Prove that $wp \ = \ wn \ \sqrt{1- 2 \xi^2}$

$\frac{x}{X_{st}} \ = \ \frac{1}{\sqrt{(1-r^2)^2 \ + \ (2 \xi r)^2}}$

For X to be max $(1-r^2)^2 \ + \ (2 \xi r)^2$ is minimum

$\therefore$ different this w.r.t r and equate to zero.

$\frac{d}{dr} \ [ (1-r^2)^2 \ + \ (2 \xi r)^2] \ = \ 0$

$2(1-r^2)(-2r) \ + \ 2(2 \xi r) . \ 2\xi \ = \ 04$

$- 4r [1- r^2] \ = \ -4 \xi \ r \ = \ 2 \ \xi^2$

$1 – r^2 \ = \ 2 \ \xi^2$

$r^2 \ = \ 1-2 \ \xi^2$

$r \ = \ \sqrt{1-2 \xi^2}$

$\frac{wp}{wn} \ = \ \sqrt{ 1 – 2 \xi^2}$

$\therefore$ $wp \ = \ wn \ \sqrt{1 – 2 \xi^2}$