Moving points and constant area

Let $A = (a,0)$ and $B = (b,0)$ . The area of the region bounded by the parabola and $PQ$ is the difference between the area of the trapezoid $APQB$ and the area between the parabola and the x-axis, which is $\int_a^b x^2 dx$ . The area of the trapezoid is $\frac{(\overline{AP} +\overline{BQ})\overline{AB}}{2} = \frac{(a^2 +b^2)(b-a)}{2}$ . Since the area of that region is always $36$ :

$36 = \frac{(a^2 +b^2)(b-a)}{2} - \int_a^b x^2 dx = \frac{(a^2 +b^2)(b-a)}{2} - \frac{x^3}{3} \Bigg|_a^b = \frac{(a^2 +b^2)(b-a)}{2} - \frac{b^3-a^3}{3}$

We can simplify that expression using the factorization $b^3 - a^3 = (b-a)(a^2+ab+b^2)$ :

$36 = \frac{(a^2 +b^2)(b-a)}{2} - \frac{(b-a)(a^2+ab+b^2)}{3} = \frac{b-a}{6}(3(a^2+b^2) - 2(a^2+ab+b^2))$

$36 = \frac{b-a}{6}(a^2+b^2-2ab) = \frac{(b-a)(a-b)^2}{6} = \frac{(b-a)^3}{6}$

$(b-a)^3 = 216 = 6^3$

Since $a$ and $b$ are real numbers, it follows that $b-a=6$

By the Pythagorean theorem, the length of the line segment $PQ$ is:

$\overline{PQ} = \sqrt{(b^2-a^2)^2 + (b-a)^2} = \sqrt{(b-a)^2(b+a)^2 + (b-a)^2}$

$\overline{PQ} = (b-a)\sqrt{(b+a)^2 + 1} = 6\sqrt{(2a+6)^2 +1} = 6\sqrt{4(a+3)^2 +1}$ (using that $b-a=6$ )

Using the limit of a composite function:

$\lim_{a \to \infty}\frac{\overline{PQ}}{a} = 6\sqrt{4(\lim_{a \to \infty}\frac{a+3}{a})^2 +\lim_{a \to \infty}\frac{1}{a^2}} = 6\sqrt{4(1)^2 +0} = \boxed{12}$