Let \(S\) denote the locus of the mid-points of those chords of the parabola \(y^2=x\), such that the area of the region enclosed between the parabola and the chord is \(\frac{4}{3}\). Let \(R\) denote the region lying in the first quadrant, enclosed by the parabola \(y^2=x\), the curve \(S\), and the lines \(x=1\) and \(x=4\).
Then which of the following statements is (are) TRUE?

\(\begin{aligned} & T=S_1 \\ & k y-\left(\frac{x+h}{2}\right)=k^2-h \\ & x-2 k y+2 k^2-h=0 \\ & k^2-h <0 \quad \Rightarrow h-k^2>0 \\ & \text { For Area : interchange } x \& y \\ & y-2 k x+2 k^2-h=0 \& y=x^2 \\ & x^2-2 k x+\left(2 k^2-h\right)=0 <_\beta^\alpha \\ & |\alpha-\beta|=\sqrt{4 k^2-4\left(2 k^2-h\right)}=\sqrt{4 h-4 k^2} \\ & A=\int_\alpha^\beta\left(\left(2 k x+h-2 k^2\right)-x^2\right) d x \\ & A=\frac{\left(4 h-4 k^2\right)^{3 / 2}}{6}=\frac{4}{3} \\ & \left(4 h-4 k^2\right)^{3 / 2}=8 \quad \Rightarrow\left(4 h-4 k^2\right)=4\end{aligned}\)
\(\mathrm{h}-\mathrm{k}^2=1\)
\((4, \sqrt{3}) \in \mathrm{S}\)
\(\begin{aligned} & \mathrm{A}=\int_1^4(\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}-1}) \mathrm{dx}=\frac{2}{3}\left(\mathrm{x}^{3 / 2}-(\mathrm{x}-1)^{3 / 2}\right)_1^4 \\ & \mathrm{~A}=\frac{2}{3}(8-3 \sqrt{3}-1)=\frac{2}{3}(7-3 \sqrt{3})=\frac{14}{3}-2 \sqrt{3}\end{aligned}\)