will be under root of a square minus m square b square. Right? Minus m b. This is going to be your cot phi. So what I am going to do is I am going to replace this expression over here. Correct? Okay? So first let me square both the sides. So c square is equal to a square plus b square whole square by b square cot square phi. Now b square cot square phi from here I can say it is going to be a square minus m square b square by m square. So let me replace it over here a square minus m square b square by m square. So m square will go up. Okay? And I think we have achieved the result which I wanted to achieve which is this result. Your c square is equal to m square times a square plus b square whole square divided by a square minus m square b square which is actually your this result. Okay? And hence from here we can say that we can write our c to b plus minus m a square plus b square by under root of a square minus m square b square which means I can always write down the equation of a normal as this.