 Hi and welcome to the session. Let us discuss the following question. Question says, form the differential equation representing the family of curves given by x minus a whole square plus 2y square is equal to a square where a is in arbitrarily constant. Let us now start with the solution. We have to find the differential equation representing the family of curves x minus a whole square plus 2y square is equal to a square. Or we can say we have to find the differential equation whose solution is x minus a whole square plus 2y square is equal to a square. Now we will consider this equation. Now we know x minus a whole square is equal to x square plus a square minus 2ax. We will write this plus 2y square as it is and here we will write is equal to a square. Applying this formula to this bracket we get these three terms. Now a square and a square will get cancelled and we get x square plus 2y square is equal to 2ax dividing both the sides of this equation by x we get x square plus 2y square upon x is equal to 2a. Now differentiating both the sides of this equation with respect to x we get x multiplied by 2x plus 4y multiplied by dy upon dx minus x square plus 2y square multiplied by 1 upon x square is equal to 0. We can find derivative of this term by applying quotient rule also derivative of 2a is equal to 0. Clearly we can see 2a is a constant and derivative of a constant term is 0. Now multiplying both the sides of this expression by x square we get x multiplied by 2x plus 4y dy upon dx minus x square plus 2y square is equal to 0. Now this further implies 2x square plus 4xy dy upon dx minus x square minus 2y square is equal to 0. Multiplying x by this bracket we get these two terms and multiplying this negative sign with this bracket we get minus x square minus 2y square and here we can write is equal to 0 as it is. Now we know 2x square minus x square is equal to x square only. So we can write this further implies 4xy dy upon dx plus x square minus 2y square is equal to 0. Now adding 2y square minus x square on both the sides of this equation we get 4xy dy upon dx is equal to 2y square minus x square. Now dividing both the sides of this equation by 4xy we get dy upon dx is equal to 2y square minus x square upon 4xy. Now we know dy upon dx is denoted by y dash so we can write y dash is equal to 2y square minus x square upon 4xy. So this is our required differential equation. This completes the session. Take care and have a nice day.