$\text { Product of focal distances }=\left(a+e x_1\right)\left(a-e x_1\right)$
$= a ^2- e ^2 x _1^2= a ^2- e ^2(3)$
$= a ^2-3 e ^2=\frac{7}{4} \Rightarrow a ^2=\frac{7}{4}+3 e ^2$
$\Rightarrow 4 a ^2=7+12 e ^2$
$\ \left(\sqrt{3}, \frac{1}{2}\right) \text { lines on } \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1$
$\therefore \frac{3}{ a ^2}+\frac{1}{4 b^2}=1$
$\frac{3}{ a ^2}+\frac{1}{4\left( a ^2\right)\left(1- e ^2\right)}=1$
$12\left(1- e ^2\right)+1=4 a ^2\left(1- e ^2\right)$
$13-12 e ^2=\left(7+12 e ^2\right)\left(1- e ^2\right)$
$\Rightarrow 13-12 e ^2=7-7 e ^2+12 e ^2-12 e ^4$
$\Rightarrow 12 e ^4-17 e ^2+6=0$
$\therefore e ^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \ \frac{2}{3}$
$\therefore e =\frac{\sqrt{3}}{2} \ \sqrt{\frac{2}{3}}$
$\therefore \text { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$