Find the point on the parabolas `x^2=2y` which is closest to the point `(0,\ 5)` .

Explanation

`x^2=2y`

`D=sqrt{(x-0)^2+(y-5)^2}`

`D^2=x^2+({x^2}/2-5)^2`

`implies D^2={x^4}/4-4x^2+25`

Now,

`{dD^2}/{dx}=x^3-8x`

Put `{dD^2}/{dx}=0`

`implies x^3-8x =0`

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