The length of the chord of the ellipse $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{2}=1$, whose mid-point is $\left(1,\frac{1}{2}\right)$, is 

$\begin{array}{l}\mathrm{T}={\mathrm{S}}_1\\ \frac{\mathrm{x}\cdot 1}{4}+\frac{\mathrm{y}\cdot 1/2}{2}=\frac{1}{4}+\frac{1}{8}\\ \mathrm{x}+\mathrm{y}=\frac{3}{2}\end{array}$
solve with ellipse
$\begin{array}{l}{\mathrm{P}}_{\mathrm{ℝ}}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ =\sqrt{2}\left|{\mathrm{x}}_2-{\mathrm{x}}_1\right|\end{array}$

$\begin{array}{l}{\mathrm{y}}_2=\frac{3}{2}-{\mathrm{x}}_2\\ {\mathrm{y}}_1=\frac{3}{2}-{\mathrm{x}}_1\\ {\mathrm{y}}_2-{\mathrm{y}}_1={\mathrm{x}}_2-{\mathrm{x}}_1\\ {\mathrm{x}}^2+2{\mathrm{y}}^2=4\\ {\mathrm{x}}^2+2{\left(\frac{3}{2}-\mathrm{x}\right)}^2=4\\ 6{\mathrm{x}}^2-12\mathrm{x}+1=0\\ {\mathrm{x}}_1+{\mathrm{x}}_2=2\\ {\mathrm{x}}_1{\mathrm{x}}_2=1/6\\ \left|{\mathrm{x}}_2-{\mathrm{x}}_1\right|=\sqrt{{\left({\mathrm{x}}_2+{\mathrm{x}}_1\right)}^2-4{\mathrm{x}}_1{\mathrm{x}}_2}\\ =\sqrt{4-4/6}\\ \mathrm{PR}=\sqrt{2}\cdot 2\cdot \frac{\sqrt{5}}{\sqrt{2}\sqrt{3}}=\frac{2}{3}\sqrt{15}\end{array}$
$=2\sqrt{5/6}$