A geometry problem by Aly Ahmed

Let the parabola above be represented by $y = 2(x+1)^2.$ Pick a point $P(x_{0}, 0)$ for $x_{0} \in (-1,0)$ such that $2(1+x_{0})^2 = -x_{0}.$ Solving for $x_{0}$ yields:

$-x_{0} = 2(1+x_{0})^2 \Rightarrow 0 = 2x_{0}^{2} + 5x_{0} + 2 \Rightarrow 0 = (2x_{0}+1)(x_{0}+2) \Rightarrow x_{0} = -\frac{1}{2}, -2$ .

Since only $-1 < -\frac{1}{2} < 0$ , this is the desired value we want. The square thus has an area of $(\frac{1}{2})^2 = \boxed{\frac{1}{4}}.$