 Hello and welcome to this session. In this session we will discuss a question which says that if u0 and v20 are two given points, a point p moves so that p is 4 plus pb2 is equal to 5 pi in the locus of p. Now before starting the solution of this question, we should know some results. First is the equation of the locus is a relation between x and y which is satisfied by the coordinates and points of the locus by no others. If the coordinates of point a are given by ax1, y1 and the coordinates of point b are given by dx2y2 then by distance formula, distance a v is equal to square root of x1 minus x2 whole square plus y1 minus y2 whole square. Now these results will work out as a key idea for solving out this question. And now let me start with the solution. Now in the question a and b are given to fix points with coordinates are given to us and this condition is also mentioned here and we have to find the locus of p. That means we have to find the equation of the locus of the point b that means that relation between x and y which is satisfied by the coordinates of all points of the locus and by no others. Now it is given a moving point. Now let the coordinates of p are xy also the coordinates of a are 3 0. Then by using the distance formula we can find out the distance p a. Now distance p a is equal to x minus 3 whole square plus y minus 0 whole square as the coordinates of p are xy and the coordinates of a are 3 0. Which further implies on square root sides p a square is equal to x minus 3 whole square plus y minus 0 whole square. Which implies p a square is equal to x square plus 9 minus 6x plus y square. Which further implies p a square can be written as x square plus y square minus 6x. Now let us name this as equation number 1. Now in the question the coordinates of b are given as 2 0. So given the coordinates of p again by using distance formula distance p b is equal to minus 2 whole square plus y minus 0 whole square. Which further implies on square root sides p b square is equal to x minus 2 whole square plus y minus 0 whole square. Which implies p b square is equal to x square plus 4 minus 4x plus y square. Further implies p b square is equal to x square plus y square minus 4x plus 4 and let us name this as equation number 2. Now for forming the equation in x and y that is for finding out the equation of the locus where we use this given condition. So this is p a square and this is p b square. So given 2 plus p b square is equal to 5. So approaching the values of p a square and p b square here this implies 2 plus y square minus 6x plus 9 plus x square plus y square minus 4x plus 4 minus 10x plus 13 is equal to 5. Which implies we are getting the equation in x and y that means we are getting the equation of locus after point p by using the given condition. Hence equation of the locus is the solution of the given question and that's all for this session. Hope you all have enjoyed this session.