When parabola meets a rectangle

It does not really matter that the origin is given at $D$ . For simplicity, I choose it to be at $O$ . The line PQ has to go through $(2,-1)$ , midway between $O=(0,0)$ and $C=(4,-2)$ . And its slope must be 2, because it is perpendicular to line OC which has slope $-1/2$ . Our line is given by $y=2x-5$

P is an intersection point of the line with parabola $COD$ , given by $y=-x^2/8$ . Filling in $2x-5=-x^2/8$ leads to $x^2+16x-40=0, x=-8 \pm \sqrt{104}$ . The x-coordinate of point P is the positive value: $x_P=2\sqrt{26}-8$

Q is an intersection point of the line with parabola $BOC$ , given by $x=y^2$ . Filling in $x=(2x-5)^2$ leads to $4x^2-21x+25=0, x=\frac{21 \pm \sqrt{41}}{8}$ . The x-coordinate of point Q is the smaller one of these values: $x_Q=\frac{21 - \sqrt{41}}{8}$

Let's consider the difference in the x-coordinates of P and Q $Δx = x_P-x_Q= \frac{\sqrt{41}+16\sqrt{26}-85}{8}$ Because P and Q are on a line with slope 2, we have $Δy=2Δx$ , and the distance is $d=\sqrt{(Δx)^2+(Δy)^2}=\sqrt{5}Δx$

$d=\sqrt{5}Δx=\frac{1}{8}\sqrt{205}+2\sqrt{130} -\frac{85}{8}\sqrt{5}\approx 0.835$

Honestly, this is not a level 5 problem.