$P(x, y) \quad\left|\frac{\sqrt{2} x+y-1}{\sqrt{3}}\right|\left|\frac{\sqrt{2} x-y+1}{\sqrt{3}}\right|=\lambda^2$

$\left|\frac{2 x^2-(y-1)^2}{3}\right|=\lambda^2, C :\left|2 x^2-(y-1)^2\right|=3 \lambda^2$

$\text { line } y =2 x +1, RS =\sqrt{\left( x _1- x _2\right)^2+\left( y _1- y _2\right)^2}, R \left( x _1, y _1\right) \text { and } S\left( x _2, y _2\right)$

$y _1=2 x _1+1 \text { and } y _2=2 x _2+1 \Rightarrow\left( y _1- y _2\right)=2\left( x _1- x _2\right)$

$\text { RS }=\sqrt{5\left( x _1- x _2\right)^2}=\sqrt{5}\left| x _1- x _2\right|$

solve curve $C$ and line $y=2 x+1$ we get

$\left|2 x^2-(2 x)^2\right|=3 \lambda^2 \Rightarrow x^2=\frac{3 \lambda^2}{2}$

$\text { RS }=\sqrt{5}\left|\frac{2 \sqrt{3} \lambda}{\sqrt{2}}\right|=\sqrt{30 \lambda}$$=\sqrt{270} \Rightarrow 30 \lambda^2=270 \Rightarrow \lambda^2=9$

$\perp$ bisector of $RS$

$T \equiv\left(\frac{ x _1+ x _2}{2}, \frac{y_1+y_2}{2}\right)$

Here $x _1+ x _2=0$

$T=(0,1)$

Equation of

$R^{\prime} S^{\prime}:(y-1)=-\frac{1}{2}(x-0) \Rightarrow x+2 y=2$

$R^{\prime}\left(a_1, b_1\right) S^{\prime}\left(a_2, b_2\right)$

$D=\left(a_1-a_2\right)^2+\left(b_1-b_2\right)^2=5\left(b_1-b_2\right)^2$

solve $x+2 y=2$ and $\left|2 x^2-(y-1)^2\right|=3 \lambda^2$

$\left|8(y-1)^2-(y-1)^2\right|=3 \lambda^2 \Rightarrow(y-1)^2=\left(\frac{\sqrt{3} \lambda}{\sqrt{7}}\right)^2$

$y-1= \pm \frac{\sqrt{3} \lambda}{\sqrt{7}} \Rightarrow b_1=1+\frac{\sqrt{3 \lambda}}{\sqrt{7}}, b_2=1-\frac{\sqrt{3} \lambda}{\sqrt{17}}$

$D =5\left(\frac{2 \sqrt{3} \lambda}{\sqrt{7}}\right)^2=\frac{5 \times 4 \times 3 \lambda^2}{7}=\frac{5 \times 4 \times 27}{7}=77.14$