 Next, we will follow it up with another simple question. For what value of lambda? For what value of lambda? Does the line y equal to x plus lambda touch or touches 9x square plus 16y square is 144. Let me call it what values it may have multiple values. So please type in your response in the chat box plus or minus 5, absolutely correct. So you can write this as x square by 16, y square by 9 equal to 1 and you know that c square is equal to a square m square plus b square which is 9. So x lambda is plus minus 5, great. So moving on next which is the concept of equation of tangents in various forms. Equation of tangents in various forms, in various forms. So a little while ago we saw the equation of a tangent in slope form that was y equal to mx plus minus under root a square m square plus b square. This was called the slope form, okay. Second is the point form, point form is basically a case where you have been given a point and you have been asked to sketch or draw the equation or write the equation of a tangent at any point. Let's say this is the point x1 comma y1 and you are sketching a tangent at this point and you have to write the equation of this tangent. Let's take the equation to be the standard form itself, okay. By the way this only works for the standard form and in this case the equation will become xx1 by a square plus yy1 by b square equal to 1. Please note that this expression would be referred to by the symbol t henceforth, okay. So this is called the point form of the equation of a tangent. Remember there is a generalization that is going on over here. We normally substitute x square with xx1, y square with yy1, xy with xy1 plus x1y by 2. We generalize x with x plus x1 by 2, y with y plus y1 by 2 and c remains c. So even if your equation of the ellipse is a standard, is a general form like this, even if the equation is a general form like this then you have to find the equation of a tangent at xx1 point, sorry tangent as x1y1 point then it will be replaced in this fashion, alright. So this is the parametric form, it is y of theta, guys can you see the screen and b side thing about when I was discussing with you the equation of the chord connecting to eccentric points theta and phi and this form is called the parametric form and it is the most widely used form when we are referring to the equation of the tangent to a parabola at a parametric point, alright. So let us take questions on these, so done the various form of the equation of a tangent to the parabola, sorry tangent to the ellipse. So let us take up some questions, product of the, from the foci, from the foci of any tangent of any tangent to an ellipse, to an ellipse is equal to the square of the semi minor axis is equal to the square of the semi minor axis, okay and the feet of these perpendicular and the feet of these perpendicular lie on the, it has to be remembered as a theorem itself because it is very very important, the question says product of the perpendicular is from the foci on any tangent of the ellipse is equal to the square of the semi minor axis and the locus of the feet of these perpendicular lie on the auxiliary circle, so guys I am sure if you talk about this standard case of an ellipse, you know the auxiliary circle for this, the auxiliary circle for this is basically x square plus y square is equal to a square, auxiliary circle is a circle which is drawn as, sorry drawn as the major axis as the diameter, which I already discussed with you while I was discussing the eccentric form. Alright guys, so I am going to just exhibit on geogebra, how the locus of the feet of the perpendicular drawn from the foci is basically going to be the auxiliary circle, okay so I will just type in any equation of an ellipse which is let us say x square, x square divided by 9 plus y square divided by 4 equal to 1, okay so let us say this is the ellipse that you have in front of you, yeah, okay so let us locate the foci, so focus of this conic C I want, so there is the focus for you A and B, right, now I will choose any point on the ellipse, let us say I choose a point C and I draw a tangent, I draw a tangent to this ellipse, okay, next I sketch a perpendicular, okay as you can see this is a perpendicular and this is the point of, D is the point where the perpendicular from the foci meets the tangent at C, right, now what I am going to do is I am going to simply move this point, okay but I am moving this point you see D is also moving, right, now in order to better locate D I would actually show you the trace of that, okay so let me start moving this C slowly and slowly and slowly just watch the moment, just watch the movement of the point D on your screen you would realize that the path traced by D is just going to be a circle, okay and that circle is going to be such a way that it is going to have the diameter as the the major axis of the ellipse which we actually call as the auxiliary circle which we actually call as the auxiliary circle, look how beautifully it is coming up, okay wow, right, so do you see an auxiliary circle being formed which is x square plus y square is equal to a square and that is what we have to mathematically prove as well, we have to prove that mathematically as well, so first let me do the last part of the question that is the feet of these perpendicular is going to be the auxiliary circle, so we all know for x square by a square plus y square by b square equal to 1, we can write a tangent as y equal to mx plus, let me take any one tangent, okay, which means y minus mx is under root of a square m square plus b square, now let me ask you what would be the equation of a line, equation of a line through a e comma zero and perpendicular to perpendicular to y equal to mx plus under root a square m square plus b square, so if I ask you this your response will be plain and simple it's going to be y minus zero is equal to minus m that is the negative reciprocal of the slope x minus a e which means x plus my is equal to a e, now from these two equations equation number one and two I need to eliminate m, so let's eliminate one let's eliminate m from equation number one and two let's eliminate m from equation number one and two, so guys just watch carefully what I'm going to do I'm going to just square one and square two and add them, so one square and two square and I'm going to add them, so one square is going to be y minus mx square and we have got this term and on the right hand side we are going to get a square m square plus b square plus a square e square, okay so let's expand this up, let's expand this up so you get y square m square x square minus two mxy and here we get x square m square y square plus two mxy and this is going to be a square m square plus b square plus a square e square, so these two terms will get cancelled off, so these two terms will get cancelled off, you just take y square common and x square common, you get one plus m square x square plus y square and on the right hand side I get a square m square plus b square, now please note that all of you please remember that a square e square is actually nothing but a square minus b square, right remember in your school days used to call this as c square equal to a square minus b square in your ncrd syllabus, correct, so I'm going to use the same expansion for this term at least, so this will be a square minus b square as you can see b square and b square gets cancelled off and this is what I'll be left with, guys what is important is not the result, what is important is how you deal with these locus questions, I'm again repeating time and again that J loves locus, so how you know how the concept with respect to locus formation should be very very clear in your mind, right, so this is the equation of the auxiliary circle of the ellipse, this is the equation of the auxiliary circle of the ellipse, right, now our next part of the question was we have to prove that the product of the slopes we have to prove that the product of the slopes is going to be b square, sorry product of the perpendicular is going to be b square, so again again let's take y equal to mx plus under root a square m square plus b square and let us take the point to be a e comma 0, okay, so we have two points a comma 0, s1 and s2 and from these two points I have to draw perpendicular to this tangent and find the product of the perpendicular, so we'll use the distance of a point from a line, so let me take the y on the other side, okay, so let me write it like this, this goes off minus y, okay, yeah, so when you substitute a comma 0 you get to see mod a square m square b square plus mae by under root of 1 plus m square and p2 would be mod under root a square m square plus b square minus mae by under root 1 square plus m square, right, so let's take the product p1 into p2, p1 into p2 that would give you mod of, remember inside is x plus y, x minus y, so it becomes x square minus y square x square minus y square by 1 plus m square, okay, yes or no, now again using the fact that a square e square is actually a square minus b square, so it becomes a square m square plus b square minus m square a square minus b square mod by 1 plus m square, if you open the brackets, if you open the brackets you get a b square 1 plus m square mod, mod doesn't have any meaning over here because every term is positive, so it's actually b square which is nothing but, which is nothing but the semi-minor axis length the whole square, very, very important a theorem over here, is that fine guys, any doubt whatsoever please type it in the chat box, if no doubt please type no doubt