 Let us try to do some questions which are based on the conjugate diameters show that show that the tangents at the ends of conjugate diameters of the ellipse x square by a square plus y square by b square equal to 1 intersect on intersect on another ellipse whose equation is x square by a square plus y square by b square equal to 2 so very simple guys you should be able to do it within 2 minutes I guess is that done so guys this concept is quite easy let's say you want to draw the tangent at okay let's go back to the previous page or let me draw it here itself let's say these are the two conjugate diameters you are planning to make tangents at the ends of these diameters so I have to find the locus of this point h comma k so guys pretty simple let's say this point is a cos phi comma b sin phi then this point here would be minus a sin phi comma b cos phi okay let's write on the equation of the tangent at this point so this tangent equation would be x by a cos phi plus y by b sin phi equal to 1 right and for the other point you just don't have to do anything else you just have to replace your x1 with minus a sin phi so it will become minus sin phi plus y by b cos phi equal to 1 so guys from these two equations my aim is to eliminate my aim is to eliminate phi right so if I want to eliminate phi the process is pretty simple let me call this as one let me call this as two so I will do one square plus two square so when I do that I'll get x square by a square cos square phi plus y square by b square sin square phi plus 2 x y by AB sine phi cos phi plus x square by a square sin square phi plus y square by B square cos square phi minus 2x y AB sin phi cos phi is equal to 1 square plus 1 square that's 2 and if you see these two terms will get cancel off these two terms will get cancel off and you can club these two terms together you can club these two terms together and you can also club these two terms together thereby giving you the required locus of their point of intersection right that's going to be two so this is another ellipse so the locus of the point of intersection of tangents drawn at the extremities of or at two ends of a pair of conjugate diameters is going to be an ellipse so guys next concept that we are going to discuss now is the concept of director circle is the concept of director circle okay so how do you define a director circle so it's the locus of the point of intersection locus of the point of intersection of two perpendicular of two perpendicular tangents drawn to the ellipse so again assuming this is your ellipse so the locus of a point h comma k from where if you draw two tangents to the ellipse these two tangents will be at right angles so this locus would be a circle which is called the director circle and prove that the director circle for the standard ellipse this the equation of the director circle is the equation of the director circle is x square plus y square is equal to a square plus b square so like all of you to please prove this quickly and type done if you are done with that yes anybody has any idea alright so we will try to use the fact that the equation of the tangent in the slope form we all know that the equation of a tangent in the slope form is y equal to mx plus minus under root of a square m square plus b square right and since this point satisfies this line I can say k is equal to mh plus minus under root of a square m square plus b square that means k minus mh whole square is a square m square plus b square that is k square plus m square at square minus 2 mh k is equal to a square m square plus b square that means m square at square minus a square minus 2 h km plus k square minus b square equal to 0 now this quadratic equation has two roots M1 and M2 are the roots of this quality equation. And if M1 and M2 are the roots, we all know that since the slopes are, since the lines are perpendicular, the product of the slope M1 and M2 should be minus 1. So we know product of the roots is equal to C by A. So K square by B square by H square minus A square that is equal to minus 1, which is nothing but K square minus B square is A square minus H square, which means H square plus K square is equal to A square plus B square. And if you generalize over here, if you generalize over here, we get X square plus Y square as A square plus B square as the equation of the director circle as the equation of the director circle. Is that fine guys? Any question with respect to this? So let us take a question on the director circle. Tangents at right angles, tangents at right angles are drawn to the ellipse X square by A square plus Y square by B square equal to 1. Show that the locus of the midpoints of the chord of contact, show that the locus of the midpoint of the chord of contact is the curve X square by A square plus Y square by B square whole square equal to X square plus Y square divided by A square plus B square. Is there any idea? If done, please type done. Alright, so let us take the situation through a diagram. So let us say this is the chord and the midpoint is H comma K. This is the chord of contact made by tangents from point X1, Y1 and these tangents are at 90 degree. So first of all, can I write down the equation of a chord whose midpoint is known to me? Of course we can write down the equation of a chord whose midpoint is known to me as T equal to S1. So if you write that you are going to get X H by A square plus Y K by B square is equal to X square by A square plus K square by B square. And in a similar way I can also write down the equation of a chord of contact drawn from X1, Y1 that is nothing but T equal to 0. So X X1 by A square plus Y Y1 by B square equal to 1. Now, ideally speaking in these two equations, these two equations stand for the same thing, AB. So these two are actually the equations of, these both are actually the equations of AB and hence they are same. And hence they are same. So if they are same, I can compare the coefficients of X, Y and the constants on both the sides. So if they are same comparing the coefficients of X, Y and constant, we get something X1 by H, Y1 by K is equal to 1 by H square by A square plus K square by B square. Which means X1 is H by H square by A square plus K square by B square, Y1 is K by H square by A square plus K square by B square. Now guys, clearly speaking this point X1, Y1 over here, this lies on the director circle. This lies on the director circle. You cannot deny that because it's already given that this angle here is a 90 degree. So this has to lie on a director circle. And if it lies on the director circle, then you would not deny that the director circle equation should be satisfied by these two points. Sorry, satisfied by these two coordinates. So these two will satisfy, these two should satisfy X square plus Y square is equal to A square plus B square. Which means H square plus K square by H square by A square plus K square by B square whole square should be equal to A square plus B square. So if we just simplify this a bit, we'll get H square by A square plus K square by B square whole square will be equal to H square plus K square by A square plus B square. And now we can generalize this. We can generalize this. And when you generalize this, you get X square by A square plus Y square by B square whole square is equal to X square plus Y square by A square plus B square, which is the required locus and hence root. So guys, any question, please feel free to ask me.