The point (0, 5) is closest to the curve {x^2 | vedclass

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(D) Let a point on the curve be $P(h, k)$.Since the point lies on the curve ${x^2} = 2y$,we have ${h^2} = 2k$ ... $(i)$.The distance $D$ between the p

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None of these

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(D) Let a point on the curve be $P(h, k)$.Since the point lies on the curve ${x^2} = 2y$,we have ${h^2} = 2k$ ... $(i)$.The distance $D$ between the point $(0, 5)$ and $P(h, k)$ is given by $D = \sqrt{(h - 0)^2 + (k - 5)^2} = \sqrt{h^2 + (k - 5)^2}$.Substituting ${h^2} = 2k$ from $(i)$,we get $D = \sqrt{2k + (k - 5)^2} = \sqrt{2k + k^2 - 10k + 25} = \sqrt{k^2 - 8k + 25}$.To minimize $D$,we minimize $f(k) = k^2 - 8k + 25$.Taking the derivative,$f'(k) = 2k - 8$.Setting $f'(k) = 0$,we get $2k = 8$,which implies $k = 4$.For $k = 4$,${h^2} = 2(4) = 8$,so $h = \pm 2\sqrt{2}$.The points on the curve closest to $(0, 5)$ are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$.Since these points are not listed in options $A$,$B$,or $C$,the correct option is $D$.

The point $(0, 5)$ is closest to the curve ${x^2} = 2y$ at

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