P is a point which moves in the xy plane such that the point P nearer to the centre of a square then any of the sides. The four vertices of the square are $(\pm a, \pm a)$ . The region in which P will move is bounded by parts of parabolas of which one has the equation

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$y^{2} + 2 a x = a^{2}$

All of these

$y^{2} = a^{2} + 2 a x$

$x^{2} = a^{2} + 2 a y$

answer is D.

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Detailed Solution

$\begin{array}{l} I f P \equiv (x, y) t h e n \sqrt{x^{2} + y^{2}} < | a - x |, \sqrt{x^{2} + y^{2}} < | a + x | \\ \sqrt{x^{2} + y^{2}} < | a - y |, \sqrt{x^{2} + y^{2}} < | a + y | \\ S q u a r i n g t h e n w e g e t t h e r e g i o n b o u n d e d b y t h e c u r v e s \\ x^{2} + y^{2} = {(a - x)}^{2}, x^{2} + y^{2} = {(a + x)}^{2} \\ x^{2} + y^{2} = {(a - y)}^{2}, x^{2} + y^{2} = {(a + y)}^{2} \end{array}$