 The next concept that we are going to talk about is the concept of conjugate diameters Is the concept of conjugate diameters? So what is a conjugate diameter? So let me show you through a diagram. So let's say this is a system of you know parallel chords Okay, and this is a diameter Which bisects is parallel chords? Okay, so basically I can say this to be bisected. Okay, and now if I start making all chords Which are parallel to this? Diameter itself. Let's say I start making all the chords which are parallel to the diameter of this. Okay Then the line Let's say this red line which bisects which bisects these chords parallel to this line That is a red line Okay, so what I'm doing first all these white lines This is the diameter Okay, so let's say diameter one and these yellow lines all these yellow lines Including the diameter itself is bisected by this diameter which is called diameter two Okay, so I'll write down the definition Two diameters are said to be conjugate two diameters Are said to be conjugate are said to be conjugate? when each bisects when each bisects All chords parallel to the other when each bisects all chords parallel to the other So the red and the blue are conjugate diameters Now, please please please do not take this angle to be 90 degree This angle is not 90 degree. I'm writing it specifically over here Okay By the way, this will become a 90 degree had this ellipse been a circle. So let us take some properties Let us take some properties with respect to the conjugate diameters properties of Conjugate diameters the first property is The eccentric angle of the ends of a pair of conjugate diameters the eccentric angle of the ends of a pair of Conjugate diameters in an ellipse differ by a right angle Differ by 90 degree. Okay, what I want to say is very very clear Focus on this diagram. So let's say this this angle over here is let's say Theta and this angle over here is let's say 5 So these are the ends. Okay, let me name it as P Q So this is P. This is Q and this is C and let's say this is D Okay, so let's say the eccentric angle of point P is theta and the eccentric angle of point C is 5 Then you have to prove that Then you have to prove that these angles differ these angles differ Right, you can write theta minus 5 or 5 minus theta that depends upon But what is the location of theta and 5 with respect to each other? They differ by 90 degrees Guys, it doesn't mean this angle is 90 degree again I'm repeating this because the eccentric angles are taken with respect to the auxiliary circle. Am I clear? This doesn't mean this doesn't mean the angle between the conjugate diameters is 90 degrees You can anybody prove this? All right guys, so let's discuss this first Before I want before you prove anything else, I would like you to just You know try and prove this Let's say the slope of diameter M1 diameter 1 is M1 and slope of diameter 2 is M2 Okay, try proving that M1 M2 is b square by a square. That's the number one step prove that the product of the Product of the slopes of The conjugate diameters is minus b square by a square and Remember since b square and a square are not equal M1 M2 will not be minus 1 and if M1 M2 is not minus 1 the conjugate diameters cannot be perpendicular to each other Yes, they would be perpendicular But that would be a case when the ellipse has changed itself to a circle. Okay, so it's very simple guys how do we prove this Let's say the blue diameter is bisecting the The chords whose slope are M2 so I can write the equation of diameter 1 as Equation of diameter 1 as Y is equal to minus b square x by a square M2 right Because it is bisecting those chords which are having a slope of M2 Right and you doubt about that and the slope of this line itself if you see this is the slope of this line And the slope of this line itself is M1 which clearly means which clearly means M1 into M2 could be easily proved as minus b square by a square Right. So number one step is accomplished Now, let us move on to prove the requirement of this property We have to prove that the difference of the eccentric angles at one end of the conjugate diameters differ by 90 degree right Now this is origin. We all know that both the conjugate diameters will pass through the origin correct, right So what is the slope of OP? the slope of OP is nothing but B sine theta by a cos theta in other words, it's B by a tan theta Correct and this is nothing but M2 for you. This is your M2 right In a similar way What will be the slope of OC? Now without waste of time will say B by a tan phi and that is going to be M1 And we already know this property that M1 into M2 is minus b square by a square So I can say b square by a square tan theta tan phi is equal to minus b square by a square which clearly implies that Tan theta tan phi is equal to minus 1 which means 1 plus tan theta tan phi is equal to 0 right and If you try to use this formula tan theta minus phi You know that in this expression you Get the required expression to be this and since this fellow is 0 over here It makes this expression go 0 correct, so it becomes undefined Which clearly implies that Theta minus phi is going to be a multiple of 90 degree and you can just call it as a 90 degree Is that fine guys any question with respect to this? Okay, and because of this property and Because of this property if you say your point P I'll draw the figure again just for the sake of clarity So let's assume that these two are your conjugate diameters So if this is your point P and this is your point Q This is your point C and this is your point D Then point P could be written as a cos theta comma B sin theta then point C could be written as minus a sin theta comma B cos theta Okay, and Leason is to tell you how could D be written just increase theta with pi So it becomes minus a cos theta minus B sin theta and This could be written as a sin theta minus B cos theta So for the sake of clarity, I have just written down the the coordinates of the extremities of the conjugate diameters for you There is next property, which is your property number two From this figure prove that OC square plus OD square is going to be a square plus B square Now this is very obvious from the property number one So just do it quickly and tell me if you're done So just type done on the chat box when you're done I don't think so it should take so much of time. It's very simple OP is nothing but Sorry, this will be OP. Yeah, I'm sorry So OP and OC square. This does nothing but the distance of them. That means it's going to be a square sin square theta plus B square cos square theta minus 2 AB sin theta cos theta and OP square is a square cos square theta Plus B square sin square theta plus 2 AB sin theta cos theta These two terms get cancelled off and you'd be left with you'd be left with a square So these two terms will get you a square and these two terms will give you B square and hence proved Is that fine guys? Any questions? Please please feel free to ask me right now Next property the area of the parallelogram the area of the parallelogram formed by tangents at the extremities of the pair of conjugate diameters at the extremities of the pair of conjugate diameters is 4 AB So please prove this So I've tried to draw it as accurate as possible So let's say this is a cos phi comma B sin phi and this point is minus A sin phi comma B cos phi and this is the center of the ellipse again P O C So here we have to give the area of this parallelogram M N Let's say S T So you have to prove that area of the parallelogram M S T is for AB Square units Guys if you see any parallelogram Okay Any parallelogram if you construct we know that area of this parallelogram is Base into height right so how can we use this concept over here to solve this problem? Okay, so first of all I would like to know from you that is CD parallel to the tangent at P. So my question to you is is CD parallel to the tangent at P if yes, please type yes and is P Q is P Q Parallel to the tangent at C if yes, please type yes on the chat box chat box So guys is very simple to prove this. What is the slope of the tangent at P? The slope of the tangent at P Let me call it as slope of the tangent at at P Okay So the slope of the tangent at P. No problem Upish. That's fine So slope of the tangent at P would be nothing but B cos Phi by minus a sin Phi That's going to be minus a cot Phi and Slope of OC slope of OC is going to be B cos Phi by minus a sin Phi that is y2 minus y1 Right treat this as 0 0 and you just have to do y2 minus y1 by x2 minus x1 and of course they all match So these two are same. It implies that the answer to these questions is yes Yes, they are Right So if they are it is something like this I mean you have a line like this which is parallel to this line and You have a line like this which is actually parallel to this line Right Okay, and they're also passing through the point of intersection of the diagonals that is your origin Which clearly implies That these four areas would be equal Correct me if I'm wrong this all area, which I'm shading this area will be equal to this area Will be equal to this area and Will be equal to this area So all these four areas will be equal to each other, right? So it's like saying that all these areas over here Okay, I'm shading with different different slopes so that you can identify Okay These all four areas will be equal. This will be equal. This will be equal. This will be equal So if I just manage to find any one of them my job will be done So let us focus on finding out the area of this parallelogram If I'm able to find the area of this green parallelogram over here that is C O P M parallelogram and I just for it four times do four times that area my job will be done So let us focus on finding the area of that parallelogram. So let's Drop a perpendicular from the origin. Let's drop a perpendicular from the origin on to this parallelogram Okay, let me choose some other color pink Let me drop a pink perpendicular from the origin on to cm so if I multiply O P or Or cm so area of Palo gram C O P M Can be written as let's say a drop-up perpendicular called. Let's say O L So O L into C M Which is actually as good as O L into O P Now O P is easy to find out O P is under root of A square cos square phi plus B square sin square phi. So no problem in finding O P now O L O L is nothing but the perpendicular of perpendicular distance of origin from the tangent drawn at C Guys, just remember one thing the equation of a tangent drawn at any point on the ellipse is going to be x x 1 by a square plus y y 1 by B square equal to 1 and Here I'm drawing it at x 1 which is equal to Your x 1 y 1 point are actually Minus a sin phi Comma B cos phi So if I use this equation the tangent equation would be tangent at C would be x by a square into minus a sin phi plus y by B square Into B cos phi is equal to 1 so cross the factor of a it will become minus x sin phi by a Plus y by B sin phi minus 1 equal to 0. So now I want the distance of this point from the origin So the distance of this point from the origin So the distance from origin or distance of origin I can say distance of origin from tangent at C is equal to 1 by under root of Sin square phi by a square plus Oh, this is going to be cos. I'm sorry just a typo error So cos square phi by B square, which is actually a B by a square cos square phi plus B square sin square phi under root right So when you put these two things over here, that is this is nothing but your O L Okay, so when you substitute it over here this and this you get your area as your area as a B by under root a square cos square phi plus B square sin square phi into under root of a square cos square phi plus B square sin square phi Which gets cancelled off and we get the area as a B Now as I already discussed with you the area of the required parallelogram would be four times the area of the parallelogram copm so your answer would be Four times the area of parallelogram Copm so your answer is going to be four times a B Okay, and hence proved is that fine guys