Question 6

Let the midpoint of the chord be $(h, k)$. The equation of the chord is $T = S_1$, which gives $yk - \frac{1}{2}(x+h) = k^2 - h$, or $x - 2yk + 2k^2 - h = 0$. The area enclosed between the parabola $x = y^2$ and this chord is given by $\frac{1}{6}(y_2 - y_1)^3$, where $y_1, y_2$ are the roots of $y^2 - 2yk + 2k^2 - h = 0$. Since $(y_2 - y_1)^2 = (2k)^2 - 4(2k^2 - h) = 4(h - k^2)$, the area is $\frac{1}{6} (2\sqrt{h - k^2})^3 = \frac{4}{3}(h - k^2)^{3/2}$. Given the area is $\frac{4}{3}$, we have $(h - k^2)^{3/2} = 1 \Rightarrow h - k^2 = 1$. Thus, the locus $S$ is $x - y^2 = 1$. For option (A), $(4)^2 - (\sqrt{3})^2 = 4 - 3 = 1$, so $(4, \sqrt{3}) \in S$ is true. For the region $\mathcal{R}$ in the first quadrant, it is bounded by $y = \sqrt{x}$ (upper), $y = \sqrt{x-1}$ (lower), $x=1$, and $x=4$. Area $= \int_1^4 (\sqrt{x} - \sqrt{x-1}) dx = [\frac{2}{3}x^{3/2} - \frac{2}{3}(x-1)^{3/2}]_1^4 = \frac{2}{3}(8 - 3\sqrt{3}) - \frac{2}{3}(1 - 0) = \frac{16}{3} - 2\sqrt{3} - \frac{2}{3} = \frac{14}{3} - 2\sqrt{3}$. Thus, (C) is true.