 Now, time to take up some problems based on this, first question is, find the point of intersection of tangents drawn at parametric points, At1 square comma 281 and At2 square comma 282. Alright, so let's discuss this, very simple, it's just solving of simultaneous equations. So equation of a tangent drawn at this point is T1y is equal to x plus At1 square, correct? And equation of tangent drawn to this point is T2y is equal to x plus At2 square, correct? Now, from 1 and 2, I have to get x and y, so subtract it. So T1 minus T2y is equal to At1 square minus T2 square, okay? So A becomes, y becomes At1 plus T2, correct? And put it in any one of them or do one thing, multiply with T1, T2 and subtract again. So multiply this with T2, multiply this with T1 and subtract. So you get 0 is equal to x T2 minus T1 plus At1 square T2 minus At1 T2 square, which means x times T1 minus T2 is At1 T2 T1 minus T2, T1 T2 cancels off, why I can cancel T1 T2 because I know I can cancel T1 minus T2 because I know T1 minus T2 is not 0 because T1 and T2 are not the same points, okay? That means the point of intersection of these two tangents, point of intersection is going to be At1 T2 comma 2A, sorry, comma At1 plus T2, exactly, focal distance of a point coordinates of ends of a focal cord, okay? Now let's say we have again a standard case of a parabola y square is equal to 4ax and I give you a point, I give you a point x1 comma y1 on this or let's say I give you a parametric point on this, let's say At square comma 2At or At1 square comma 2At1, okay? Now I ask you, what is the distance of this point P from this focus? Focus is A comma 0, what will your answer be? So if you are trying to solve this by finding the distance between these two points, let me tell you there is a shorter way to do that. The distance sp is equal to same as the distance Pm, remember the basic Locust definition of a parabola, so sp is equal to Pm, so sp which is nothing but your focal distance could also easily found out by finding Pm, so Pm will be nothing but this is your A and this is your At square, right? So Pm will be A plus At square, right? Okay, so no question with respect to this. Now I have another question, let's say I draw a focal chord, okay? Focal chord is connecting At1 square comma 2At1, let's say point P to a point Q here which is At2 square comma 2At2, okay? And of course it is passing through A comma 0, that's the focus, okay? Prove that T1 into T2 is minus 1, prove that T1 into T2 is minus 1, so just remember this, this result is very very helpful, I will be using it in some other question as well. Distance of a focus from any point or distance of any point from the focus is A plus At square, so please prove that T1 into T2 will be minus of 1, alright? So this can be very easily proved, if you see this is the same line, I can say slope of sp is equal to slope of sQ, right? What is slope of sp? Slope of sp will be 2At1 minus 0, that is y2 minus y1 by x2 minus x1, that's going to be 2T1 by T1 square minus 1, correct? What is the slope of sQ? In a similar way I can say 2T2 by T2 square minus 1, correct? Now since both slopes are equal because they belong to the same line which is the focal chord, I can equate 2T1 by T1 square minus 1 with 2T2 by T2 square minus 1, so 2 and 2 gets cancelled, cross multiply, so T1 times T2 square minus 1 is equal to T2 times T1 square minus 1, expand it, this is what you get, yes? Now group the terms together, bring it this side, bring it to other side, so it becomes T1 T2 square minus T1 square T2 is equal to T1 minus T2, right? Get T1 T2 common, you get T2 minus T1 is equal to T1 minus T2, cancel out, cancel out this with this leaving you with a minus 1, so that's a very, very important result guys which tells you the relationship between the parameters at the extremities of the focal chord. So what it means is that, if you know one end of the focal chord, let's say, this is my parabola, if I know one end of the focal chord, let's say I know this end is AT square comma 2AT, then the other end I can easily predict as A by T1, sorry, A by T square comma minus 2A by T, correct? Because of this relation, the relation is T2 would be negative reciprocal of T1. So here if it is T, here it will be minus 1 by T, so it is as we were saying A times minus 1 by T square comma 2A minus of 1 by T. Is that clear? This is a very important relationship. Now based on this, I will do a question with you, the tangents drawn at the extremities of a focal chord will meet on the directrix at right angles. Let me explain you the question so that you can try it out. Question says, if you have a focal chord, let's say this is the focal chord, okay, passing through A comma 0 and you are drawing tangents at the extremities of the focal chord, prove that it meets on the directrix, prove that they meet on the directrix X equal to minus A at right angles, at right angles. Any idea guys? How to do this? Again, let's say this point is A T1 square comma 2A T1, okay? This point is A T2 square comma 2A T2, correct? Now we know that the point of intersection of tangents drawn at any two parametric points A T1 square comma 2A T1 and A T2 square comma 2A T2 meets at A T1 T2 comma A T1 plus T2. That's why it's very important to remember this result, okay? Now since they are extremities, since P and Q are extremities of a focal chord, of a focal chord, it implies T1 into T2 is minus of 1, right? That means this point will become A into minus 1 comma something, right? Which is minus A comma something, correct? That means the X coordinate of the point of intersection of these two tangents will always be minus A, which implies the tangents will meet on the directrix. This implies the tangent will meet on the directrix for sure, okay? So the first part of the question is done where we had to prove that the focal chord, the tangents drawn at the focal chord will meet on the directrix. Now the second point that we have to prove is that they are at right angles. How do we prove they are at right angles, okay? That means I have to see the product of the slope of these two tangents, right? So let's say I call this as tangent T1 and I call this as tangent T2, okay? Now tangent T1, you know the equation is going to be T1Y is equal to X plus A T1 square, correct? So slope, if I ask you the slope of this, you would say slope of T1 is going to be 1 by T1, right? In a similar way, in a similar way, I can say slope of T2 will be 1 by T2, correct? Now let us multiply these two slopes. Let's multiply T1 and T2 slopes. That is 1 by T1, T2. And we know if it is at the focal, if T1 and T2 are the extremities or T1 and T2 are the parameters at the extremities of the focal chord, T1 into T2 will actually become minus 1. So it becomes 1 by minus 1, which is minus 1 itself. And that's how we can prove that the tangents are at right angles to each other. Tangents are at right angles to each other. So guys, just extra information that I would like to give you here is that there is something called the concept of director circle. There is something called the concept of director circle, which again I'm going to deal with you when I'm going to finish off the normal school syllabus for you with respect to each of the conics. So I'll be telling you something extra about all the conics. So there is one concept which is called the concept of director circle. Now I would not spend too much time on it. I'll just like to quickly define it. Director circle is defined as locus of such points. From where tangents drawn to any conic are at 90 degrees to each other. In case of a circle also the same concept is there. In case of a parabola is also there. In case of ellipse also you'll find this concept. In case of hyperbola also you'll find this concept of director circle. So meanwhile just because I discussed this problem I would like you to know that director circle in case of a parabola the director circle is actually your directrix. You must be wondering how come a line is a circle. Always remember line is a circle of infinite radius. Okay so guys with this we'll close this chapter of parabola and actually already covered up the school part of it. Next class after your UTS probably we'll start with ellipse. I think one class that will also take and one class for hyperbola and once we have completed sequence series and progressions and statistics. An introduction to 3D geometry will come back again to this chapter okay. This chapter and theory of equations which is actually a quadratic equation also will be involved in that. That will be doing it in details okay. So guys with this we come to the end of today's session. Thank you so much for coming online okay over and out from Centrum Academy.