$\text{Let}\mathrm{ S}\text{ denote the locus of mid-points of those chords of the parabola}{\mathrm{ y}}^2=x,\text{such that area of the region}$

$\text{enclosed between the parabola and the chord is }\frac{4}{3}.\text{Let}\mathrm{ R}\text{ denote the region lying in the first quadrant,}$

$\text{enclosed by the parabola}{y}^2=x,\text{the curve}S,\text{and the lines}x=1\text{and}x=4.\text{Then which of the following}$

$\text{statements is (are) TRUE?}$

To find the locus $S$ of the mid-points of the chords of the parabola $y^2=x$, let the mid-point of a chord be $(h,k)$.

The equation of a chord of a conic with a given mid-point $(h,k)$ is represented by $T=S_1$.
For the parabola $y^2=x$, we have:
$yk-\frac{x+h}{2}=k^2-h$
Multiplying the entire equation by 2, we get:
$2yk-x-h=2k^2-2h$
Rearranging for $x$:
$x=2ky-2k^2+h$

Now, we find the points of intersection of this chord with the parabola $y^2=x$ by substituting $x=y^2$:
$y^2=2ky-2k^2+h$
Which gives the quadratic equation in $y$:
$y^2-2ky+2k^2-h=0$

Let the roots of this quadratic equation be $y_1$ and $y_2$. The sum and product of the roots are:
$y_1+y_2=2k$
$y_1{y}_2=2k^2-h$

The difference between the roots is:
$|y_1-y_2|=\sqrt{(y_1+y_2{)}^2-4y_1{y}_2}=\sqrt{4k^2-4(2k^2-h)}=2\sqrt{h-k^2}$

The area of the region enclosed between the parabola and the chord is given by:
$\mathrm{Area}={\int }_{y_1}^{y_2}(x_{\mathrm{chord}}-x_{\mathrm{parabola}})dy={\int }_{y_1}^{y_2}(2ky-2k^2+h-y^2)dy$
Using the standard integration formula for a quadratic function between its roots:
$\mathrm{Area}=\frac{1}{6}|y_1-y_2{|}^3=\frac{1}{6}{\left(2\sqrt{h-k^2}\right)}^3=\frac{4}{3}(h-k^2{)}^{3/2}$

We are given that this area is $\frac{4}{3}$:
$\frac{4}{3}(h-k^2{)}^{3/2}=\frac{4}{3}\Rightarrow h-k^2=1\Rightarrow h=k^2+1$

Replacing $(h,k)$ with $(x,y)$, the equation of the locus $S$ is:
$x=y^2+1\Rightarrow y^2=x-1$

Let us verify the statements:
1. Check if $(4,\sqrt{3})\in S$:
Substituting $x=4$ and $y=\sqrt{3}$ into the equation of $S$:
$(\sqrt{3}{)}^2=4-1\Rightarrow 3=3$, which is true. Thus, $(4,\sqrt{3})\in S$ is TRUE.

2. Calculate the area of the region $R$ in the first quadrant, bounded by $y^2=x$, the curve $S$ ($y^2=x-1$), and the lines $x=1$ and $x=4$:
In the first quadrant, the upper curve is $y=\sqrt{x}$ and the lower curve is $y=\sqrt{x-1}$.
$\mathrm{Area}\left(R\right)={\int }_1^4\left(\sqrt{x}-\sqrt{x-1}\right)dx$
Integrating term by term:
$\mathrm{Area}\left(R\right)={\left[\frac{2}{3}{x}^{3/2}-\frac{2}{3}(x-1{)}^{3/2}\right]}_1^4$
$=\left(\frac{2}{3}(4{)}^{3/2}-\frac{2}{3}(3{)}^{3/2}\right)-\left(\frac{2}{3}(1{)}^{3/2}-0\right)$
$=\left(\frac{2}{3}\left(8\right)-\frac{2}{3}\left(3\sqrt{3}\right)\right)-\frac{2}{3}$
$=\frac{16}{3}-2\sqrt{3}-\frac{2}{3}=\frac{14}{3}-2\sqrt{3}$

Thus, the area of $R$ is indeed $\frac{14}{3}-2\sqrt{3}$.