 Hello students, let's learn the second question of paper 1 of JADVANCE 2023. It is from the chapter conic sections and it is indeed an easy question of JADVANCE because all you need to know to solve this question is equation of tangent in slow form to both the curves. So, you know that equation of tangent to parabola is y is equals to mx plus a by m for the parabola y square is equals to 4ax. If you compare it to the 12x then the value of a here is 3. So the equation of tangent in slow form to the parabola will be y is equals to mx plus 3 by m. Alright, now let us look at the ellipse. For the ellipse the equation of tangent in slow form is y is equals to mx plus minus square root of a square m square plus b square where the equation of ellipse is x square upon a square plus y square upon b square is equals to 1. If you compare you will get the value of a square and b square is 6 and 3 correspondingly. So the equation of tangent will be y is equals to mx plus minus square root of 6 m square plus 3. Now if it is a common tangent right you want to find the common tangent to the ellipse and the parabola then you just equate the y intercept because rest everything is the same. So we have 3 by m is equals to plus minus under root of 6 m square plus 3 or if you cross multiply and square you will get 9 is equals to m square times 6 m square plus 3. If you solve this you will get 6 m raised to the power 4 plus 3 m square minus 9 is equals to 0. So you can break it up as 9 and minus 6 alright 6 m raised to the power 4 plus 9 m square minus 6 m square minus 9 m minus 9 is equals to 0. So either you can take common as 3 m square. So you get 2 m square plus 3 or you can take a minus 3 common and again 2 m square plus 3 is equals to 0. Now of course this factor cannot be 0 for any real x and the remaining factor 3 m square minus 3 can be equals to 0 for m is equals to plus minus 1. So the real values of m that I got is 1 or minus 1 and if you substitute these values in any of these equations it is easy to put them in this equation you get the equation of both the tangents. So let me write the equations of both the tangents alright. So the equations of tangent 1 and tangent 2 are tangent 1, tangent 1 will turn out to be y is equals to if you put m equals to 1 then you will have x plus 3 and t2 can be y is equals to if you put m equals to minus 1 then minus x minus 3. Now you also want to find the corresponding points of context because let us say for tangent t1 the corresponding points are a1 and a2 at parabolic ellipse respectively and tangent t2 cuts the parabolic ellipse at a4 and a3 respectively. Now you have to look at the order very carefully because if you do not look at the order correctly then you will have trouble ok. So a1 and a4 are the corresponding points of context for parabola. Now for parabola if you go about a parametric form then the relation between t and m of a tangent is t is equals to 1 by m alright. So the corresponding t1 and t2 parametric coordinates of parabola will be 1 and minus 1 corresponding to these values of m. And the corresponding points then will be at1 square comma 281 and at2 square comma 282 ok. So if you put the value of a as 3 and t as 1 and minus 1 so you will get 3 comma 6 and 3 comma minus 6. So you can call them a1 and a4 this point is a1 this point is a4. Similarly if you want to find the corresponding points on ellipse what you can do is you can write the equation of tangent of ellipse in point form so it will be xx1 by a square which is 6 plus yy1 by b square which is 3 is equals to 1 and compare it with the equation let us say x minus y so xy3 minus yy3 is equals to 1 alright x upon you can say minus 3 and minus y sorry plus y upon 3 is equals to 1 alright. So if you compare it with xx1 upon 6 plus yy1 by 3 you will get x1 as minus 2 and y1 as 1 so that is one point of contact let us call it what a2 and the next point of contact will turn out to be minus 2 comma minus 1 ok. If you compare it with the equation that x upon minus 3 plus y upon minus 3 is equals to 1 with xx1 upon 6 it will be minus 2 comma minus 1 so a3 will be minus 2 comma minus 1. Now you can draw all of these points on the coordinate plane and then find whichever the area it is asking about and point of intersection of tangents if you want to find then you can simply equate the y's and you will get 2x as minus 6 so x is minus 3 comma 0 so the option number c is correct ok it was a simple one and if you want to find the area of quadrilateral a1 a2 a3 a4 let me just mark these points in the coordinate plane ok so we have 3 comma 6 and 3 comma minus 6 and then minus 2 comma 1 and then minus 2 comma minus alright so let us say minus 2 comma 1 and minus 2 comma minus if you join these four points alright you will get this you know trapezium actually this figure this quadrilateral is turning out to be a trapezium whose coordinates of end points are minus 2 comma minus 1 ok minus 2 comma 1 this is 3 comma minus 6 and this is 3 comma 6 you can see that these two sides are parallel they are vertical and the distance between these lines the distance between the parallel lines is 3 plus 2 that is 5 this side is 1 parallel side is 6 plus 6 12 another parallel side is 1 plus 1 that is 2 and what is the area of trapezium half into sum of parallel sides that is 2 plus 12 into height that is 5 so it will be 7 5s or 35 minutes which is your option number 8.