$$\begin{aligned} &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \cdot 1}{4}+\frac{\mathrm{y} \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8} \\ & \mathrm{x}+\mathrm{y}=\frac{3}{2} \end{aligned}\\ &\text { solve with ellipse }\\ &\begin{aligned} \mathrm{PR} & =\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2} \\ & =\sqrt{2}\left|\mathrm{x}_2-\mathrm{x}_1\right| \end{aligned} \end{aligned}$$

$$\begin{aligned} & y_2=\frac{3}{2}-x_2 \\ & y_1=\frac{3}{2}-x_1 \\ & y_2-y_1=x_2-x_1 \\ & x^2+2 y^2=4 \\ & x^2+2\left(\frac{3}{2}-x\right)^2=4 \\ & 6 x^2-12 x+1=0 \\ & x_1+x_2=2 \\ & x_1 x_2=1 / 6 \\ & \left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2} \\ & \quad=\sqrt{4-4 / 6} \\ & P R=\sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15} \end{aligned}$$