Let the product of the focal distances of the point $\left(\sqrt{3},\frac{1}{2}\right)\text{ on the ellipse }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,(\mathrm{a}>\mathrm{b})\text{, be }\frac{7}{4}\text{. }$
Then the absolute difference of the eccentricities of two such ellipses is

Product of focal distances = (a + ex₁) (a – ex₁) 
$\begin{array}{l}={\mathrm{a}}^2-{\mathrm{e}}^2{\mathrm{x}}_1^2={\mathrm{a}}^2-{\mathrm{e}}^2\left(3\right)\\ ={\mathrm{a}}^2-3{\mathrm{e}}^2=\frac{7}{4}\Rightarrow {\mathrm{a}}^2=\frac{7}{4}+3{\mathrm{e}}^2\\ \Rightarrow 4{\mathrm{a}}^2=7+12{\mathrm{e}}^2\\ \mathrm{\&}\left(\sqrt{3},\frac{1}{2}\right)\text{ lines on }\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \therefore \frac{3}{{\mathrm{a}}^2}+\frac{1}{4{\mathrm{b}}^2}=1\end{array}$
$\begin{array}{l}12\left(1-{\mathrm{e}}^2\right)+1=4{\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)\\ 13-12{\mathrm{e}}^2=\left(7+12{\mathrm{e}}^2\right)\left(1-{\mathrm{e}}^2\right)\\ \Rightarrow 13-12{\mathrm{e}}^2=7-7{\mathrm{e}}^2+12{\mathrm{e}}^2-12{\mathrm{e}}^4\\ \Rightarrow 12{\mathrm{e}}^4-17{\mathrm{e}}^2+6=0\\ \therefore {\mathrm{e}}^2=\frac{17\pm \sqrt{289-288}}{24}=\frac{17\pm 1}{24}=\frac{3}{4}\mathrm{\&}\frac{2}{3}\\ \therefore \mathrm{e}=\frac{\sqrt{3}}{2}\mathrm{\&}\sqrt{\frac{2}{3}}\\ \therefore \text{ difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2\sqrt{2}}{2\sqrt{3}}\end{array}$