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)

Answers (1)

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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