 Hello and welcome to the session. In this session, we discuss the following questions with face. A point P with coordinates x, y moves so that the sum of its distances from the point S with coordinates 4, 3 and S dash with coordinates minus 4, 3 is 10. Find the equation of its locus and show that it is an ellipse. Before moving on to the solution, let's see the general equation of an ellipse which is given by A x square plus 2 h x y plus B y square plus 2 g x plus 2 f y plus C equal to 0. This is the general equation of the ellipse. It is a second degree equation and the constants A, h and B are such that x square is less than AB. This is the key idea that we use in this question. Let's now proceed with the solution. In the equation, we have a point P with coordinates x, y and a point S with coordinates 4, 3, point S dash with coordinates minus 4, 3. Now, modulus of S P is given as square root of, that is, we find the distance between the points P and S and this would be given as square root of x minus 4 d whole square plus y minus 3 d whole square. Then, modulus of S dash P that is distance between the points S dash and P is given as square root of x plus 4 d whole square plus y minus 3 d whole square. Often the question is given that this point P moves such that the sum of its distances on the points S and S dash is given as 10. So, we can say modulus of S P plus modulus of S dash P is equal to 10. This means that square root of minus 4 d whole square plus y minus 3 d whole square plus the square root of d whole square plus y minus 3 d whole square is equal to 10. Now, this could be written as square root of x minus 4 d whole square plus y minus 3 d whole square is equal to 10 minus square root plus 4 d whole square plus y minus 3 d whole square. Now, next we get minus 4 whole square plus y minus 3 whole square is equal to 10 square that is 100 plus d square of the square root it would be x plus 4 d whole square plus y minus 3 d whole square minus which is 20 into square root of d whole square plus y minus 3 d whole square. Now, this y plus 3 whole square and y plus 3 whole square cancels and further we have x square minus 8x plus 16 is equal to 100 plus x square plus 8x plus 16 minus 20 into square root of x plus 4 d whole square plus y minus 3 d whole square. x square cancels with x square 16 cancels with 16 so we now have 20 into square root of d whole square plus y minus 3 d whole square is equal to 100 plus 8x plus 8x so we now have 20 into square root of d whole square plus y minus 3 d whole square is equal to 100 plus 16x. Now, further 20 into 20 is into square of the square root is d whole square plus 0 to 100 plus 16x d whole square which is 100 square gives us 30 into square x square plus 8x plus 6 is equal to 100 square that is 1000 plus 3200x 3200x 100 plus 400 y square 2400 y 600 is equal to x square plus x and 3200x cancels 6400 plus 3600 is 10,000 this cancels with this 10,000 x square 2400 y is equal to 0 8 400 y is equal to 0 now we take 16 common on the left hand side so we have 16 into 9x square plus this whole is equal to 0 this gives us plus 25 y 50 y is equal to 0 and this is the required equation plus 25 y square minus 150 y is equal to 0 that with the general equation of the ellipse which is ax square plus dy square plus plus 2s y plus c is equal to 0 now comparing these two equations we get that a is equal to 9 b is equal to 25 this equation we don't have any term containing x y so this means that h would be 0 in this case and also we find that 2f is equal to minus 150 which means that we have f equal to 75 here we have already stated that this is the general equation of the ellipse when the constant x square is less than a b so we now check minus a b which is equal to 0 minus 9 into 25 that is equal to f which is less than 0 that is less than 0 therefore b equation 9x square plus 25 y square minus 150 y is equal to 0 is the equation is the required locus and this is the equation of an ellipse hence we at 9x square plus 25 y square minus 150 y equal to 0 is the required equation it is an equation so we have understood the solution of this question