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

1
\(\frac{3-2 \sqrt{2}}{3 \sqrt{2}}\)
2
\(\frac{1-\sqrt{3}}{\sqrt{2}}\)
3
\(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)
4
\(\frac{1-2 \sqrt{2}}{\sqrt{3}}\)

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