 Alright, so we will move on to the equation of normal now. So equation of normal to the ellipse at x1, y1. So again, let's take our standard case of an ellipse, okay. So let's say at a point x1, y1, you are sketching a normal, okay. So please prove that, I am not going to do this very simple, it's basically associated with your application of derivatives. So prove that the equation of the normal drawn to the standard form x square by a square plus y square by b square equal to 1 at point x1, y1 is given by a square x by x1 minus b square y by y1 equal to a square minus b square and please remember this result. So please prove this and please remember this result as well. Just one minute to do that guys, just one minute. Every side done if you are done with the derivation. So you can directly take a clue from the equation of the tangent which is xx1 by a square yy1 by b square equal to 1 and you know the slope of the tangent is going to be, the slope of the tangent is going to be minus b square x1 by a square y1, so slope of the normal is going to be a square y1 by b square x1. So writing down the slope point form which is y minus y1 is equal to a square y1 b square x1 x minus x1 and further simplifying it, you get b square x1 y minus y1 is a square y1 x minus x1. So it becomes b square x1 y minus a square y1 x is equal to b square x1 y1 minus a square x1 y1. So now divide throughout with x1 y1, when you divide throughout with x1 y1 you get something like this b square y by y1 minus a square x by x1 is equal to b square minus a square just flip the position of both the terms. So this is what you realize which is the equation of the normal. This is called the equation of the normal in the point form, equation of the normal in the point form, right? Let's discuss about the equation of the normal in parametric form. So a square x by x1 minus b square y by y1 equal to a square minus b square is the point form. If you replace your x1 as a cos theta and y1 as b sin theta, that means you're looking for the equation of the normal at the parametric point, you get a square x by a cos theta minus b square y by b sin theta equal to a square minus b square which is nothing but a square which is nothing but ax c theta minus by cos theta equal to a square minus b square. So please, please, please remember this expression for the parametric equation of the equation of the normal. So this is called the parametric form. This is called the parametric form of the normal. Is that fine, guys? Okay. So if this is understood, we can start solving a problem based on this question is question is if y equal to mx plus c, if y is equal to mx plus c is a normal is a normal to the standard form of the ellipse that's x square by a square plus y square by b square equal to a square then prove that c square is equal to m square a square minus b square whole square by a square b square m square too many squares. So prove this, guys. I think this should be very simple. This is called by the way condition of normality. This condition is called the condition of normality, the condition of normality. And once you're done, please type done on the chat box, guys, pretty simple. Just again, you have to be very, very clear with your approach. Just now we learned that the equation of the normal drawn at any parametric point a cos theta comma b sin theta normal at this point is given by ax c theta minus by cos theta equal to a square minus b square, right? So what I'm going to do next is I'm going to just write it as y equal to mx plus c, okay? So I'm going to write it as y equal to mx plus c. So for that I need to write it like ax c theta plus b square minus a square, okay? So y is equal to a by b, c theta by cos theta will become tan theta x plus b square by a square by b sin theta, sorry, b cos theta. So sin theta will go up, right? So no problem in that. So guys, here you can say that your m, just try to compare these two. Just compare these two. Let's compare these two. So when you compare these two, you would realize m is actually a by b tan theta and c is actually b square by a square by b sin theta, right? Yes or no? So from here I get a tan theta as mb by a, which means sin of theta is going to be mb by under root of a square minus m square b square, right? And from here I get sin theta as bc by b square minus a square, bc by b square minus a square. Now these two expressions must be the same. These two expressions must be the same. So 1 and 2 must be the same. That means bc by b square minus a square is mb by under root of a square m square plus b square, cancel off a factor of b. So we can write this as c is equal to m times b square minus a square by under root of a square m square plus b square. And if you square both the sides, and if you square both the sides, you get c square as m square b square minus a square whole square by a square m square plus b square. Just note that the order of a square and b square doesn't matter because they are under the influence of a square and hence proved and hence proved. This is the condition for normality. This is the condition for normality. Any question guys? Please feel free to type in in the chat box. If there's no question, please type no question so that I can move ahead with the next problem. Alright, so next question goes like this. If the normal at an end of a lattice rectum, if the normal at the end of the lattice rectum of the ellipse of the ellipse x square by a square plus y square by b square equal to 1 passes through passes through one extremity of the minor axis passes through one extremity of the minor axis passes through one extremity of the minor axis, then show Show that, show that, eccentricity square is under root 5 minus 1 by 2. Show that the eccentricity square is under root 5 minus 1 by 2. So please type done if you are done, whenever you are done please type it. I hope you are aware of the end of the latter spectrum, the end of the latter spectrum has the coordinates of a comma b square by a. Remember the latter spectrum is aligned perpendicular to the axis and passing to the focus so the x coordinate will always be a and since its length is 2 b square half of it will be below so it's a comma b square by a so you're drawing a normal at this point and it happens to pass through one extremity of the minor axis one extremity of the minor axis so this is passing through 0 comma minus b 0 comma minus b if done please type done on your chat box done no not done so the bus has done so far what about others all right so let's do this quickly so we know the equation of the normal drawn at any point is given by a square x by x 1 minus b square y by y 1 equal to a square minus b square now here your x 1 y 1 themselves are given to you as a e and b square by a so it's going to be a square x by a e minus b square y by b square by a to a square minus b square that is given that it is passing through 0 comma minus b it's passing through 0 comma minus b so here I can say 0 will become the a square right now can I square both the sides if I square both the sides of this term I get a square b square is equal to a square minus b square which itself will square so we can also write b square as a square 1 minus e square and this side we have e to the power square will get cancelled so this brings me to a box e squares now you can apply the Shridharacharya formula over here treating e square to be x it is minus b square minus 4 ac by 2 a that's going to be minus 1 plus minus root 5 2 remember it cannot be minus so it has to be minus 1 is right hence sure hope you don't have any code so one more question will take up and then we'll take a small break after that prove that the straight line lx plus my plus n equal to 0 is a normal is a normal to the standard form of an ellipse to the standard form of an ellipse if a square by l square plus b square by m square is equal to a square minus b square whole square by n square alright shall we discuss this okay guys I've just now sent you some mock papers so please do that before you go for your Friday exam so two mock papers I've given you make sure you solve one today and one tomorrow okay again I will be just using the parametric form of the equation of a normal which is this and we'll try to compare it with your lx plus my plus n equal to 0 okay so we'll compare it with lx plus my is equal to minus n so when we compare this we can write a c theta by l is equal to minus b cos theta by m and is equal to a square minus b square by minus n because these two equations are the same equations right so I will get a cos of theta as correct me if I'm wrong minus n a by l a square minus b square and I'll get sine of theta as sine of theta as n b by m a square minus b square right correct and we know the famous Pythagorean identity cos square theta plus sine square theta is going to be one right which brings me to this form n square a square by l square a square minus b square plus n square b square by m square a square minus b square equal to 1 which means which means a square by l square plus b square by m square could be written as a square minus b square whole square by n square because these two terms can be taken common from both the places and taken it to the right hand side okay