 Next concepts that we are going to talk about is the concept of pair of tangents Now from a point external to a parabola from a point external to a parabola You can always catch two tangents to the parabola One is like this another is like this Okay, so let's say this point is h comma k And you can draw two tangents to the parabola. Let's say P t1 and P t2 Okay, then the combined equation of these two tangents is called the pair of tangents Okay, and we had already learned about this in circle as well that the combined equation of The pair of tangents is given by t square is equal to ss1. So nothing changes even in case of a parabola What is t in this case? Let's say if I'm talking about y square is equal to 4ax parabola. What is t in this case? T will be y y1 y1 will be k. This is your x1 y1 Okay, y1 will be k Equal to 2ax plus x1 Okay, what will be s? What will be s s will be y square minus 4ax What will be s1 s1 will be k square minus 4ah So you can write down the combined equation as t square which is this equal to s s1 So that will give you the pair of tangent equation pair of tangents combined equation pair of tangents Combined equation, okay So nothing new here. It is very much similar to the concept which we had learned before as well Okay Now chord of contact chord of contact. What is chord of contact? Chord of contact is nothing but the line connecting the points of contact of these two tangents So this line is called the chord of contact And just like we learned in case of a circle chord of contact equation is still t equal to zero that means yk minus 2ax plus h equal to zero Okay Let's take a question on this concept prove that the area of the triangle The area of the triangle Formed by the tangents drawn from x1 comma y1 to the parabola y square is equal to 4ax and their chord of contact and their chord of contact is y1 square minus 4ax1 to the power of 3 by 2 divided by 2a So please take few minutes to work on it. Very simple problem In case you're done with the problem, please type done on this on the chat box so that I can start the discussion Yeah, so we need to find the area of this triangle pqr that you can see on your screen This point is x1 y1 Now if anybody is trying to attempt this problem, of course, uh method that would come in your mind is Okay, we we can find out the length of the normal drawn from this point Okay Uh, let me call it as m And we can find the area of the triangle by using half into base into height So half into qr into pm Now finding pm is not difficult because you know the equation of this line Right the equation of this line is yy1 is equal to 2ax plus x1 And you know this point p so you can always find out the distance of the point p from the line qr Which is actually your chord of contact But how do I find this is easy? Okay, but how do I find qr? That is the length of the chord of contact How do I find that? See for that you have to go the other way round. Let's say this point is a t1 square comma 2 a t1 And this point is let's say a t2 square comma 2 a t2 Okay Now the meeting point of these two tangents we all know from our previous formula It's actually a t1 t2 comma a t1 plus t2 remember Goa correct That I discussed in the session the other day Right Now According to a given problem a t1 t2 Will be equal to x1 And a t1 t2 will be equal to y1 correct Which implies t1 t2 is x1 by a And t1 plus t2 is equal to y1 by a Okay Now these two in piece of information is very important for me because it will help me to write the Length of this tan length of this chord of contact in terms of x1 and y1 So if you see qr is nothing but Under root of x1 minus x2 square Plus y1 minus y2 square Okay And if you try to simplify it Right by taking your some terms common First of all a could be taken out common So mod a will come out From inside I can take t1 minus t2 whole square also common Okay, and I'll get t1 plus t2 whole square Plus four So that will become mod a t1 minus t2 will come out in fact mod of t1 minus t2 will come out And within the brackets I'll get t1 plus t2 square Plus four Now t1 minus t2 mod is as good as saying t1 plus t2 whole square minus four t1 t2 Yes or no So this term could be written like this Mod t1 minus t2 is nothing but t1 plus t2 whole square minus four t1 t2 So if I decide to use t1 plus t2 as y1 square by a square This is what I will get Correct and here I will get y1 square by a square plus four Okay, we can simplify it And if you simplify it I'll get a mod a mod a out. So you'll have one by mod a Under root of y square minus four a x1 Into y1 square plus four a square Okay, so this will be your qr. What about pm What about pm? So for pm, you will directly use the distance of a point from a line. So you'll say y1 y1 minus 2a x1 plus x1 mod By under root of y1 square four a square Which is going to be y1 square minus four a x1 mod Under root of y1 square plus four a square Okay, now I can use the two formulas that is half qr into pm That will give me half one by mod a under root of y1 square minus four a x1 times y1 square plus four a square into y1 square minus four a x1 mod by under root of y1 square plus four a square So this term and this term will get cancelled So you will get the answer as one by two mod a now mod a is as good as a because a is always positive in a parabola So it's one by two a y1 square minus four a x1 to the power of three by two and that's proved Okay So quite a lengthy problem, but I'm sure you would have understood the concept now moving on to the next concept That is the equation of the chord bisected at a given point equation of a chord bisected at x1 y1 So please note that the equation just like in case of circle becomes t equal to s1 that means If I'm talking about the parabola y square is equal to four a x t will be y1 minus two a x plus x1 Is equal to s1 s1 will be k square minus four a sorry Not k square It will be y1 square minus four a x1 Okay So without wasting much time, let us look into problems which have been asked in this case find the locus of the find the locus of the midpoints of normal chords of the parabola y square is equal to four a x I'll make this question as show that so that you can prove it Show that the locus of the midpoints of the normal chords of the parabola y square is equal to four a x is y to the power four minus two a x minus two a y square Plus eight eight to the power four equal to zero Again, please remember locus questions are very very important because j simply likes locus questions So, please think about it very simple concept any idea guys okay, first of all You have been given that you have to find out the locus of the midpoints of normal chords So if I just ask you what will be the locus let's say the point The midpoint is h comma k. Okay Let's say the midpoint of the chord is h comma k Then what would be the equation of the chord? Equation of the chord will be y k minus two a h plus x plus h Is equal to h square plus k square is equal to k square minus two a Sorry k square minus four a h So basically this is nothing but t. This is nothing but s one t equal to s one is the equation of the chord Whose midpoint is given to you? correct Now you are claiming that Now you're claiming that this chord is also the normal This chord is also the normal to the Parabola right that means We can compare this equation with the equation of the normal And we know the equation of the normal is y plus tx is equal to two at plus at cube correct So this is the equation of the normal And both the equations are actually the same equations Both the equations are actually the same equations. Correct. So can I compare their coefficients? So when I compare their coefficient, I can say one by k Let's say this is the one one divided by k Is equal to uh By the way, I have to rearrange and write it in a proper way first Just give me a moment So this I have to write it as y k minus two ax Is equal to k square minus two a h Okay Now we can compare these two So comparing Let me call it one and two So I get one by k is equal to t by minus two a is equal to two at plus at cube By k square minus two a h Remember I have to eliminate t because t is something which I introduce in the problem and that's a parameter So we have to eliminate t which can be easily done through this equation So t is minus two a by k correct And now I can use these two equations Okay, so from these two equations I can say t by minus two a is equal to t times two a plus t square By k square minus two a h t and t gets cancelled Just cross multiply so you get k square minus two a h Let the minus two a be over here that is equal to two a plus at square So this will become k square minus two a h by minus two a is equal to two a plus a times Four a square by k square. Let us simplify this So k square minus two a h is equal to minus two a times Two a k square plus four a cube Into k square. So it becomes k to the power four minus two a h k square And on this side I get minus four a square k square Minus eight eight to the power four collect the terms So it'll become k to the power four Minus two a Uh, we'll have minus two a k square. You can take common. So it'll become h minus two a Plus eight eight to the power four equal to zero Okay, now just generalize this So when you generalize this We get y to the power four minus two a y square x minus two a Plus eight eight to the power four equal to zero And that resembles with the required equation over here as you can see This is what we obtained down over here as well. Okay So let's move on to the next concept now any question so far guys Please feel free to type it in the chat box