Co-Ordinate Geometry

Question

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:

Answer 1

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 $y = \frac{x^2-a^2}{2a}$.

Answer 2

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 $y=-x$.