 So let's take a quick question on condition of tangency and then we will make some more analysis on it. Question is prove that Lx plus My plus N equal to 0 touches the parabola y square is equal to 4ax if ln is equal to am square if ln is equal to am square. Please prove this. If done, please let me know. Okay, so let's discuss. So again just now we have discussed that c should be equal to a by m. Okay, now for this line what is the slope? First of all you have to write my is equal to minus Lx minus N then y will be equal to minus L by Mx minus N by M correct. So note that this will be acting as your slope. Now don't confuse between this M and the M given in the problem. Okay, so just replace c this is going to play the role of c. C is equal to a by M. Okay, so cross multiply. So minus N into minus L is going to am square which becomes nL equal to am square is the required condition for this line to be a tangent to this parabola. Okay, next question show that show that x cos alpha plus y sin alpha equal to p touches y square is equal to 4ax if p cos alpha plus a sin square alpha is equal to 0 and also show that the point of contact point of contact is a tan square alpha, minus 2 a tan alpha. So all this and let me know if you are done. So I think first part would be easy you would be able to do the first part very easily. Okay, so getting this condition is not difficult. You are going to have to use c is equal to a by M since your parabola is y square is equal to 4ax you can use this condition again over here. Okay. Now what is c in this case? If you write it in the y equal to Mx plus c form if you write this parabola in y equal to Mx plus c form we get something like this correct. So M is minus cot alpha and c is p by sin alpha correct. So c which is p by sin alpha this will convert to c which is p by sin alpha is equal to a by M. A by M by M will be minus of tan alpha and minus of tan alpha could be written as minus a sin alpha cos alpha cross multiply. So it becomes p cos alpha is equal to minus a sin square alpha. So p cos alpha plus a sin square alpha equal to 0. So first condition is proven. Now this is done. The next is point of contact. How do I find point of contact? First let us go back to the case where we had y square is equal to 4ax and the line y equal to Mx plus c was touching it. Okay. Then what is your point of contact in this case? Can somebody tell me the point of contact in terms of a M and c? C is already a by M right? So you can change this to a by M. So in terms of a and M can somebody tell me what would be this point of contact coordinates be? Can you type it on your screen on the chat box? Okay. So it is very simple. You just have to simultaneously solve it. So y square which is Mx plus am whole square is equal to 4ax. Okay. If you expand it you are going to get M square x square. Okay. Plus 2ax minus 4ax plus a square M square. If you simplify it this is what it appears which can further be simplified to Mx minus a by M whole square equal to 0. Right? As you can see this is going to be a perfect square. This is going to be a perfect square. Okay. So here Mx will become a by M which means x would become a by M square. And if x becomes a by M square we can easily find out y as well because y is nothing but Mx plus a by M. So y will become M into a by M square plus a by M that's going to give you 2a by M. Okay. So this point of contact is am square comma 2a by M. Right? Now we have to prove now in this case that it is a tan square alpha minus 2a tan alpha. Now remember your M role is being played by minus cot alpha. Isn't it? Your M, I think we had written it somewhere over here. Yeah, here. M role is being played by minus of cot alpha. So your point which is poc which is a by M square comma minus sorry a by M square comma 2a by M could be written as a tan square alpha minus comma minus 2a tan alpha and hence proved. Exactly. Correct Rima. Okay. Now since you have started talking about the equation of a tangent, so we'll talk about equation of tangents in other forms, in other forms. By the way, y is equal to Mx plus a by M. This is called the slope form of the equation of a tangent. Right? Because in this case you're just given the slope and the parabola and you're asked to get the equation of a tangent. Right? For example, if I ask you, find the equation of a tangent, find the equation of a tangent to y square is equal to let's say 8x with a slope of 5. Then what would be the answer for that? Answer would be y equal to Mx plus a by M. So this would be your answer. Now the second form of the equation of a tangent that we are going to look at that is called the point form. Now what is the point form? Let's say I give you the standard case of a parabola y square is equal to 4ax. Okay? And I give you a point on this. And I give you a point on this. Okay? And I have to draw the equation of a tangent to this parabola. Okay? So what would be the equation of a tangent to this parabola at this point? Can you tell me the expression? At x1, y1, what would be the equation of a tangent to the parabola? Yes, these equations are only for y square is equal to 4ax rm. So as the parabola changes, the equation will undergo a lot of changes. But why we are dealing with this case is because it's the most frequently asked type of parabola. The standard form is most commonly asked. That's very simple in this case. For finding the equation of this tangent, you would need the slope. Right? And you have all learned that slope can be obtained by finding the derivative of the function at x1, y1. Correct? Yes or no? So when you have been given 4ax, let's find dy by dx at x1, y1. Now, when we find dy by dx, it means 2y dy by dx is equal to 4a. Correct? So I differentiated both sides with respect to x. That means dy by dx is 2a by y. Correct? That means dy by dx at x1, y1 would be 2a by y1. Okay? So using the form of the equation of a line in point slope form, I can write it as y minus y1 is equal to slope x minus x1. Which means yy1 minus y1 square is 2ax minus 2ax1. Correct? Which means yy1 is equal to 2ax minus 2ax1 plus y1 square. Now we know from here that y1 square will be 4ax1 because x1, y1 lies on the parabola. Right? As x1 comma y1 lies on the parabola. Okay? So yy1 is equal to 2ax minus 2ax1. And instead of y1 square, I can write it as 4ax1. Which makes the equation of the parabola as 2ax plus 2ax1. Which is yy1 is equal to 2ax plus x1. So please remember this as the slope form or sorry, as the point form of the equation of the tangent. Point form of the equation of the tangent. Clear? Now the third form that we are going to deal with is called the parametric form. The third form that we are going to deal with is called the parametric form. This is one of the most important of all the forms. You will be heavily using it in problem solving in case of comparative exams. Now in parametric form if I ask you, find the equation of a tangent at some point at square comma 2at. Okay? So find the equation of the tangent at atat square comma 2at. Then all I need to do is replace my y1. Replace my y1 over here with 2at and replace my x with at square. Okay? So 2a and 2a gets cancelled. So ty is equal to x plus at square becomes the equation of a tangent in parametric form. Please remember these results. So please remember this. Remember this. Okay? So this is called the equation of a tangent in parametric form.