 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the point on the curve x square is equal to 8y which is nearest to the point 2,4. Let us now start with the solution. Let the point on the curve be x,y. Now the distance between the points x,y and 2,4 is given by d which is equal to square root of x minus 2 whole square plus y minus 4 whole square. Here we have applied distance formula to find the distance between these two points. Now squaring both the sides of this expression we get d square is equal to x minus 2 whole square plus y minus 4 whole square. Now let us assume that s is equal to d square which is further equal to square of x minus 2 plus square of y minus 4. Now we know equation of the curve is x square is equal to 8y. Now we can write given curve is x square is equal to 8y. Now this further implies x square upon 8 is equal to y or we can simply write y is equal to x square upon 8. Let us name this equation as 1 and this equation as 2. Now substituting this value of y in this equation we get s is equal to x minus 2 whole square plus x square upon 8 minus 4 whole square. Now differentiating both the sides of this equation with respect to x we get ds upon dx is equal to 2 multiplied by x minus 2 plus 2 multiplied by x square upon 8 minus 4 multiplied by 2x upon 8. Now this is further equal to 2x minus 4 plus 4x cube upon 64 minus 16x upon 8. Now we know 4 multiplied by 16 is equal to 64 and 8 multiplied by 2 is equal to 16. So we are left with 2x minus 4 plus x cube upon 16 minus 2x. Now plus 2x and minus 2x will get cancelled and we get ds upon dx is equal to x cube upon 16 minus 4. Now we have to find this point such that it is nearest to this point or we can say the distance between these two points is minimum. Now for maximum or minimum we put ds upon dx is equal to 0. Now this implies x cube upon 16 minus 4 is equal to 0. This further implies x cube upon 16 is equal to 4. Adding 4 on both the sides of this equation we get x cube upon 16 is equal to 4. Now multiplying both the sides by 16 we get x cube is equal to 64. Now this further implies x cube is equal to 4 cube. Now taking cube root on both the sides we get x is equal to 4. Now we will substitute x is equal to 4 in equation 2 to find the value of y. We get y is equal to 4 square upon 8 which is further equal to 2. Now we have to show that the point 4 comma 2 is nearest to the point 2 comma 4. Or we can say we have to show that the distance between the points 4 comma 2 and 2 comma 4 is minimum. Now let us recall that x is equal to c is a point of local minima if f dash c is equal to 0 and f double dash c is greater than 0. Now clearly we can see ds upon dx is equal to 0 and x is equal to 4. So value of c is equal to 4 here. Now we will check if d square s upon dx square is greater than 0 at x is equal to 4. Now we know ds upon dx is equal to x cube upon 16 minus 4. Now differentiating both the sides of this expression with respect to x we get d square s upon dx square is equal to 3x square upon 16. Now clearly we can see even power of x implies that this term will be positive for any value of x. So we can say d square s upon dx square is greater than 0 for any value of x. So we can say distance between the points is minimum also point 4 comma 2 lies on the given curve and it is nearest to the point 2 comma 4. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.