# Co-Ordinate Geometry

P is a point which moves in the x-y plane, such that the point P is nearer to the centre of a square than any of the sides. The 4 vertices of square are (+/-a,+/-a). The region in which P will move is bounded by parabolas of equation:

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Are you asking a question here, or did you not complete what you wanted to post? – Ron Gordon Jan 14 at 17:45
Its a question. We have to tell the equation of parabolas – Ravi Jan 14 at 17:46
Looks similar to a Putnam Exam question (1989 B1): "A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge." – Ron Gordon Jan 14 at 17:48
Yea,somewhat you can say similar. So basically we have to find the locus of the mid point of the square? – Ravi Jan 14 at 17:52
Uhhh...I'm not following you. I'd think you'd be more interested in the locus of points that comprise the boundary of the region that contains all points closer to the center of the square than to any of the sides of the square. But I could be wrong. – Ron Gordon Jan 14 at 17:54

Use the directrix definition of a parabola, where the distance from a line (the square's edge) to a point (the center) are equal. There are 4 parabolas and they're symmetric, so I'll just find one of them.

We have the center at $(0,0)$ and the line segment $y = -a$. Then, a point $(x,y)$ lies on the parabola if the distance to the focus is equal to the distance to the directrix.

That is, $\sqrt{x^2 + y^2} = y + a$ and so $x^2 + y^2 = (y+a)^2$ and so $y = \frac{x^2-a^2}{2a}$.

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Let's look at only the side $x=-a$. Setting the point-center and point-side distance equal to each other,

$$x+a = \sqrt{x^2+y^2}.$$

Squaring both sides gives $$x^2+2xa+a^2 = x^2 + y^2$$ or $$x = \frac{y^2}{2a}-\frac{a}{2},$$ which is one piece of your curve. A similar calculation can compute the other three pieces. By symmetry, they will meet at the lines $y=x$ and $y=-x$.

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