Question

$P$ is a point which moves in the $XY$ plane such that the point $P$ is nearer to the centre of the square than 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 parabola of which the equation is

• A. $y^2=a^2-2ax$
• B. $y=a^2+2a{x}^2$
• C. $x=a^2-2a{y}^2$
• D. $x^2=a^2+2ay$

Answer 1

Answer: (A), (D)

If $P(x,y)$, then according to the question,
$\sqrt{x^2+y^2}<|x-a|,\sqrt{x^2+y^2}<|x+a|$
$\sqrt{x^2+y^2}<|a-y|,\sqrt{x^2+y^2}<|a+y|$
On simplification, the region bounded by these curves becomes
$y^2=a^2-2ax,y^2=a^2+2ax,x^2=a^2-2ay\text{ and }{x}^2=a^2+2ay$