 Hi, and welcome to the session. I am Deepika here. Let's discuss a question. If x minus a whole square plus y minus b whole square is equal to c square for some c greater than 0, prove that 1 plus dy by dx square raised to power 3 by 2 upon d2y by dx square is a constant independent of a and b. So let's start the solution given whole square plus y minus b whole square is equal to c square. Let us give this as number equation as number one differentiating this with respect to x we have d by dx of x minus a minus b into d by dx of y minus b is equal to 0 is equal to minus a now d by dx of x minus a is 1 to y minus b into d by dx of y is dy by dx equal to 0 this implies equal to minus y minus b dy by dx because we have divided this equation by 2 so We have x minus a is equal to minus y minus b into dy by dx So let us give this equation as number two on again differentiating this equation with respect to x we have dx of y minus b into dy by dx equal to 0 this implies now derivative of x minus a with respect to x is 1 plus here we will use a product rule so this is 1 plus y minus b into d2 y by dx square plus dy by dx into Again derivative of y with respect to x that is dy by dx is equal to 0 So this is equal to 1 plus y minus b d2 y by dx square plus dy by dx whole square is equal to 0 this implies 1 plus dy by dx whole square is equal to minus y minus b d2 y by dx square this implies y minus b is equal to plus dy by dx d2 y by dx square or substitute Now substitute y by y minus b equation 2 x minus a is equal to Now equation 2 was x minus a is equal to minus of y minus b into dy by dx So x minus a is equal to substitute the value here we get minus minus 1 plus dy by dx square upon d2 y by dx square into dy by dx So this is equal to x minus a is equal to now 1 plus dy by dx square upon d2 y by dx square into dy by dx So this is a value of x minus a now substitutes value of values of x minus a and y minus b In equation 1 we get Now equation 1 was x minus a whole square plus y minus b whole square is equal to c square So we get x minus a square that is 1 plus dy by dx square upon d2 y by dx square This is x minus a square into dy by dx whole square plus y minus b square Now y minus b whole square is 1 plus dy by dx square upon d2 y by dx square is equal to c square So this implies 1 plus dy by dx square upon d2 y by dx square If we take this as common we get inside we get dy by dx square plus 1 is equal to c square So on multiplying we get 1 plus dy by dx square raised to power 3 upon d2 y by dx square is equal to c square This implies c is under root of this that is 1 plus dy by dx square raised to power 3 into 1 by 2 that is 3 by 2 upon under root of this that is d2 y by dx square So this implies 1 plus dy by dx square raised to power 3 by 2 upon d2 y by dx square is a constant independent of A and B. Hence we have proved our question. I hope the question is clear to you. Bye and have a nice day