 Next property which is the fifth property, if S be the focus of the parabola, tangent and normal at any point P meet its axis in T and G respectively, so tangent meets at T normal meets at G, then S T is equal to S G is equal to S P. Let us prove it, it is very very simple you just have to make proper diagram and write the equation of tangents and normal and see where they are meeting the axis of the parabola. So tangent and normal they meet at T and G and this is your S and this is your P. So we have already seen S T is A plus A T square and S P is also A plus A T square, that is the focal distance. Now what is S G? So we know the equation of the normal is y equal to minus T x plus A T plus A T cube put y as 0. So T x will become this, so x will become 2 A plus A T square correct and this is already A. So this distance has to be A plus A T square again, which clearly implies S T is equal to S G is equal to S P. That means with S as the center and S T, S G and S P as the radius we can actually draw a circle like this, very bad circle but this is a circle. And this is going to be 90 degrees over here. So S is going to be the midpoint of, S is going to be the midpoint of T and G is the midpoint of T and G. Next is the sixth property guys, so let us note this down also. Sixth property says if S be the focus, if S be the focus, S H be perpendicular to the tangent at P then S H square is O S into S P, S H square is O S into S P. So let us say this is a tangent at a point and this is your S, this is your O. So you have to prove that S H square is O S into S P. So I will give you one minute, please type done on the chat box if you are done with it. If done please type done on your chat box if you are done with it, done. It is pretty simple, P is let us say A T square comma 2 A T and tangent will be T Y equal to X plus A T square. Let us find the coordinates of H. So where it meets H, X will be 0. So put this as 0, so T Y is equal to 0 plus A T square, so Y is equal to A T. So this point will be 0 comma A T and A will be A comma 0, sorry S will be A comma 0. So S X square is nothing but A minus 0 square A T minus 0 square which is actually A square plus A square T square. O S is nothing but A, S P is nothing but A plus A T square. So you can clearly see that when you multiply O S with S P you get A into A plus A T square which is nothing but A square plus A square T square which matches with this result. Done. So guys here we come to the end of this chapter. Now I will solve some problems on it because I do not want to start with ellipse right now. So next class we will be starting with an ellipse but presently we will take some problems. So let us start with this question very basic ones. The ones which are based on the locus definition of a parabola. Find the equation, find the equation of the parabola whose focus is minus 6 comma minus 6 and vertex is minus 2 comma minus 2. So if you make a diagram for this let us say minus 2 comma minus 2 is your vertex minus 6 minus 6 is your focus okay. So we know the line connecting the vertex and the focus is the axis correct. So this will be your axis of the parabola and also we know that directrix will be aligned perpendicular to the axis and it will meet axis at such a point which will be exactly in such a way that V is the midpoint of S and N. So these two distance will be equal. So it is very simple to find the equation of this line. This line equation is y equal to x correct right. So the equation of the directrix would be y equal to minus x plus some lambda correct. Now to get this lambda I need to know this point. Let us say this point is a comma let us say alpha comma beta. So alpha minus 6 by 2 is minus 2 and beta minus 6 by 2 is minus 2 again that means alpha is going to be minus 4 plus 6 which is going to be 2. So beta is also going to be 2. So put alpha beta in place of x and y to get the value of lambda. So 2 is equal to minus 2 plus lambda. So lambda is equal to 4 correct. So we get the equation of the directrix as x plus y equal to 4. Now once the equation of the directrix is known and the position of the focus is known we can easily write the equation of the parabola right. So let us clear this up. So x plus 6 the whole square y plus 6 the whole square under root is equal to distance of any point from this line by root 2 okay. So square both the sides and multiply with 2 you will get this equation okay and if you simplify this you are going to get x square plus y square minus 2xy plus 32x plus 32y plus 144 minus 16 128 equal to 0. This becomes your answer yeah okay. Let us take up next question now from a point A common tangents are drawn to the circle x square plus y square is equal to a square by 2 and the parabola y square is equal to 4x. Find the area of the quadrilateral find the area of the quadrilateral formed by the common tangents the chords of contact of the point A with respect to the circle and the parabola. So read the question very very carefully from a point A common tangents are drawn to the circle and the parabola. Find the area of the quadrilateral formed by the common tangents the chord of contact of the point A with respect to the circle and the parabola. So this is the situation actually so this is your circle also and this is a point A through which you are drawing common tangents to these two okay. So this is a common tangent okay. So now let us say this is the point of contact this is a point of contact here and you are connecting these by straight lines and making a quadrilateral from there. So the quadrilateral is p q r s this is your point A and you have been asked to find out this area any idea anybody okay. So let us discuss this without wasting much time. So first of all let's say I take this point q to be at square comma 280 okay and I write the equation of the tangent drawn at q that will be ty is equal to x plus at square correct. Okay now this tangent is also a tangent to this circle correct. That means the distance of origin from this tangent should be equal to the radius correct. So let me write it in the generalized form. So distance from the origin will be mod at square by under root of 1 plus t square and this should be equal to the radius. Radius is radius is a by root 2 radius is a by root 2. So if you cancel out the factor of a and square both the sides you will get 2 t to the power 4 is equal to 1 plus t square okay. So just square both the sides cancel out a squares you will get this. Now clearly a root here is t is equal to plus minus 1 right. That means I can say p point could be either be a comma 2 a or a comma minus 2 a correct. That means it has to pass through the focus itself right and clearly it is the lattice rectum. So q r has to be the lattice rectum right. Now I want to see what is p and s what is p and s from this diagram. What will be these two points? So when you put now the value of t as 1 let's say I take a t as 1 over here so I get y is equal to x plus a okay. Let's say consider this point to be x 1 comma y 1 correct. So the equation of the tangent to the circle to the circle at x 1 comma y 1 is nothing but x x 1 plus y y 1 equal to a square by 2 okay. And I try to compare this equation with this. So basically I am trying to compare these two equations. So this equation and this equation I am trying to compare okay. So when I do that I get I get x 1 by 1 is equal to y 1 by minus 1 is equal to a square by minus 2 a. That means x 1 is minus a by 2 and y 1 is a by 2. So it's very clear that this point p has now coordinates minus a by 2 comma a by 2. And similarly I can say s will have coordinates of minus a by 2 minus a by 2 because they would be symmetrically located they would be symmetrically located about the x axis. So this has to be your x axis the line connecting a to m has to be your x axis okay. So p s will be nothing but the distance between p and s which is going to be a okay. Moreover we can say that p s will be parallel to q r both will be parallel to the y axis correct. So indirectly your quadrilateral p q r s is basically a is basically a trapezium okay p q r s is basically a trapezium correct. Now what is the area of the trapezium? Area of the trapezium will be half of sum of parallel sides. So sum of p q plus r s into the distance between them okay. So please note that this distance will be 3 a by 2. So it becomes half of 4 a plus a into 3 a by 2 which will become which will become 1 fourth of 15 a square. So answer is 15 by 4 a. Any questions with respect to this solution? Rohan, Atmaesh. So we will take one more question before we end this session. If t p and t q are any two tangents or any two tangents to a parabola and the tangent at a third point r and the tangent at a third point r cuts them in p dash and q dash then prove that t p dash by t p plus t q dash by t q is going to be 1. So we have a tangent like this we have a tangent like this and we have a tangent like this let's say so let's say this is t p q p dash q dash this is r okay. So let's let's try to solve this question. So let's say point p is corresponding to a parameter t 1 point q is corresponding to a parameter t 2 and point r is corresponding to a parameter t 3 okay. Now we know that the tangents at the tangents drawn at p and q will meet at a t 1 t 2 comma a t 1 plus t 2 correct. So let's say this point is a t 1 square comma 2 a t 1 and let's say this is t p dash is to t p is let's say lambda is to 1 okay. Let's say it is lambda is to 1 and by the way this point is going to be the meeting point of tangent at p and tangent at r which is going to be a t 1 t 3 comma a t 1 plus t 3. So now t p dash by t p is going to be the distance between these two points divided by the distance between these two points okay which I'm calling as lambda is to 1. So let's find out the ratio of the distance between these two points. So t p dash will be a t 1 t 3 minus a t 1 t 2 whole square minus a t 1 plus t 3 minus t 1 minus t 2 whole square under root okay which is going to be a take t 1 a square t 1 square common t 3 minus t 2 whole square plus a square t 3 minus t 2 whole square under root which is going to be a t 3 minus t 2 under root of t 1 square plus 1. In a similar way if I find t p if I find t p what will I get under root of a t 1 square minus a t 1 t 2 whole square plus 2 a t 1 minus a t 1 plus t 2 whole square. And again if I am not wrong that will give me a t 1 minus t 2 under root of t 1 square plus 1 correct. So this ratio will be so this ratio will be t p dash by t p will be actually t 3 minus t 2 by t 1 minus t 2. Can I say similarly similarly t q dash by t q would be t 1 minus t 3 by t 1 minus t 2 correct. Now let's add them let's add them so t p dash by t p plus t q dash by t q will be t 3 minus t 2 by t 1 minus t 2 plus t 1 minus t 3 by t 1 minus t 2 you can take t 1 minus t 2 as common you will have t 3 minus t 2 plus t 1 minus t 3 t 3 and t 3 gets cancelled and you will left with t 1 minus t 2 by t 1 minus t 2 which becomes 1 and hence proved okay. So quite a hectic problem but if you know the road map you will be able to easily navigate through this problem okay. So guys with this we come to the end of your parabola chapter we are left with ellipse and hyperbola. So next class also we'll keep it online and I'll start with the ellipse okay. Best of luck for your coming exams so over and out from Centrum Academy thank you for coming online.