The nearest point on the curve x^2=2y to the | vedclass

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(D) Let the point on the curve $x^2=2y$ be $P(x, y)$. Since $x^2=2y$,we have $y = \frac{x^2}{2}$.Thus,the point $P$ is $(x, \frac{x^2}{2})$.The distan

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$(2, 2)$

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(D) Let the point on the curve $x^2=2y$ be $P(x, y)$. Since $x^2=2y$,we have $y = \frac{x^2}{2}$.Thus,the point $P$ is $(x, \frac{x^2}{2})$.The distance $D$ between $P(x, \frac{x^2}{2})$ and $(0, 5)$ is given by $D^2 = (x-0)^2 + (\frac{x^2}{2}-5)^2$.Let $f(x) = D^2 = x^2 + (\frac{x^2}{2}-5)^2 = x^2 + \frac{x^4}{4} - 5x^2 + 25 = \frac{x^4}{4} - 4x^2 + 25$.To find the minimum distance,we differentiate $f(x)$ with respect to $x$ and set it to $0$:$f'(x) = x^3 - 8x = x(x^2 - 8) = 0$.This gives $x = 0$ or $x^2 = 8$,so $x = \pm 2\sqrt{2}$.For $x=0$,$y=0$,$D^2 = 25$,$D=5$.For $x^2=8$,$y = \frac{8}{2} = 4$. Then $D^2 = 8 + (4-5)^2 = 8 + 1 = 9$,so $D=3$.Since $3 Comparing with the given options,the correct point is $(2\sqrt{2}, 4)$.

The nearest point on the curve $x^2=2y$ to the point $(0,5)$ is . . . . . . .

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