 So let's take another question. The question is the polar of a point with respect to the parabola y square is equal to 4Ax, sorry with respect to the parabola y square is equal to 4Bx touches x square is equal to 4Ay, x square is equal to 4Ay. Then show that the locus of this point, this point whichever is the point mentioned over here, the locus of this point is xy equal to minus 2AB. In other words, it is a rectangular hyperbola. Take locus questions very very seriously because Jay is going to ask you at least one question on locus, definitely. Type done if you are done with this problem so that I can start the discussion for that. It's very simple problem. Just try to understand the pulse of locus questions. So let the point be h comma k. Let the point be h comma k. So what will be the polar for this point with respect to this parabola y square is equal to 4Bx. So again you will say t equal to 0. So t equal to 0. This will be your t equal to 0. And this is supposed to touch x square is equal to 4Ay. That means this line is going to behave as a tangent to x square is equal to 4Ay. Now try to recall what is the condition for this line y equal to mx plus c to be tangent to be tangent to x square is equal to 4Ay. What is the condition? So it is very simple. We had already learned that c is equal to minus a m square. Now let us see here which expression is behaving as your c. So in this if you write it properly it will be 2B by kx plus 2Bh by k. So this is your m. This is your c. So this expression will be m. This expression will be c. So you will have c is equal to minus a m square. So if you want you can cancel out of terms over here minus a 2B by k and 2B by k. So 2B by k will be cancelled. So it will be hk is equal to minus 2ab. Now we can generalize this and we get xy is equal to minus 2ab as the desired locus which is clearly a case of a rectangular. It is a case of a rectangular hyperbola. Next we are going to talk about some very important properties. Some standard properties of the parabola. Some standard properties of the parabola. Please remember them because they are very very useful in problem solving. First property is that you have already done in the class so we do not need to prove it again. The tangents at the extremities of a focal cord intersects at right angles on the directrix. So directrix is nothing but your director circle of the parabola. So this we already know we do not have to prove it again. Second property is the portion of the of a tangent intercepted between the directrix and the curve subtends a right angle at the focus. So let us quickly prove this. It is a very very simple proof we will not take much time. So it is that let us say there is a tangent like this and this is your directrix. So the portion of the tangent which is intercepted between the curve and the directrix this will subtend a 90 degrees over here. This would subtend a 90 degrees over here. It is very simple to prove. You know this point let us say it is a t square comma 2 a t. This point is a comma 0. So if you write a tangent it will be t y is equal to x plus a t square and here you know x coordinate is minus a. So put a minus a in place of x. So y becomes a t minus a by t. So this point is a t minus a by t. Okay. Now let us say slope of this line is m1 slope of this line is m2. So m1 is going to be 2 a t minus 0 by a t square minus a which is nothing but 2 t by t square minus 1. m2 would be nothing but a t minus a by t divided by 2 a. That is nothing but t square minus 1 by 2 t. Now clearly when you multiply these two, so this will be minus 2 a. Now clearly when you multiply these two m1 into m2 is going to give you minus 1. That means a right angle is subtended at this point. This is a very very important property. Please remember this. The tangent at any point p of a parabola, the tangent at any point p of a parabola bisects the angle between, bisects the angle between the focal chord through p, the focal chord through p and the perpendicular and the perpendicular from p to the directx. Let me just show you on the diagram what do I mean by this property. So this property says that the tangent at any point will bisect this angle. That means this angle would be same as this angle theta theta. So tangent at p bisects the angle between the focal chord through p and the perpendicular from p on the directx and the perpendicular from p on the directx. Now it is very easy to prove. Let us say I can say this point to be point t. So can we find out the coordinates of t? It is very simple. We know the equation is ty is equal to x plus at square and the point where it meets the x axis, y is going to be 0. So put 0 it will be at square. So x will be minus of at square. So I can say st, st will be a plus at square. Isn't it? Because from o to s it is a and from o to t it is minus at square. So or you can say mod of that. Okay. So the sum of the distance will be a plus at square. And we also know that the focal distance of any point is also a plus at square. Correct. So indirectly what I have proven that st is equal to sp. That means triangle stp is an isosceles triangle. Triangle stp is an isosceles triangle. So this length and this length are equal. Okay. So I will draw it in a proper way so that it's quite legible to you all. So this distance and this distance are same. So this angle and this angle are the same. Correct. And if I drop a perpendicular I can say this angle is also this angle. Yes or no? So this angle and this angle are same because they are isosceles. So let me name it. Angle stp would be same as angle spt because it is isosceles. And angle spt is same as angle stp stp is same as angle tpm. Okay. Because they are vertically opposite. So indirectly these two angles are also equal. So spt and tpm they will be equal to each other. In other words the tangent is bisecting the angle between the focal chord and the perpendicular dropped onto the directrix. Next property, property number four, the foot of the perpendicular the foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at the vertex. Lies on the tangent at the vertex. So remember the question the property says foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at the vertex. So what is trying to say is that if I draw any tangent at a point then the foot of the perpendicular dropped from the focus. The foot of the perpendicular dropped from the focus would be meeting the tangent at the vertex. So remember y axis is nothing but y axis is the tangent at the vertex. Can you prove this quickly? Prove that the foot of the perpendicular from the focus on any tangent to the parabola lies on the tangent at the vertex. So it's very simple. Let's solve this. So let's say this point of tangency was at square comma 280 okay. So we know that the equation of the tangent is ty equal to x plus at square. Now let's draw a line which is passing through a comma zero and the slope of it is negative reciprocal of the slope. So if you can see the slope of this is 1 by t right. So let's write the equation of a line with slope as minus t and passing through a comma zero. That means y is equal to minus tx plus at. So let's solve one and two simultaneously. Let's solve one and two simultaneously. Solving one and two simultaneously. So let's substitute y as this in the other equation. You will automatically get minus t square x plus at square is x plus at square. At square at square gets cancelled. Which means t square x plus x is equal to zero, which means x times one plus t square is equal to zero, which implies x equal to zero. That means they meet where x is zero. That means they meet on the y axis, which is happening to be the tangent at the vertex. Is that fine guys? Please type CLR if it is clear to you. Okay.