 Hello friends. Good morning to all of you. So hope you all are doing good. Practicing sincerely for the coming exams this board and J exams and need exams. So, we were discussing conic sections and we have covered our circles and parabola right. So, today we will start the next conic that is ellipse. So, without wasting any time. Let's start. So, yeah, this is a exercise one of our ellipse chapter. Okay. So here comes the first question. It is saying that the length of major axis of the ellipse x square upon a square plus y square upon b square equal to one is three times the length of the minor axis. Then it's eccentricity is so the given ellipse is our standard ellipse that is x square upon a square plus y square upon b square is equal to one. Now it is saying that the length of major axis. Okay. So I am taking suppose I'm taking this a is greater than b. So what will be the length of major axis the length of major axis will be two way. And it is given to be three times the length of minor axis three times length of minor axis. This is what is provided in the question. Now we have to find the eccentricity of this ellipse. So from here if you see. b is coming out to be 3b right. And what is our eccentricity we know that e square is equal to one minus b square upon a square. So, one minus b square upon a square in place of a square I can write nine b square so this b square b square got cancelled. And we will be left with nine minus one that is eight upon nine. So this is our e square. Now what will be e what will be e it will be root eight upon three. Okay root eight we can write it as two root two upon three. Okay it cannot be negative because the eccentricity is the ratio of distance right. It's the distance ratio of the distance so eccentricity cannot be negative. So this will be our answer. Now if anyone wants that this b why I have taken a is greater than b you can take b is also you can take b greater than a and then also you can apply the given condition and you will be having the same eccentricity as answer. Remind me in that case our major axis will be 2b and our minor axis will be 2b right. So here our answer is c two root two upon three. Okay so moving to the next question. This is question number two the equation x square upon ten minus a plus y square upon four minus a equal to one represents an ellipse if. Okay so the question is saying this curve x square upon ten minus a plus y square upon four minus a is equal to one this is representing an ellipse. So for this curve to represent an ellipse this ten minus a if you compare it with our standard ellipse that is x square upon a square. Plus y square upon b square is equal to one so this denominator and this the things in the denominator this must be greater than zero this must be greater than zero this b square should also be greater than zero right. So I will apply the same condition here this ten minus a this ten minus a should be greater than zero so from here what we get ten is greater than a or we can say a is less than ten right. And from here if you see this four minus a this should also be greater than zero so this four should be greater than a or we can say a is less than four. Now we have to take the intersection of these two values of a so a is less than a okay and from here you see a is less than ten. So what you can say a should be less than four taking the intersection taking the intersection we can say the value of a should be less than four. Is it okay. The value of a should be less than four for this curve to represent an ellipse so this is our answer. Right. Now let's take the next question. Question number three, it is saying the eccentricity of an ellipse x square upon a square plus y square upon b square equal to one whose lattice rectum is half of its major axis. So let me write the ellipse that is x square upon a square plus y square upon b square is equal to one. What I will assume I will assume this a is greater than b. Again if you want to assume b greater than a you can do not an issue. So considering this what will be the length of lattice rectum for this ellipse. What will be the length of lattice rectum for this ellipse it will be two times b square upon a right two times b square upon a. And what will be the major axis for this ellipse what will be our major axis since a is greater than b the major axis will be to a. Now as per given condition this two times b square upon a this lattice rectum is half of its major axis half of its major axis. So in this condition we have to find the eccentricity. So from here if you see we got a square I am cross multiplying so a square is equal to two b square. Right. And we know for this ellipse what is our eccentricity eccentricity square is equal to one minus b square upon a square. So one minus b square I am writing as it is and I am writing a square as two b square. So this b square b square term will get cancelled out and from here we will get e square is equal to half. Okay. So our eccentricity will be one by root two our eccentricity will be one by root two. Now suppose if you want to take b greater than a yeah some student may say sir why you are taking a greater than b. So let me take b greater than a so x square upon a square plus y square upon b square is equal to one. Now this time I am taking b is greater than a. So in this situation what will be our lattice rectum our lattice rectum will be two times a square upon b mind it. Here it was b square upon a here it will be two times a square upon b and in this condition what will be our major axis. It will be since b is greater than a so major axis will be two b. Now as per given condition it in the question it is two into a square upon b is half of the major axis. Right. So here what we get here we get p square is equal to two a square. Okay. Now in this situation our eccentricity will be one minus a square upon b square for this ellipse it was one minus b b square upon a square for this ellipse it will be 1 minus a square upon b square that is 1 minus a square let it be as it is and our b square will be 2 times 2 times a square. So, this will be got cancelled it will be 1 by 2. So, from here also we get eccentricity as 1 by root 2. So, 1 by root 2 will be answered to this question right. Now, let's take this next question question number 4. If the eccentricity of an ellipse is 1 by root 2 then its lattice rectum is equal to eccentricity for ellipse is given as 1 by root 2 and we have to find its lattice rectum. We have to find its lattice rectum ok. So, what is lattice rectum basically? Lattice rectum is 2 times b square upon a right and eccentricity what we can write eccentricity as or you can square it a square is equal to 1 by 2 and e square we can write it as 1 minus b square upon a square is equal to 1 by 2. So, from here we get a square minus b square is equals to a square and this thing whole it will be multiplied by 2 right. So, 2 a square minus 2 b square is equal to a square or we can say a square is equal to 2 times b square this is the relation what we got. So, now putting the value here we will put the value here we will get 2 times b square upon a. So, b square in place of b square what we can say this this is nothing but we can write b square as a square upon 2 right. So, put this value. So, b square upon a or I am putting this b square as a square upon 2 and already a 1 a was here. So, a will get cancelled out this 2 and 2 will get cancelled out. So, we are left with a. So, our latest victim is equals to a and what is a it is the semi major axis semi major axis right semi major axis because here we have considered a is a is greater than b for this image. Therefore, that is why the we have calculated the length of lattice rectum as 2 times b square upon a. So, our lattice rectum will be equal to a semi major axis that is option a ok. Now, moving to the next one question number 5. If the distance between the foci of an ellipse is equal to minor axis then its eccentricity is if the distance between foci of the ellipse means distance between foci is given to us and it is given to be equal to minor axis ok. So, what is the distance between the focuses or distance between foci distance between foci of ellipse is equals to 2 a a right. If you let me draw let me draw a rough diagram for this. So, suppose I am taking this ellipse ok and let me draw the major and minor axis for this ok. So, this is our suppose this is our first foci we already know we all know that ellipse has two focuses right and we call it as this if we call it as s we normally call this focus as s dash and the coordinates of s is what a e comma 0 right and the coordinates of s dash is minus a e comma 0. So, what is the distance between foci it is equal to 2 a e this distance we are talking about this distance. So, this will be 2 a right if the distance between foci of an ellipse is equal to its minor axis ok minor axis. So, here we are considering a is greater than b ok. So, minor axis is this thing basically this this right this will be our minor axis and it is this its length is basically 2 b. So, in as per question it is given that this 2 a e the distance between 2 foci is equal to its a minor axis and what is minor axis that is 2 b. So, this 2 2 got cancelled out we are left with a e is equal to b now we have to find the eccentricity right we have to find eccentricity. So, this will be what can we write here squaring it what we will get a square e square is equal to b square now a square in place of e square I will write 1 minus b square upon a square is equal to b square. So, this a square and this a square minus b square is equal to b square. So, this a square a square which will get cancelled out and we will have a square is equal to 2 b square this one relation we got as per given information in the question right. Now, what I will do further we need to find the eccentricity. So, eccentricity of the ellipse will be 1 minus b square upon a square this is our e square. So, 1 minus b square b square let it be as it is and a square we can write it as 2 b square. So, our eccentricity square will be equal to 1 minus 1 by 2 that is 1 by 2. So, our eccentricity will be 1 by root 2 is it okay? So, same similar type of questions we are getting okay. So, this will be b right. So, I hope everyone has got the answers correctly okay like before seeing this video I assume that all have all the students have tried the questions by themselves first then we then they have come to this video. Okay. So, let us take the next question this is question number 6 it is saying the eccentric angle of a point on the ellipse this whose distance from the center of ellipse is 2 is okay eccentric angle of a point. So, we have to draw a diagram for this. So, let me say this is our ellipse okay and I will draw one auxiliary circle for this ellipse right okay. So, this is our auxiliary circle for this ellipse now I will also draw this axis of the ellipse right. Now, there is one catch in this question which I think many students make mistake on that. So, okay let me first write the names. So, this given ellipse is x square upon 6 plus y square upon 2 is equal to 1. So, here a is greater than b. So, it will be a horizontal axis and its center will be at 0 comma 0. So, this will be our y coordinate and this will be our x coordinate. Now, one point is given on the ellipse whose center from whose distance from the center is 2 okay. So, let me assume that point as this point this point as p okay and its distance is given out to be 2 from the from the center. So, what will be center for this what will be center for this ellipse center will be our origin right it is in the standard form the given ellipse is of a standard form only. So, this distance let me show you first which distance I am talking about or the question is talking about. So, this distance is given to be 2. Now, many students make a mistake and they say that the eccentric that the eccentric angle of this point suppose I am taking it as theta okay. So, this will be our theta the angle made between this OP this I am writing it as angle P O X P O X many students assume this to be theta but it is wrong it is wrong this is not the eccentric angle of the point basically what we will do we will extend this point P okay we will extend this point P and wherever this point will meet the auxiliary circle that angle will be called as theta this angle will be called as theta right. So, let me show you in this diagram. So, this if the eccentric angle of this point is theta. So, the angle this this angle let me write it as Q. So, this angle angle Q O X this will be equal to theta this is the right interpretation this is the correct interpretation this is wrong interpretation. So, please keep it in mind and this OP as per question if you see this OP is given to be equal to 2 it is it is this information is given in the question the distance from center of ellipses 2 right. Now, we have to find this theta basically we have to find this theta. So, if you see this coordinates of P this coordinates of P we can write it as A cos theta and what is it here if you see A square is equal to 6 and this V square is equal to B square is equal to 2 right. So, any point on the ellipse I can assume it as A cos theta that means from here we get A as root 6 and from here we get B as root 2. So, this will be a root 6 or let me write it as A cos theta and A sin theta sorry A cos theta comma B sin theta as of now okay and we will put the value later on if required. So, our OP if you see this OP. So, this is a distance where I am squaring it so that that is square root term will be gone right. So, OP square will be equal to 4 now what will be OP square this will be A cos theta minus 0 chi square I am writing that also and B sin theta minus 0 chi square that will be equal to 4 right. So, from here we get A square cos square theta plus B square sin square theta is equal to 4 right. Now, we know the value of A square that is 6 sorry 6 cos square theta okay and what is the value of B square that is 2 2 and instead of sin square theta what I will do I will write 1 minus cos square theta right 1 minus cos square theta and that minus 4 I am taking 4 also here. So, minus 4 is equal to 0. So, we are left with 6 cos square theta minus 2 cos square theta right. So, that will be 4 cos square theta and plus 2 plus 2 from here and minus 4 that is minus 2 or you can say 4 cos square theta is equal to 2 and here we got cos square theta is equals to 1 by 2. So, this is what we got cos square theta is equals to 1 by 2. So, what will be the value of cos theta? So, cos theta will be equal to plus minus 1 by root 2 this is our cos theta. So, this theta is coming out to be plus minus when does cos theta gives 1 by root 2 that is theta will be equal to cos inverse 1 by root 2 that will be pi by 4 on cos on pi by 4 we get the value of cos theta as 1 by root 2. Now, this Q point can be here also know that is why this is coming out to be now this Q point can be in the fourth quadrant also right. So, in this case this will be our minus theta hence we will get the two values of eccentric angle that is plus minus pi by 4 ok. So, this option A and option B both are correct in this case right hope it is clear to all and you are aware like you got the this theta thing like what is the eccentric angle of a point if that point is lying on the lips. So, that point doesn't make like the line joining the point and center and the angle between that what you say that line and the positive direction of x axis doesn't give the this eccentric angle. However, we extend this point P right we extend this point P and we allow to cut it as point cut it at a point on the auxiliary circle. Now, when this Q point is joined with the center of the ellipse or the center of the auxiliary circle since both these are auxiliary circle and this ellipse are concentric. So, the angle between this OQ and positive direction of x axis or angle between this OQ and OX will give us the eccentric angle right. So, this was for this question now let's take the next one question number 7 ok. So, if tan alpha tan beta is equal to minus A square upon B square then the chord joining the two points alpha, beta on the ellipse x square upon A square plus y square upon B square equal to 1 will subtend a right angle at ok like one chord is given here right. So, let me draw one diagram for this. So, this is our chord let me draw auxiliary circle also for better understanding. So, basically if you see we are given with one chord. So, let me take this chord as PQ ok. So, I am taking this chord as PQ now the eccentric angle right the eccentric angle of this point P is alpha and the eccentric angle of this point Q is beta and in question it is given as tan alpha into tan beta is equal to minus of A square upon B square. Again here also if you see what will be angle alpha basically this will be angle alpha right this will be angle alpha just for sake of understanding I am showing you and similarly for beta if you extend this point ok and if you join this point then this will be beta. So, hope everyone is clear on this right. So, we will use this information wherever required, but in question it is given that this tan alpha into tan beta is equal to minus A square upon B square right. So, let me draw it neatly ok everything is getting developed off ok then leave it leave it we will see if required we will see. Now, this chord is making then the chord joining two points like we have to find this chord is making a right angle at which point like this is our center of the circle sorry center of the circle and center of the ellipse or center of auxiliary circle same thing because both this auxiliary circle and this ellipse are concentric ok and end of major axis like this and this is the end of minor axis. So, these four options are given here and we have to check where we are where this chord is making an angle of 90 degree or subtaining an angle of 90 degree. So, let me check with this let me check with point O ok. So, the coordinates of P will be it is a centric angle is alpha. So, it will be A cos alpha right and B sin alpha and what will be the coordinates of Q it is a centric angle is beta. So, it will be A cos beta comma B sin beta is it ok now let me join these two points let me join these two points this O and P ok and let me join these two points O and Q ok. So, now what will I will do I am trying to find the slope of O P right the slope of O P will be basically y 1 minus y 2 that is B sin alpha minus 0 upon A cos alpha is it clear. So, it will be nothing but B upon A sin alpha upon cos alpha will give me tan alpha. Now what will be slope of O Q what will be the slope of O Q it will be B sin beta right upon A cos beta is it ok to everyone. So, it will be B by A upon tan into tan beta right now what we can do we can multiply both these slopes slope of O P into slope of O Q because it will contain tan alpha into tan beta and we are having the value of tan alpha minus tan beta. So, if you see this slope of O P into slope of O Q that will be equal to what B upon A tan alpha into B upon A tan beta right. So, it will be B square upon A square and tan alpha into tan beta what is this value equal to as per question it is equal to minus of A square upon B square. So, this will be B square upon A square into minus of A square upon B square right. So, this B square B square got cancelled this A square A square got cancelled and this is equal to minus 1. So, the slope of O P into slope of O Q what we got we got it as minus 1. So, this angle basically this angle is equal to 90 degrees. So, what does it imply basically what does it imply this O P is perpendicular to O Q and what is O? O is the center of the circle O is the center of the circle. So, it is making an angle of 90 degree at the center of the ellipse or the center of the circle O is the center of ellipse right. So, this will be our answer this option B is correct right. So, question number 7 is done we take the next one question number 8. So, if the centricity of two ellipse x square upon 169 plus y square upon 25 equal to 1 and x square upon y square plus y square upon B square equal to 1 are equal ok. So, two ellipse equations are given to us and it is saying that the centricity of both ellipses are equal then we have to find this ratio A upon B ok. So, let us go with whatever is given in the question. So, our first ellipse is given as what x square upon 169 plus y square upon 25 is equals to 1 right this is our first ellipse and our second ellipse is what we write here only x square upon E square plus y square upon B square is equals to 1. Now, what will be the centricity for this ellipse what will be the centricity for this ellipse let me represent it as a small e 1. So, here A is greater than B right here A is greater than B it means the centricity we can give its centricity as e 1 square as 1 minus B square upon A square. So, this will be 1 minus what is B square 25 and what is A square 169. So, 169 minus 25 how much 144 54 64. So, it will be 144 upon 169 right. So, we got the value of e 1 as 12 upon 13 is it ok we got the value of e 1 as 12 upon 13. Now, what will be the centricity for this ellipse it will be 1 minus B square upon A square is anything known upon for this A and B no. So, same thing we can write it as A square minus B square right under root of A square minus B square upon A is it ok now in the question it is given given as both this both the centricities are equal. So, given that in the question it is given that e 1 is equal to e 2 now square it also it will give the same. So, I am writing this e 1 square is equal to e 2 square ok. So, e 1 square we got as 144 upon 169 and what is our e 2 square e 2 square why I have put in why I have put this square root here. So, basically this is what we got e 2. So, anyhow I will take e 2 square only I do not want to deal with this square thing. So, this will be basically e 2 square will be our A square minus B square upon A square is it ok or we can write it as or we can write it as this this thing same thing we can write it as 144 upon 169 is equal to 1 minus B square upon A square is it ok 1 minus B square upon A square that we can further write it as B square upon A square is equal to 1 minus 144 upon 169 that will be 169 and 169 minus 144 will be equal to 25. So, our ratio B upon A will be equal to under root of Pachis upon 169 that is nothing but B upon A is equal to 5 upon 30 5 upon 30 ok. So, till now we are getting just basic questions formula based questions right. So, it is not that much difficult I have seen many students like who skip this ellipse and hyperbola part because till parabola they like they like do the questions do the practice the questions but after this they lift like they leave this coordinate geometry part right. So, it is not like that it is this ellipse and hyperbola part is also very easy. So, do not worry we will take our journey to this ellipse and hyperbola also. So, you just keep on practicing the questions and if any difficulty you can ask in the class or in the center itself. So, we can discuss this and this Arihant's solution I am providing you all. So, it is not that much tough this ellipse part right. So, now let us see the next one. The ratio of the areas of triangle inscribed in the ellipse x square upon A square plus y square upon b square equal to 1 to that of the triangle formed by the corresponding points on the auxiliary circle is 1 by 2 ok. Now, here the question is asking something on auxiliary circle and it is saying that the eccentricity of the ellipse ok. So, let me plot the diagram for this ok. So, so, this is our ellipse let me draw the auxiliary circle for this ellipse ok ok a little bit small yeah it will work. So, this is our ellipse oh sorry sorry take a straight line. So, this is the axis of the ellipse this will be basically coordinate axis only ok because the center of this ellipse is origin. So, this is our y axis this is our x axis. Now, now let us look at the triangle the ratio of the areas of the triangle inscribed in the ellipse ok. So, we have to draw a triangle which is inscribed in the ellipse ok and further we have to draw one more triangle which is inscribed in the auxiliary circle because the ratio of the areas of both these triangles are given and we are asked to find the eccentricity of the ellipse ok. So, we have to draw one triangle that is inscribed in the inscribed in the ellipse. So, can you figure it out like which will be the most simplest case because I don't want to go in the difficulties. So, what I will do I will draw the triangle sorry I will draw the triangle which is inscribed in the ellipse ok which is inscribed in the ellipse I will take this points this vertices of this ellipse as the two vertices of triangle and I will take this point. So, let me first write this. So, for this ellipse if you see this will be let me name it as a and this is our a dash. So, this a a dash is the major axis of the ellipse right and let me write it as point B and B dash ok and what I am trying to convey what I am trying to do is I will be making triangle taking these points a a dash and B as the vertices of the triangle like why I am taking this because we we know like let me draw this thing also because we know the vertices of these three points is it ok. So, this ellipse if you see this ellipse is x square upon a square ok plus y square upon b square is equal to 1. What I am assuming I am assuming this to be a horizontal ellipse that means the coordinates of this b this coordinates of a will be a comma 0 is it ok and what will be the coordinates of this a dash it will be minus a comma 0 it will be minus a comma 0 and what will be the coordinates of this point B it will be 0 comma B and we can either take this B dash also its coordinate is 0 comma minus B. So, that triangle can be formed with this b a a dash taking as the vertices or b dash a a dash taking as the vertices. So, it is not like that you can take a point anywhere on the ellipse and corresponding point on the auxiliary circle but for the sake of simplicity I have taken this point as B. So, I hope it is clear. So, this is one triangle ok which is inscribed in the ellipse this let me write it as triangle this triangle a dash a dash b a that triangle is inscribed in inscribed in ellipse right this triangle is inscribed in ellipse. Now, we have to draw one more triangle ok and that triangle should be that point on the auxiliary circle should be corresponding point of B. So, what will be the corresponding point of B it will be some way here. So, let me change the color change the color of this triangle. So, that corresponding point will be here right. So, let me name it as name this as C ok. Now, what will be the coordinates of this C this is the center this is the center of the ellipse and center of the and what you say auxiliary circle. So, if you observe this O C this O C will be equal to radius radius of the auxiliary circle radius of the auxiliary circle and it will be equal to O A and it will be equal to O A that will be nothing but equal to A. So, this O C will be equal to A and this triangle this triangle was inscribed in ellipse and this triangle A dash A dash C A this triangle if you see this triangle is inscribed in the this triangle is inscribed in the auxiliary circle right. So, as per given information in the question the ratio of the areas of these two triangles right. So, let me write the area of this triangle right area of triangle A dash B A. So, it will be half times base into height. So, what is the base base is 2 A and what is the height height is O B height is O B and that height will be equal to B right. So, it will be equal to AB and what will be the area of triangle A dash C A A dash C A that will be half into base base is same 2 A and what will be height here height in this case will be equal to A that is equal to A square. Now, as per given information what we can do we can write this area of triangle A dash B A that is AB upon A square is equal to 1 by 2 ok. So, from here what we got we got this ratio B upon A is equal to 1 upon 2 ok. Now, we have to find the eccentricity of this triangle this ellipse. So, we very well know that the eccentricity squared is equal to 1 minus B square upon A square that will be 1 minus B by A whole square that is 1 by 4 B by A whole square will be equal to 1 by 4 or that will be equal to 1 minus 1 by 4 will be 3 by 4 hence our eccentricity we get as root 3 upon 2. So, the eccentricity for this ellipse will be equal to root 3 upon 2 right. So, this question involved the concept of this auxiliary circle. So, what will be this corresponding point of B? It will be moving parallel to the axis right it will be moving parallel to the axis and wherever this point joins the auxiliary circles that will that will be the corresponding point of this B. So, hope it is clear to all. So, now take let us take the next one oh my god I think I forgot to paste the questions ok no worries just give me just give me a few seconds I will add it right I will add it yeah. So, here we go I have added the questions right. So, after question 9 ok I have added the full questions of this exercise. So, yeah please take please take this question number 10 it is saying if P sq is a focal chord of the ellipse 16x square plus 25y square equal to 400 such that sp is equal to 16 then the length of sq. So, basically it is asking the length of the the segments of the focal chord. So, for better understanding let me draw this ellipse ok let me draw this ellipse and I will draw the axis also for this ellipse ok and let me draw one focal chord. So, I am drawing this focal chord pq ok. So, this is our focal chord pq. So, what does this focal chord means means this pq will pass through the will pass through the focus one of the focus of this ellipse. Now, let us try to write rewrite this ellipse in our standard form. So, basically if you see the 16x square plus 25y square is equal to 400 right. So, dividing both sides by 400 what I will get x square upon 16 to 32 16 5s are 80 plus y square upon 16 right is equals to 1. So, basically the given ellipse is nothing but this ellipse where ok. So, here a square is equal to 25 right I hope this is this all things are clear to everyone. So, b square for this ellipse will be equal to 16 ok I have written it we will use it whenever required in the question ok. Now, this sp length now this sp length is given out to be this sp length is given out to be 8 if I am not wrong no it is given out to be 16. So, this sp is 16 units and we have to find the length of sq we have to find the length of sq. Now, one very important property one very important property of this ellipse let me remind you all that this pq this pq is the focal chord what does it mean what does it mean the semi lattice rectum semi lattice rectum will be the harmonic mean harmonic mean of a segments of focal chord segments of focal chord focal chord. This is the property I am talking about this is the one of the important property of ellipse the length of semi lattice rectum will be will be the harmonic mean of the segments of focal chord. Now, what does it mean basically the semi lattice rectum if you observe ok let me write the details for this ellipse. So, this is our x axis this is our y axis ok and what will be the length of semi lattice rectum ok semi lattice rectum if you see our lattice rectum for this for this ellipse since a is greater than b it will be 2 times b square upon a right. Therefore, the length of semi lattice rectum will be half of this 2 times b square upon a this 2 and 2 we got cancelled out it will be equal to b square upon a right. So, as per given property as per this property what can we say as per this property what can we say this b square upon a and or you can say you can write in this way this sp ok this sp the segment of the focal chord sp this b square upon a and this sq these 3 things will be in harmonic progression these 3 things will be in harmonic progression. The same property the same property what I have mentioned here I have written the same thing, but in a different way. So, from here if you see this b square upon a will be equal to 2 times 2 times sp into sq upon sp plus sq is it ok is it ok to everyone. So, suppose this a b c is in harmonic progression. So, what we used to write we used to write this b as 2 a c upon a plus c 2 a c upon a plus c right this we have already learned in our sequence sensory chapter. So, this b is the harmonic main of this a and c and its value is given as 2 a c upon a plus c same thing I have written here. So, if you see here we know the value of this sp. So, this will be ok let me write here only let me write in the next line ok. So, this b square ok we can put the value of b square and a also. So, b square you see b square is nothing but 16 ok and what is a a is equal to 5 and 2 this will be 2 times sp sp is given out to be 16. So, put that value 16 this will be sq and sq what we need to find it out. So, sp again 16 plus sq hope it is clear. So, this 16 and this 16 will get cancelled out. So, we will have this 16 plus sq 16 plus sq cross multiplying I am cross multiplying and this will be 10 times of sq is it ok. So, from here what we get 10 minus that is 9 times of sq 9 times of sq is equal to 16 or our sq length will be equal to 16 upon 9. So, this will be the length of sq 16 upon 9. So, option b is correct. So, what is the property the semi-letters rectum the semi-letters rectum of ellipse will behave as a harmonic mean of the segments of any focal chord of the given ellipse right. So, in this case pq is the focal chord right and this sp and sq are the segments of that focal chord. So, using the property we have written this thing sp b square upon a and sq will be in the harmonic progression right. So, I advise all of you to remember this property sometimes j directly asks questions based on the properties itself ok. So, yeah now let us take this question let p be a variable point on the ellipse this ok. So, it will be a horizontal ellipse because a is greater than b here with foci s and s test if a be the area of triangle p is s test then the maximum value of a then the maximum value of a ok. So, let me draw the ellipse first ok. So, this is our ellipse these are the says of the ellipse and p is a variable point on the ellipse right p is the variable point on ellipse. So, let me take this point as p and s and s test. So, we have to make a triangle with the point p and both the focuses ok. So, let me assume this focus to be here right so this is our focus for one focus and this is our second focus ok. So, we have to basically we have to find the maximum value the area of this triangle right. So, this is our point p this is our first focus is and this is our second focus is dash right and we are talking about area of this triangle and p is any variable point on the ellipse. So, what is our ellipse our ellipse is x square upon 25 plus y square upon 16 is equals to 1 right and p is any point. So, normally what we do for assuming any point on the curve we take we normally take the parametric form right. So, here if you observe our a square is equal to 25 or you can say a is 5 and b square is 16. So, b is equal to 4 right. So, we take this point p as a cos theta that is 5 cos theta 5 cos theta and b sin theta that is 4 sin theta. Now, let me remind you what is the theta this theta is not this angle from center to point p this will be corresponding point on the auxiliary circle when we join that point from auxiliary circle to the center then the angle form will be our theta. So, ok don't start that point here, but since I have written this in parametric form I thought to remind you. So, this will be our coordinates of this point ok. Now, I have to find I have to find the this area of this triangle. So, let me drop one perpendicular here on this major axis of this triangle right and it is 90 degree here. So, if you see if you see what will be the area like what will be the area of triangle of triangle p s s dash right and this is let me write this area mention this area as a it will be equal to half times base. So, base means what s into s dash and into height. So, you can name it you can name this point as point m right. So, this I am writing it as p n ok and what will be this s into s dash distance between the focuses distance between the two focuses for this triangle it will be basically 2 times a e normally what we write the coordinates of s is basically a e comma 0 right a e comma 0 and the coordinates of s dash is minus a e comma 0 hope this is known to you. So, the s s dash is equal to 2 times of a e now a is 5, but what is e what is e. So, we know a square is equal to 1 minus of b square upon a square here in this case a is greater than b right this a is greater than b. So, 1 minus b square means what 16 upon a square is 25. So, this will be 25 minus 16 9 right. So, basically the value of e is 3 by 5 ok. So, the distance between the focuses will be equal to 2 a what is a a is 5 what is e e is 3 by 5. So, 5 5 will get cancelled out this will be 6 right. So, our area triangle a will be equal to half into 6 into p m this height right now what is this height this will be the y coordinate this p m will be equal to y coordinate right and what is the y coordinate of this point p this is 4 sin theta right. So, now I am writing p m as 4 sin theta. So, this will be nothing but 2 3 are 6. So, 3 into 4 will be 12 sin theta. So, this is the area of triangle 12 times sin theta 12 times sin theta right. Now, we need to maximize this area we need to maximize this area. So, what should I do I will differentiate this area with respect to theta. So, d a upon d theta will be equal to 12 times of cos theta right and for maximum or minimum what we do we put this d a upon d theta equals to 0. So, basically what we do cos theta is equals to 0 for that we have theta is equal to 90 degree is it okay. Now, this can be the point of maxima or this can be the point of minima. So, what I will do I will double differentiate it. So, this will be d square a upon d theta square this will be equal to minus 12 sin theta now put the value of theta here. So, it will be at theta is equals to 90 degree this is coming out to be negative right. This will come out to be at theta equal to 90 degree our double differentiation will be negative. Hence, this value is this point is the point of maxima this will be point of maxima and this is what we need. So, at theta equal to 90 degree at theta equal to 90 degree a will be maximum the area will be maximum right. So, basically what does it mean this theta has to be on the minor axis right this point p this point p has to be on this minor axis. But let me remind you once again this is not the theta this point o o p angle o p x angle o p and s you can say o p s is not theta what is theta the point on corresponding point of p on the auxiliary circle. So, at theta 90 degree a will be maximum and what will be the value of a what will be the value of a we know a is equals to 12 sin theta. So, area will be or we can say maximum area will be 12 sin 90 that is equal to 12 square unit right this will be our final answer. So, maximum area will be at theta equal to 90 degree and the value the numerical value of that area will be 12 square units right. So, this is our option a its option a is correct for this one. Now, coming to the next one let's read what this question is saying if s and h dash are the foci of an ellipse of major axis of length 10 units and p is any point on the ellipse such that the perimeter of the triangle p s s is 15 units then the eccentricity of the ellipses ok. So, this is also a nice question and we will be seeing one another definition of ellipse through this question we can figure it out ok. Let me first draw let me just draw one rough sketch for this oh yeah this is our ellipse ok and what I have to do ok. So, only this major axis I need I need only this major axis ok now what is the point s and h dash are the foci. So, this is our point s ok this is our point s and suppose this is our point s dash these are the foci of the ellipse right and p is any point on the ellipse. So, let me assume this point let me assume this point p on the ellipse right such that the perimeter of p s s ok means the question is asking to make one triangle joining the points this p s s right ok. So, let us join this points. So, this this and this point. So, this is triangle ok. So, basically if you see p is any point on the ellipse and the perimeter of this triangle right the perimeter of triangle. So, let me write what is the given information in question. So, a perimeter of triangle p s s dash is given to be 15 units ok this one information is given and the length of major axis is also known to us. So, a length of major axis. So, this length of major axis is 2 a and that is also known to us that is 10 units. So, these two informations we are provided with and we have to find the eccentricity of the ellipse we have to find the eccentricity of ellipse. So, if you know the sum like this sum of this sp sum of sp means joining the point on the ellipse with one focus and joining the same point with another focus right that is the second focus that s dash p s dash p is a constant for any constant this is valid or true you can say true for all the ellipse true for all the ellipse and this constant this constant is equals to the value of this constant is equals to major axis right this value of constant is equals to major axis. So, this if this thing if this information is known to you you will be able to crack this question. So, whether you know it or not I don t know, but in the same way we can define the ellipse also. So, what is an ellipse like ellipse is the locus of this point p locus of point p right whose sum of whose distances from the two points sum of whose distances from the two fixed points is equal to constant and that constant should be greater than the distance between the two fixed points like this constant this constant should must be this constant should must be greater than this constant should must be greater than s s dash. So, I am talking about the distance right this constant should must be greater than s s dash right this distance okay. So, in this way we can define the we can give the another definition of ellipse right. So, hope it is clear to all now in the question it is saying in the question it is saying the perimeter of triangle p s s right. So, if you see this s p this is one of the side of the triangle this h dash p right this h dash p and plus s h dash should be this is given to be equal to 15 right. Now, this s p plus h dash p as per given information what we have written here this should be equal to major axis means what this will be equal to two way and what is this distance s distance between this s and h dash this will be two times a right the distance between two focuses this is equal to 15 now we can take two way common from here and we will be left with one plus a and that will be equal to 15. So, now we know the value of this two way also. So, let us put it out. So, from here if you see one plus e will be equal to 15 upon two way that is 10 right. So, from here we get e is equals to three by two right minus one. So, one by one point five minus one that is point five that is e is equals to one by two right. So, but we are not having this option one by two. So, please check it out whether we have done some mistake or what. So, taking two a common okay okay it is there no it is there in option a so option a is correct two way we have taken common one plus e one plus e will be equal to 15 upon 10 that is three by two minus one yeah it is okay. So, option a is correct. So, this question number 12 is done okay now we have entered into this subjective part right. So, let me have some water first yeah. So, please have a look at this find the lattice rectum eccentricity coordinates vertices axis center of ellipse okay. So, one ellipse is given here and we have to find all the these information about this ellipse. So, let me write it first. So, it is basically four x square plus nine y square minus eight x minus 36 y plus four equal to zero. So, my target will be to make it in form of standard ellipse like a standard form of ellipse I will try to make it in the standard form. So, I will take common from these x terms what can we take common from here we can take four common right and this will be x square x square minus two x is it okay and from here we can take nine common. So, we will be left with nine is nine y square and minus four y is it okay and this plus four I will take it to right hand side. So, this will be minus four right now what I will do I will try to make these things as the perfect square right I will try to make these things as the perfect square then only I can approach to standard form right. So, this will be four times x square minus two x plus one right and this will be nine times y square minus four y plus four so what we have added here we have added generally actually four here not one it will be multiplied by four no so minus four was there from last step as it is plus four and here we have added 36 so 36 plus four we have actually added 40 so we have added 40 in the right hand side also now if you observe here it becomes our x minus a whole square plus nine into y minus two whole square and that will be equal to this plus four minus four will get cancelled out this will be 36 only right now divide both sides by 36 what we will get we will have x minus a whole square upon nine plus y minus two whole square upon four is equal to one right so from this equation we have changed the form of ellipse and we got it as a standard form right now we can say now we can compare it with our standard form of ellipse that is x square upon a square plus y square upon b square is equal to one right now we will compare it compare it with our standard ellipse right so let me bifurcate it and so one thing if you observe one thing is very common like very uh usual what is the center what is the center for this ellipse this will be one comma two right because this for this circle this x minus one should be equal to zero so from here we got x is equal to one and y minus two should be equal to zero so from here we get y is equals to two so this is the this is the center of ellipse center of ellipse i think center is asked for this question right so we uh got this center of ellipse okay now what can we find next comparing it with our standard one we see here a square is equal to nine right and b square is equal to four so here you see a is greater than b so it will be basically a horizontal ellipse horizontal ellipse okay so what next we can find the center city right we can find the center city so e square will be equal to one minus b square upon a square so one minus what is b square b square is uh four and what is a square a square is nine so this will be basically five upon nine so e square is uh nine minus four five upon nine so we got eccentricity as uh root of five root five upon three so this will be our value of e that is eccentricity right now we got the center we got the eccentricity what else we can figure it uh now so uh should we draw the ellipse first okay uh let me let me draw that so this will be our horizontal ellipse and this will be our axis right and so this is our eccentricity where will be our focus our focus will be somewhere here our one focus and our other focus will be somewhere here now what is the center for this center is if you see the center of this ellipse is at which point one comma two right the coordinates of center is basically one comma two so this major axis if you observe this is our major axis now since this is the horizontal ellipse because a is greater than b so basically our major axis says y equal to two y equal to two is will be our major axis right so the y coordinate of both focuses will be two y coordinate of both the focuses will be equal to two that is sure now we have to figure out the x coordinate of the focuses now how to find it if you know for any standard ellipse the distance between this the distance between focus and center is a right is a so if you see the coordinates of s the coordinates of s will be one plus a comma the y coordinate will be common y coordinate will be two why because this the focus will lie on the major axis major axis itself and what is the major axis major axis is y equal to two in this case in this particular case now we know the value of a we know the value of e so we can easily figure figure it out the coordinates of s now why one is added that if you see the center itself is one no the center itself is one if it if it would have been 0 comma 0 it would have been a but since it is a one the x coordinate of center is one so it will be added on so basically this s will be one plus a right comma two and what will be our s dash s dash will be nothing but one minus a comma two right now what is a if you see what will be the value of a so our a is three so a e will be basically three into root five upon three so this will be equal to root five only okay so we got the coordinates of s s we got the coordinates of s s one plus root five comma two and the second focus coordinates will be one minus root five upon comma two so this is our focus coordinates of focus so we got the center we got the centricity coordinates of foci we got coordinates of vertices coordinates of vertices right so vertices basically this a and a dash is called the vertices like where the ellipse is intersecting with the major axis that points are called as vertices right so the coordinates of a will be for our standard case for a standard case the coordinates of a used to be a comma zero right and a dash used to be minus a comma zero right so in this case what will be the case so basically this distance this c a distance no this c a if you observe this c a should be equal to a this a c a should be equal to a now what is a here c a will be this distance will be equal to a and we know the value of a as we know the value of a as three from here we already know this value of a as three so this will be three this c a distance will be three so our coordinates of a will be the coordinates of a will be one plus three right so this will be four comma two and what will be the coordinates of a dash again this c a c a dash should also be equal to three right the c a dash should also be equal to three so a dash will be minus two right x coordinate will be minus two and y coordinate will be two because y coordinate is not changing y coordinate is the focal major axis and that is y equal to two so we got the coordinates of vertices also right we got the coordinates of vertices also now what what we are left out with if you see latest rectum no okay so latest rectum if you say latest rectum will be two times b square upon a so this will be two times what is b square b square is four and what is a a is three so latest rectum the length of latest rectum will be equal to eight by three units okay so latest rectum also we got eccentricity is root five by three coordinates of pokai we have given coordinates of vertices we got length of the axis okay so length of axis if you see major axis major axis the length of major axis will be equal to two times a that will be equal to two into three that is six units and our minor axis will be equal to two two b so two into b what is b b is equal to two right so two into two four units so this will be the length of axis and center we have initially find it out so for the given ellipse we have achieved all the required values so hope this is clear to all what I have done I have first changed like I have first changed the equation into a nice looking form or we can say a standard form so that we can compare with our with our standard form of circle a standard form of ellipse right so this is what okay you we have four or five questions more okay anyhow I will complete it by today only so next question question number 14 it is saying the distance between the pokai of an ellipse is 10 okay and it's let us rectum length is 15 so find its equation referred to its axis as axis of the coordinates okay so the axis of the ellipses our coordinate axis it is mentioned in the question itself so let me write let me assume let me assume let the ellipse be let the ellipse be x square upon a square okay plus y square upon b square is equal to one because it is given that the coordinate axis are the axis for this ellipse now what are the given informations the distance between pokai of an ellipse so distance between pokai is 2 ae and it is given to be 10 so this is one information what we got and the other information is its lattice rectum the length of lattice rectum is two times b square upon a is given to be 15 right and we have to find the equation of the ellipse so okay what can we do this is if you see here this ae is equal to 5 right and ae we can write it as a square minus b square this is equal to 5 now square it it will be a square minus b square is equal to 25 right and from this equation what we get 2 b square 2 b square is equal to 15 a 2 b square is equal to 15 a and okay so basically we can make a quadratic in a or b so this a square minus 25 will be equal to b square what i will do i will put this value here i will put this value here so this will be two times what is b square this is a square minus 25 minus 15 a is equals to zero so this becomes 2 a square minus 15 a and minus of 50 equal to zero right so 15 to 200 20 and 5 we can split the middle term so 2 a square minus 20 a plus 5 a minus 50 equal to zero so 2 a a minus 10 plus 5 into a minus 10 is equals to zero so a minus 10 into 2 a plus 5 equal to zero so from here we get two values of a a is equals to 10 and a is equals to minus of 5 by 2 right so this is what we got the values of a so say when a is equals to 10 when a is equals to 10 what will be our b square so from here we give we take 2 b square is equal to 15 into a a means 10 so from here we get b square is equal to 5 15 into 5 that is 75 so from here we get b square is equal to 75 when a is equals to 10 and when a is equals to minus 5 by 2 we get b square is equals to 2 times b square is equal to 15 into a that is minus 5 by 2 that is we get b square is equals to minus 75 upon 4 that is not possible so rejected so b square cannot be negative no so basically this value will be rejected so what will be our equation so our equation becomes this this is what we supposed right initially the equation of the parabola so this will be x square upon a square so if a is 10 what will be our a square it will be 100 and plus y square upon b square is 75 that is equals to 1 right so this will be our answer this will be our answer if you further want to expand it you can do what will be the lcm of 100 and 75 it will be 300 so check lcm 300 it will be 3 x square plus 4 y square is equals to 1 or you can say 3 x square plus 4 y square is equal to 300 same both these equations represent the same circle same ellipse so you can live here also or you can expand it's up to you so this will be our equation of the ellipse this will be our equation of ellipse now coming to the next question it is saying find the equation of the ellipse okay again we have to find the equation of ellipse whose axis are parallel to the coordinate axis having its center at the point is 2 comma minus 3 one focus at this point and one vertex at this so this time if you notice here the coordinate axis are not the axis of the ellipse now the axis of the ellipse are parallel to the coordinate axis okay so like let me draw one ellipse okay for better understanding of this question so basically this is the ellipse given here and since it's coordinate axis are parallel no it's coordinate axis means coordinate axis are parallel to this and its focus is 2 comma minus 3 right 2 comma minus 3 so basically its center will lie in fourth quadrant so I can draw coordinate axis in this way is it okay so if you see what I have done this is our x axis okay and this is our y axis or you can say y dash axis and this is our fourth quadrant basically this is our fourth quadrant why this will be fourth quadrant because its center of the ellipse is given to be 2 comma minus 3 right so let me write this center as O and its coordinate is given as 2 comma minus 3 and since its axis is parallel right its axis is parallel to the coordinate axis it will be if you observe the where does this focus line like one focus is given here no the one focus whose coordinates is 3 comma minus 3 and one vertices is also given here so let me write it as a its vertices the coordinates is given as 4 comma minus 3 so basically if you observe this line is this line is major axis because this vertices this vertices this vertices and we take the vertices of ellipse as the point of intersection of ellipse and major axis one more hint is given here the focus lies on the major axis itself the focus should lie on the major axis itself hence we can conclude that it will be a horizontal ellipse so i hope it is clear to all so our major axis is nothing but y is equals to y is equals to minus 3 what is this line this is y is equals to minus 3 so this will be our major axis now what is asked in the question we have to find the equation of the ellipse we have to find the equation of the ellipse so basically this will be an ellipse with shifted origin this will be ellipse this will be ellipse with shifted origin right now for shifted origin how can we write the equation of ellipse it will be x minus 2 whole square right upon a square and plus y minus this thing so y plus 3 whole square y plus 3 whole square upon b square and that should be equal to 1 i have written it in the extended form itself right so this will be our equation of ellipse now if we can find the value of this a and b or a square and b square we can easily do the question we can easily write the equation so if you see if you observe the this distance what is this distance always the x coordinate of o is 2 and it is 3 so this distance is 1 right this distance is one unit and if you observe this thing this distance between s and a the x coordinate of s is 3 and the x coordinate of a is 4 so this is also one unit basically right so what can we say what can we conclude from here this oa if you see this oa is major axis this oa is major axis and what is this distance oa this will be 2 right this one unit plus one unit this will be equal to 2 and this will be equal to length of major axis okay basically it is not major axis it is semi major axis or we can say half of major axis half of major axis right why because this a a dash will be major axis this a a dash will be major axis and but this oa will be half of major axis so major axis will be 2a into half so basically we got the value of a the value of a is 2 right the value of a is 2 now we found this value a we have to find the value of b now how can we derive how can we get the value of b so what can we do what can we do here do we know the eccentricity if you see this o s no this o s distance will be equal to a right and what is this o s distance this is one and a is 2 e means 2 e a we got this value e so from here we got eccentricity has 1 by 2 right now what will be the eccentricity eccentricity is squared is equal to 1 minus b square 1 minus b square upon a square so e square will be 1 by 4 okay 1 by 4 is equals to 1 minus b square upon a square a square we can write it as 4 so from here if you see b square upon 4 will be equal to 1 minus 1 by 4 will be 3 by 4 okay so from here we get b square is equal to 3 so if this is known we can easily write the equation what will be our equation of ellipse x minus 2 whole square upon a square a square is 4 this is a we need to find we need to write a square and this will be y plus 3 whole square upon b square b square is 3 so this will be equal to 1 so this will be the required equation of ellipse now again if you want to expand you want to open the squares and write it you can do but I am living here only I am living here only further you can do it by itself yourself okay so this ellipse is with the shifted vertex shifted origin right generally we deal with center as the center or a origin as the center but this we will be having with shifted origin and for shifted origin this will be our approach to write the equation of ellipse so question number 15 is done now let's take the next one find the equation of ellipse again we have to write the equation of ellipse whose foci are this and this and whose semi minor axis is root 5 okay so basically distance between distance between foci are given so what will be the distance of foci so this is two way distance between the focus and what will be the distance if you observe the y coordinates of both the foci are three right so from here you can see y equal to 3 will be your major axis since y coordinate is not changing for both the focuses it will be horizontal it will be horizontal horizontal ellipse we can see so we can write it as we can assume the equation of ellipse as x square upon a square plus y now whether the center is origin or not that we have to see so should I draw okay let me let me draw this rough sketch for this ellipse okay I think this will be also the case of shifted center why because this one focus is given as this is our focus first focus it's it is given as 2 comma 3 right it's it will this focus will be in the first quadrant and this is our a dash right and whose coordinates are minus 2 comma 3 minus 2 comma 3 so this is our major axis y equal to 3 right this line y equal to 3 will be our major axis now let me take this point as oh which is the center of this ellipse so this oh should be the midpoint of this right oh should be oh will be the midpoint midpoint of s s dash right midpoint of s s dash so can we find can we find the coordinates of oh so the coordinates of oh will be minus 2 plus 2 upon 2 and 3 plus 3 upon 2 so the coordinates of oh will be basically 0 upon 3 right so the coordinates of oh will be 0 upon 3 so we are having this thing so we know the center of the ellipse right and one more information is given here one more information is given here from here if you see semi minor axis semi minor axis so 2 b is our minor axis and what will be semi means it's half so half of 2 b is given to be root 5 so basically we know the value of b b is equals to root 5 okay and here if you see here if you see this oh s oh s will be equals to oh s the distance oh s is equal to 2 right and this will be equal to a times e a times e so should can we find the value of e so okay a e oh s the distance oh s will be a e okay and that is equal to 2 so but how can we find here like what information we can take out from here this oh s is given to be 2 so our a e will be equal to 2 okay so one okay let me so one thing is sure the center is no no so let me write the equation of a ellipse okay let me first write the equation of ellipse so x minus 0 whole square that will be x square upon a square I don't know the value of a till now now what will be the y coordinate of center it is minus 3 sorry 3 so we have to write this y minus 3 whole square and upon b square okay b square we know b square is equal to 5 it is given in the question itself so y minus 3 whole square upon 5 is equals to 1 so only thing where we are stucking is this a so how to find the value of a how to find the value of a so we got the center okay and this a e is equals to 2 okay this is y equal to 3 no so this oh a let me take this point as a okay so this will be oh a will be equal to basically oh a will be equal to a right but how to find the coordinate of a then a basically actually actually this video is getting longer no so I mean okay just let me I have some water first okay got it I think it is almost I think it is almost two hours now for this video anyhow this a no this a we can write it as a under root of a square minus b square and that is equal to 2 okay now we square it we will have a square minus b square is equal to 4 now we know the value of b square that is 5 so a square minus 5 is equals to 4 so our a square will be equal to 9 5 plus 4 9 so finally our equation of ellipse become x square upon a square that is 9 and plus y minus 3 whole square upon b square b square is 5 that is equal to 1 so this will be our equation of ellipse now again if you want to expand open the squares and expand it it's up to you so yeah we are done with this question I think how many questions are left for this two questions we are having two more questions right for this for this exercise so what is given here so that the equation this represents an ellipse and find the eccentricity of the ellipse okay so one equation is given here and we have to show that this equation is an ellipse so let me write the equation first 10 times 10 x minus 5 whole square plus 10 y minus 5 whole square is equal to 3 x plus 4 y minus 1 whole square now what can we do again we we should take the basic basic of our ellipse like okay take this 5 common here we will have 5 square will get out means comes out of the bracket so this will be 25 5 this 2 x minus 1 whole square again we can take 5 common so after coming out of the bracket it will become 25 and 2 y minus 1 whole square is equal to 3 x plus 4 y minus 1 whole square let it the expression in the right hand side let it be as it is so again what I will do I will take 2 common from here and I will put this 25 to this right hand side so 2 if I take 2 common I will get x minus 1 by 2 whole square plus 4 into y minus 1 by 2 whole square and that will be equal to 3 x plus 4 y minus 5 this 25 I am putting here this 25 I am taking common I am putting here right so this will be basically 5 I can write it as 5 and this whole square okay so further what we can say this will be x minus half square plus y minus half square again I will take this 4 and 4 to this side so this will be 1 by 4 okay and 3 x plus 4 y minus 1 upon 5 whole square so seeing this expression you can assume that this thing the things written in the left hand side will be a distance right a distance from any point half comma half square okay and this distance will be basically any perpendicular distance from this line from this line 3 x plus 4 y minus 1 right so and this thing this ratio is less than 1 so this we can interpret it this we can interpret it as sp so s is the point okay so this will be sp square is equal to 1 by 4 okay into p m ka square okay so basically you can write in this way also 1 by 2 whole square what I mean to say there is one point okay let me draw this ellipse so this is this is one any point s okay and this is any point p this is any point p and this is any line x so what I am saying this sp square this distance so this distance this focus will be the coordinates of this focus I am assuming it as half comma half right the coordinates of this focus is half comma half then only this sp square let me assume this point p as x comma y okay let me assume this point as p as x comma y and this line as 3 x plus 4 y 3 x plus 4 y minus 1 equal to 0 and let me write this distance p m now compare this compare grow if you compare this if you compare this okay let me write it neatly so it will be basically 1 by 2 so if you compare it it represents an ellipse it represents an ellipse represents and ellipse okay and what will be its eccentricity what will be its eccentricity its eccentricity if you see here e square is coming out to be e square is coming out to be 1 by 4 okay or you can say e is equals to 1 by 2 which is less than which is less than 1 and greater than 0 which is less than 1 and greater than 0 so is this what question is asking so that the equation represents an ellipse and find the eccentricity of the ellipse okay we can further find the value what we can further find the value of a and b also so here if you see it represents it is representing an ellipse why because the distance between any point okay and the ratio if you see if you further write it as the ratio how to define how do we define know this thing this distance upon distance from a point and distance from a line is equals to e and e is less than e is less than 1 so obvious it's obvious that it will represent an ellipse so I think we are proved like we are we have done the proof for this question so we don't need to go further so since e is between 0 to 1 and this ratio of these two distances this same thing you can write it as s p square upon p m square is equal to p square same thing nothing no changes we have just squared it so okay I'm not giving much time on this so let me take this last question this will be our last question for this exercise so question number 18 it is saying find the locus of the extremities of the lattice rectum of the family of ellipses so family of ellipses is given as b square x square plus y square is equal to a square into b square so let me divide it by a square b b square throughout we will get x square upon a square right b square will get cancelled out plus y square upon a square b square is equal to 1 so this is what we got so if you compare it with our standard ellipse x square upon a square plus y square upon b square is equal to 1 now what will be what is asked we have to find the locus of the extremities of the lattice rectum so for this for this what is the lattice rectum basically lattice rectum the extremities of lattice rectum will be plus minus ae right the x coordinate and what will be the y coordinate y coordinate will be plus minus b square upon a right for this the extremities of lattice rectum will be this now for this thing for this what will be the extremities of lattice rectum the extremities of lattice rectum for this will be plus minus ae a is same here so it will be ae and plus minus b what is b b square here is a square b square actually so it will be a square b square upon a so a is a square or we can say a is same no so a so the extremities of lattice rectum and since we have to find the it's what you say locus so I will write it as h okay and I will write this thing as k is it okay now we have to find the locus of the extremity so what can we write for the h or we can say this h we can write as we can write h as plus minus a is it okay and further what we can do we can like we also need to have this k also right so what can we do and k if you see what is k this a and a will get cancelled out no one a will will be gone so this k will be basically plus minus a b square okay so this will be our a b square now from here if you see I can write the value of k or substitute the value of a whatever you want to do you can do so this is the two relations basically these are the two relations and you can easily find the locus okay you can do anything no like you can write this thing this h a square can be written as what h square we can write it as a square into e square right and further we can write this e square can we can write it as a square minus b square right a square minus b square upon a square right so a square a square will get cancelled out and this we are left out with h square is equal to this thing a square minus b square now we can what we can do since the value of k is a b square no so a b square means what we can multiply by a throughout so this will be a h square is equals to a cube minus a b square minus a b square why I am multiplying a so that I can replace this value k so this will be a h square is equal to a cube minus a b square can be written as k okay so in this way you can derive any relation I am not going in detail okay so this will be a x square is equal to a cube minus y something like this will be our locus right something like this will be our locus so I am not going in the details I am stopping here only so I think this is the last question right okay yeah thank you so we have completed this first exercise of ellipse okay and we will soon meet once again with the next exercise of this conic that is ellipse so till then take care of yourself and keep practicing a lot of questions be motivated okay so thank you all thank you