Subject: Mathematics — Parabola
Question: Consider a square with vertices at
$(1,1),( - 1,1),( - 1, - 1)$ and $(1, - 1)$. Let $S$ be the region
consisting of all points inside the square which are nearer to the
origin than to any edge. Sketch the region $S$ and find its area.
Answer: $\frac{16\sqrt{2} - 20}{3}$
Solution: Let us consider any point $P(x,y)$ inside the square such
that its distance from origin $\leq$ its distance from any of the edges
say AD\
$$\therefore OP \leq PM\text{~or~}\sqrt{\left( x^{2} + y^{2} \right)} < 1 - x
$$$\text{or~}y^{2} \leq - 2\left( x - \frac{1}{2} \right)$\
Above represents all points within and on the parabola 1. If we consider
the edges $BC$ then $OP < PN$ will imply\
$y^{2} \leq 2\left( x + \frac{1}{2} \right)$ ......(ii)\
Similarly if we consider the edges $AB$ and $CD$, we will have\
$x^{2} \leq - 2\left( y - \frac{1}{2} \right)$ ......(iii)\
$x^{2} \leq 2\left( y + \frac{1}{2} \right)$ ......(iv)\
Hence $S$ consists of the region bounded by four parabolas meeting the
axes at $\left( \pm \frac{1}{2},0 \right)$ and
$\left( 0, \pm \frac{1}{2} \right)$\
The point $L$ is intersection of $P_{1}$ and $P_{3}$ given by (i) and
(iii).
[IMAGE] $y^{2} - x^{2} = - 2(x - y) = 2(y - x)$\
$${\therefore\ y - x = 0\therefore y = x
}{\therefore\ x^{2} + 2x - 1 = 0 \Rightarrow (x + 1)^{2} = 2
}{\therefore\ x = \sqrt{2} - 1\text{~as~}x\text{~is +ve~}}$$
$\therefore\mspace{6mu} L\text{ is }\left( \sqrt{2} - 1,\sqrt{2} - 1 \right)$
$\therefore\mspace{6mu}\text{Total area}$ =
$4\left\lbrack \text{square of side }\left( \sqrt{2} - 1 \right) + 2\int_{\sqrt{2} - 1}^{\frac{1}{2}}y\, dx \right\rbrack$
=
$4\left\{ \left( \sqrt{2} - 1 \right)^{2} + 2\int_{\sqrt{2} - 1}^{\frac{1}{2}}\sqrt{(1 - 2x)}\, dx \right\}$
=
$4\left\lbrack 3 - 2\sqrt{2} - \frac{2}{2} \cdot \frac{2}{3}\left. \ \{(1 - 2x)^{\frac{3}{2}}\} \right|_{\sqrt{2} - 1}^{\frac{1}{2}} \right\rbrack$
=
$4\left( 3 - 2\sqrt{2} \right)\left\lbrack 1 + \frac{2}{3}\sqrt{\left( 3 - 2\sqrt{2} \right)} \right\rbrack$
=
$\frac{4}{3}\left( 3 - 2\sqrt{2} \right)\left( 1 + 2\sqrt{2} \right)$
= $\frac{4}{3}\left\lbrack \left( 4\sqrt{2} - 5 \right) \right\rbrack$
= $\frac{16\sqrt{2} - 20}{3}$