Answer :

It is given that x2 = 2y

For each value of x, the position of the point will be


The distance d(x) between the points and (0,5) is given by:





d’(x) = 0


x3 – 8x = 0


x(x2-8) =0


x = 0,


And, .



So, now when x = 0, then d’’(x) = < 0


And when, x = , d’’(x) >0


Then, by second derivative test, d(x) is minimum at


So, when


Therefore, the point on the curve x2 = 2y which is nearest to the point (0,5) is .

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