The point on the parabola 2y = x^2 which is n | vedclass

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(C) Let the point on the parabola be $(x, y)$. Since $2y = x^2$,we have $y = x^2/2$.The distance $D$ between $(x, x^2/2)$ and $(0, 3)$ is given by $D^

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$(\pm 2, 2)$

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(C) Let the point on the parabola be $(x, y)$. Since $2y = x^2$,we have $y = x^2/2$.The distance $D$ between $(x, x^2/2)$ and $(0, 3)$ is given by $D^2 = (x - 0)^2 + (x^2/2 - 3)^2$.Let $f(x) = D^2 = x^2 + (x^4/4 - 3x^2 + 9) = x^4/4 - 2x^2 + 9$.To find the minimum distance,we set $f'(x) = 0$:$f'(x) = x^3 - 4x = x(x^2 - 4) = 0$.This gives $x = 0$ or $x = \pm 2$.For $x = 0$,$y = 0$,distance is $\sqrt{0^2 + (0-3)^2} = 3$.For $x = \pm 2$,$y = 2^2/2 = 2$,distance is $\sqrt{(\pm 2 - 0)^2 + (2 - 3)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236$.Since $\sqrt{5} Thus,option $(c)$ is correct.

The point on the parabola $2y = x^2$ which is nearest to the point $(0, 3)$ is

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