# 27) The point on the curve $x^2 = 2y$ which is nearest to the point (0, 5) is $(A) (2 \sqrt 2,4) \: \: (B) (2 \sqrt 2,0)\: \: (C) (0, 0)\: \: (D) (2, 2)$

G Gautam harsolia

Given curve is
$x^2 = 2y$
Let the points on curve be  $\left ( x, \frac{x^2}{2} \right )$
Distance between two points is given by
$f(x)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$= \sqrt{(x-0)^2+(\frac{x^2}{2}-5)^2} = \sqrt{x^2+ \frac{x^4}{4}-5x^2+25} = \sqrt{ \frac{x^4}{4}-4x^2+25}$
$f^{'}(x) = \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}}\\ f^{'}(x)= 0\\ \frac{x^3-8x}{2\sqrt{\frac{x^4}{4}-4x^2+25}} =0\\ x(x^2 - 8)=0\\x=0 \ and \ x^2 = 8\Rightarrow x = 2\sqrt2$
$f^{''}(x) = \frac{1}{2}\left (\frac{(3x^2-8)(\sqrt{\frac{x^4}{4}-4x^2+25} - (x^3-8x).\frac{(x^3-8x)}{2\sqrt{\frac{x^4}{4}-4x^2+25}}}{(\sqrt{\frac{x^4}{4}-4x^2+25})^2}) \right )$
$f^{''}(0) = -8 < 0$
Hence, x = 0 is the point of maxima
$f^{''}(2\sqrt2) > 0$
Hence, the point $x = 2\sqrt2$ is the point of minima
$x^2 = 2y\Rightarrow y = \frac{x^2}{2} = \frac{8}{2}=4$
Hence, the point $(2\sqrt2,4)$ is the  point on the curve $x^2 = 2y$ which is nearest to the point (0, 5)
Hence, the correct answer is (A)

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