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

  • A
    $\frac{3-2\sqrt{2}}{3\sqrt{2}}$
  • B
    $\frac{1-\sqrt{3}}{\sqrt{2}}$
  • C
    $\frac{3-2\sqrt{2}}{2\sqrt{3}}$
  • D
    $\frac{1-2\sqrt{2}}{\sqrt{3}}$

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Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
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$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
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