The point on the curve x^{2}=2 y which is nea | vedclass

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Question

(C) The given curve is $x^{2}=2 y$. For any point on the curve,let its coordinates be $(x, y) = (x, x^{2}/2)$. The distance $D$ between $(x, x^{2}/2)$

Vedclass · Accepted Answer

$(2 \sqrt{2}, 4)$

Vedclass · Answer

(C) The given curve is $x^{2}=2 y$. For any point on the curve,let its coordinates be $(x, y) = (x, x^{2}/2)$. The distance $D$ between $(x, x^{2}/2)$ and $(0, 5)$ is given by $D = \sqrt{(x-0)^{2} + (x^{2}/2 - 5)^{2}}$. To minimize the distance,we minimize $f(x) = D^{2} = x^{2} + (x^{2}/2 - 5)^{2} = x^{2} + x^{4}/4 - 5x^{2} + 25 = x^{4}/4 - 4x^{2} + 25$. Taking the derivative with respect to $x$,we get $f'(x) = x^{3} - 8x$. Setting $f'(x) = 0$,we have $x(x^{2} - 8) = 0$,which gives $x = 0$ or $x = \pm 2\sqrt{2}$. Using the second derivative test,$f''(x) = 3x^{2} - 8$. For $x = 0$,$f''(0) = -8 For $x = \pm 2\sqrt{2}$,$f''(2\sqrt{2}) = 3(8) - 8 = 16 > 0$ (local minimum). Thus,the distance is minimum at $x = \pm 2\sqrt{2}$. For $x = \pm 2\sqrt{2}$,$y = (\pm 2\sqrt{2})^{2} / 2 = 8/2 = 4$. Therefore,the points nearest to $(0, 5)$ are $(\pm 2\sqrt{2}, 4)$. Comparing with the given options,the correct answer is $(2\sqrt{2}, 4)$.

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

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