Question:

Let the chord's midpoint be $(h, k)$. The chord of the parabola $y^2 = x$ is given by $T = S_1$: $$yk - \dfrac{x+h}{2} = k^2 - h \implies x = 2yk - 2k^2 + h$$ Intersect the chord with the parabola $x = y^2$: $$y^2 - 2ky + 2k^2 - h = 0$$ Let the roots be $y_1$ and $y_2$. The area enclosed is: $$\text{Area} = \int_{y_1}^{y_2} (x_{\text{chord}} - x_{\text{parabola}}) dy = \int_{y_1}^{y_2} (2yk - 2k^2 + h - y^2) dy = \dfrac{1}{6}(y_2 - y_1)^3$$ Using the root difference $y_2 - y_1 = \sqrt{4k^2 - 4(2k^2 - h)} = 2\sqrt{h - k^2}$: $$\text{Area} = \dfrac{8}{6}(h - k^2)^{3/2} = \dfrac{4}{3}(h - k^2)^{3/2}$$ We are given the area is $4/3$: $$\dfrac{4}{3}(h - k^2)^{3/2} = \dfrac{4}{3} \implies h - k^2 = 1 \implies h = k^2 + 1$$ Thus, the locus $S$ is $x = y^2 + 1$. Verify points on $S$: - For $(4, \sqrt{3})$: $(\dots\sqrt{3})^2 + 1 = 4$, so $(4, \sqrt{3}) \in S$ (True). - For $(5, \sqrt{2})$: $(\sqrt{2})^2 + 1 = 3 \neq 5$, so $(5, \sqrt{2}) \notin S$ (False). Compute the area of region $\mathcal{R}$ in the first quadrant bounded by $y = \sqrt{x}$, $y = \sqrt{x-1}$, $x=1$ and $x=4$: $$\text{Area} = \int_{1}^{4} (\sqrt{x} - \sqrt{x - 1}) dx = \left[ \dfrac{2}{3}x^{3/2} - \dfrac{2}{3}(x-1)^{3/2} \right]_1^4$$ $$= \dfrac{2}{3} \left[ (8 - 3\sqrt{3}) - 1 \right] = \dfrac{14}{3} - 2\sqrt{3}$$