The point on the curve x^2 = 2y which is near | vedclass

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(A) Let the point on the curve $x^2 = 2y$ be $(x, y)$. Since $x^2 = 2y$,we have $y = \frac{x^2}{2}$.The distance $D$ between $(x, y)$ and $(0, 5)$ is

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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