Find the point on the curve `x^2=8y` which is nearest to the point `(2,\ 4)` .

Explanation

`x^2=8y`

`D=sqrt{(x-2)^2+(y-4)^2}`

`D^2=(x-2)^2+({x^2}/8-4)^2`

`implies D^2={x^4}/64-4x+20`

Now,

`{dD^2}/{dx}={x^3}/16-4`

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

`implies {x^3}/16-4=0`

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